Mathematical modeling / English

Linearization and Differentials - Video 2 - Differentials

Rutgers Math 151 · YouTube · 4:44

YouTube · Rutgers Math 151Original video

Selected moment · 2:23 · Chapter markers are time references, not playback progress.

IN THIS VIDEO

01Independent Differential (dx)
02Dependent Differential (dy)
03Differential vs. Fraction Misconception

FROM THE LEARNING NOTES

dx

Independent Differential (dx)

SELECTED MOMENT

Recall the linearization $L(x)=f(a)+f'(a)(x-a)$

SUGGESTED PREPARATION

Derivatives
Functions
Derivative notation $f'(x)$
Tangent line / linear approximation at a point
Basic algebraic rearrangement of formulas
Meaning of a finite increment $\Delta x$
Derivative at a point
Tangent line / linearization
Function values at a base point
Signed change in the input variable

READ & KEEP

The explanation, unpacked.

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Reviewed learning material · Video analysis · English

Read the full overview

This video segment introduces the concept of differentials in calculus. It defines the independent differential dx and the dependent differential dy as $dy = f$ '(x)dx. The instructor clarifies that while $dy/dx$ resembles a fraction, it represents a derivative and should not be treated with simple algebraic cancellation. Alternative notation df is introduced. Finally, the segment begins to explain how differentials are used for linear approximation by equating dx to the change in x (Delta x). This 120-second calculus whiteboard segment connects differentials to linearization. It begins with $df=f'(x)dx$ , substitutes $dx=\Delta x$ , then recalls the linearization $L(x)=f(a)+f'(a)(x-a)$ . By identifying $x-a$ with $\Delta x$ , the speaker derives $L(x)-f(a)=f'(a)\Delta x$ , replaces $L(x)$ by its approximation $f(x)$ to obtain $f(x)-f(a)\approx f'(a)\Delta x$ , defines $\Delta f=f(x)-f(a)$ , and concludes with the central rule $\Delta f\approx df=f'(a)dx=f'(a)\Delta x$ . The clip ends by noting that an earlier $\sqrt{35.8}$ approximation example fits the same framework, though less directly. This 44-second calculus clip demonstrates linearization using differentials. The board shows the general relation Δf ≈ df = f'(a) $dx = f$ '(a)Δx, then applies it to a concrete example with base point 36, target point 35.8, derivative f'(36)= $1/12$ , and function value $f(36)=6$ . The speaker computes Δf ≈ ( $1/12$ )(-0.2) = -1/60, rewrites Δf as $f(35.8)-6$ , and explains that adding this approximated change back yields the same result as the earlier linearization method. The segment closes by framing differentials as an alternative viewpoint on linearization.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction to Differentials 0:10Defining Independent Differential dx 0:25Defining Dependent Differential dy 0:45Relationship to Derivatives and Misconceptions 1:30Alternative Notation df 1:45Application to Linearization 2:00Differential formula and the substitution

dx=\Delta x

2:23Recall the linearization

L(x)=f(a)+f'(a)(x-a)

2:43Rewrite the linearization as a change formula 2:57Approximate the true change

f(x)-f(a)

3:05Define

\Delta f

3:22Linearization with differentials:

\Delta f\approx df

3:53Connection to the earlier

\sqrt{35.8}

example 4:00General differential formula 4:04Substitute f'(36)=

1/12

and

Δx=-0.2

4:19Compute Δf ≈ -1/60 4:22Relate Δf to

f(35.8)-6

4:33Interpretation as a new linearization technique

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

Welcome to the lesson on differentials. We will explore how differentials serve as a tool for linear approximation.

Let's start with the definition. Consider a differentiable function $y = f(x)$ . We introduce a new independent variable called the differential $dx$ .

Next, we define the dependent differential $dy$ . It is calculated as the product of the derivative and the independent differential: $dy = f'(x) dx$ . This allows us to compute approximate changes in $y$ given changes in $x$ .

Notice the relationship between $dy$ , $dx$ , and the derivative. Since $f'(x) = \frac{dy}{dx}$ , we can write $dy = \frac{dy}{dx} dx$ . While this looks like fraction cancellation, remember that the derivative is a limit, not a literal fraction, so be cautious with algebraic manipulation.

We can also express this using function notation. Instead of $dy$ , we use $df$ , defined as $df = f'(x) dx$ . This is simply alternative notation for the same concept.

Finally, how do we use this for linearization? The key step is to equate the differential $dx$ to the actual change in $x$ , denoted as $\Delta x$ . This sets the stage for approximating function values.

The segment opens by revisiting the differential identity for a one-variable function. On the board, in blue, the formula $df=f'(x)\,dx$ is written, and above it the instructor notes the planned substitution $dx=\Delta x$ . The mathematical point is simple but important: once the independent-variable differential is replaced by a finite input increment, the same expression becomes $df=f'(x)\,\Delta x$ , setting up a comparison with linear approximation.

Next the lecture switches to orange and recalls the linearization of $f$ at a base point $a$ : $L(x)=f(a)+f'(a)(x-a)$ . This is the tangent-line model built from the known value $f(a)$ and the slope $f'(a)$ . The speaker then focuses on the factor $(x-a)$ and explicitly relabels it as $\Delta x$ , because it measures the change in the input from $a$ to $x$ .

With that identification in place, the linearization is rearranged into a change formula. Subtracting $f(a)$ from both sides gives the exact statement $L(x)-f(a)=f'(a)\,\Delta x$ . This step isolates what the linear model predicts for the change in output when the input moves by $\Delta x$ .

The argument then passes from the linear model to the original function. Since linear approximation means $L(x)\approx f(x)$ near $a$ , the exact change in $L$ is converted into the approximate change in $f$ : $f(x)-f(a)\approx f'(a)\,\Delta x$ . At this stage the board has shown why the tangent-line formula is not just a separate construction; it is a local estimator for the true function increment.

To compress the notation, the instructor defines $\Delta f=f(x)-f(a)$ . This names the left-hand side of the previous approximation as the actual finite change in the function value. The purpose is practical: if one knows the convenient value $f(a)$ and can estimate $\Delta f$ , then one can recover an approximation to $f(x)$ .

The final synthesis appears under the heading "Linearization with differentials." The board writes $\Delta f\approx df=f'(a)\,dx=f'(a)\,\Delta x$ . This line unifies the two threads of the clip: the differential formula from the beginning and the linearization-derived approximation from the middle. Conceptually, the differential $df$ is being used as the linearized version of the actual change $\Delta f$ .

The clip closes by linking this framework to an earlier numerical example involving $\sqrt{35.8}$ . No computation is shown here, but the speaker states that the same reasoning applies. The only caveat given is that the differential route is "not as direct" because one must first pass through the change-in- $f$ interpretation before using $df$ as the approximation tool.

The clip opens on a whiteboard already containing the linearization framework. At the top are formulas such as $L(x)=f(a)+f'(a)(x-a)$ and $\Delta f = f(x)-f(a) \approx f'(a)\Delta x$ , while the main focus is the heading “Linearization with differentials” and the displayed relation $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ . The speaker says the same idea works this way, signaling that this is an alternate formulation of the previous method.

The example begins by identifying the base point as $a=36$ . The board writes $f'(36)=\frac{1}{12}$ , and the audio confirms that this derivative value was already known. The next step is to determine the input change: moving from $36$ to $35.8$ gives $\Delta x = dx = -0.2$ . The negative sign matters because the input is decreasing.

Substituting these values into the differential approximation formula yields $\Delta f \approx df = \frac{1}{12}(-0.2)$ . The arithmetic is then simplified on the board to $-\frac{1}{60}$ . This is the approximated change in the function value, not the exact change itself.

To connect this back to the function values, the speaker rewrites the actual change using the definition $\Delta f = f(x)-f(a)$ . With $x=35.8$ and $a=36$ , this becomes $\Delta f = f(35.8)-f(36)$ . Since the clip states $f(36)=6$ , the expression shown is $f(35.8)-6$ . A colored arrow visually links this expression to the previously computed value $-\frac{1}{60}$ , indicating that $f(35.8)-6 \approx -\frac{1}{60}$ , so $f(35.8)$ can be estimated by adding the approximated change to $6$ .

The closing commentary explains the pedagogical purpose: differentials provide another way to view linearization. The speaker notes that this perspective may be easier in some contexts and harder in others, and says the next video will show cases where it is especially convenient.

Knowledge cards

Independent Differential (dx)

For a differentiable function $y = f(x)$ , dx is defined as a new independent variable representing an arbitrary change in x.

dx

Dependent Differential (dy)

The differential dy represents the corresponding change in y along the tangent line. It is defined as the derivative multiplied by dx.

dy = f'(x) dx

Differential vs. Fraction Misconception

Although $dy = (dy/dx)dx$ resembles fraction cancellation, $dy/dx$ is a derivative (a limit), not a true fraction. Treat the notation carefully.

dy = \frac{dy}{dx} dx

Function Differential (df)

An alternative notation for the differential of a function f, equivalent to dy when $y=f(x)$ .

df = f'(x) dx

Linearization Setup

To apply differentials for linear approximation, set the differential dx equal to the actual change in the independent variable, Delta x.

dx = \Delta x

Differential formula with $dx=\Delta x$

The clip begins from the standard one-variable differential identity and then substitutes the finite input increment for the differential of the independent variable. This prepares the formula for comparison with linearization, because the same derivative factor will later appear in the tangent-line approximation at a base point $a$ .

df = f'(x)\,dx = f'(x)\,\Delta x

Linearization at a base point

The linearization of $f$ at $a$ is the tangent-line approximation built from the known value $f(a)$ and the slope $f'(a)$ . It expresses the nearby behavior of $f$ by a linear function of the displacement from $a$ .

L(x)=f(a)+f'(a)(x-a)

$(x-a)$ is the input increment $\Delta x$

In the linearization formula, the factor $(x-a)$ measures how far the input has moved from the base point $a$ to the target point $x$ . The video explicitly relabels this displacement as $\Delta x$ , which allows the linearization to be rewritten in the same increment notation used for differentials.

\Delta x = x-a

Exact change predicted by the linearization

Once the linearization is written in increment form, subtracting the base value isolates the model's predicted output change. This is an exact statement about $L$ , not yet about the original nonlinear function.

L(x)-f(a)=f'(a)\,\Delta x

Approximating the true change in $f$

Because linearization means $L(x)\approx f(x)$ near $a$ , the exact change formula for $L$ becomes an approximation for the actual function change. This is the key bridge from tangent-line geometry to numerical estimation of $f(x)-f(a)$ .

f(x)-f(a)\approx f'(a)\,\Delta x

Definition of $\Delta f$

The symbol $\Delta f$ is introduced as the actual finite change in the function value between the base point and the nearby target point. Naming this quantity makes the later comparison with the differential concise and conceptually clear.

\Delta f = f(x)-f(a)

Linearization with differentials

The main conclusion of the clip is that the actual change $\Delta f$ can be approximated by the differential $df$ . Evaluating the differential at the base point $a$ and using $dx=\Delta x$ gives the practical local-approximation rule emphasized throughout the segment.

\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x

Connection to the earlier $\sqrt{35.8}$ example

The speaker notes that a previously discussed approximation problem involving $\sqrt{35.8}$ fits the same differential framework. The only difference stressed here is procedural: the differential method is less direct unless one first recognizes the quantity being approximated as the change $\Delta f$ .

Linearization with differentials

The central method shown in the clip is to approximate the actual change in a function by its differential. The board states $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ . Here $\Delta f$ is the true change from $f(a)$ to $f(x)$ , while $df$ is the linear approximation obtained by multiplying the derivative at the base point by the input change.

\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x

Meaning of Δf

The clip defines the actual change in the function as $\Delta f = f(x)-f(a)$ . In the example, this becomes $f(35.8)-f(36)$ , and because $f(36)=6$ , it is written as $f(35.8)-6$ . This quantity is what the differential method estimates.

\Delta f = f(x)-f(a)

Worked example setup

The example uses base point $a=36$ , target point $x=35.8$ , derivative $f'(36)=\frac{1}{12}$ , and known value $f(36)=6$ . The input change is therefore $\Delta x = dx = 35.8-36 = -0.2$ . These are the quantities substituted into the differential approximation formula.

a=36,\quad x=35.8,\quad f'(36)=\frac{1}{12},\quad f(36)=6,\quad \Delta x=-0.2

Computing the approximated change

Substitution gives $\Delta f \approx \frac{1}{12}(-0.2)$ . Since $-0.2=-\frac{1}{5}$ , the product simplifies to $-\frac{1}{60}$ . This is the differential estimate for how much $f$ changes when $x$ moves from $36$ to $35.8$ .

\Delta f \approx \frac{1}{12}(-0.2) = -\frac{1}{60}

Recovering $f(35.8)$ from the approximation

Because $\Delta f = f(35.8)-6$ and the approximation gives $\Delta f \approx -\frac{1}{60}$ , the clip sets up $f(35.8)-6 \approx -\frac{1}{60}$ . Solving for the nearby function value yields $f(35.8) \approx 6 - \frac{1}{60}$ . The speaker notes that this leads to the same answer as the earlier linearization approach.

f(35.8)-6 \approx -\frac{1}{60},\quad f(35.8) \approx 6 - \frac{1}{60}

Why differentials are useful here

The final commentary frames differentials as an alternative viewpoint on linearization. Instead of first writing the tangent-line formula and then subtracting, one can directly approximate the change $\Delta f$ by $df=f'(a)\Delta x$ and then add that change back to the known base value. The speaker says this can be easier in some situations and more complicated in others.

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 24

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as $y = f(x)$

Symbol

Meaning

Dependent variable in the function definition

Domain

Real numbers

$f(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as $f(x)$

Symbol

$f(x)$

Meaning

Differentiable function of x

Domain

Real numbers

dx

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as dx

Symbol

Meaning

Independent differential variable

Domain

Real numbers

dy

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as dy

Symbol

Meaning

Dependent differential variable

Domain

Real numbers

f'(x)

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as f'(x) or $dy/dx$

Symbol

f'(x)

Meaning

Derivative of f with respect to x

Domain

Real numbers

df

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as df

Symbol

Meaning

Differential of the function f

Domain

Real numbers

$\Delta x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as $\Delta x$

Symbol

$\Delta x$

Meaning

Change in x

Domain

Real numbers

df

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $df = f'(x)\,dx$ and later uses $df$ in the summary line $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ .
Audio
Observation
The speaker says that $df$ was equal to $f'$ at $x$ times $dx$ , and later refers to the differential of $f$ .

Symbol

Meaning

The differential of $f$ ; in this clip it is used as the linear approximation to the actual change in the function value.

Domain

Defined for a differentiable function $f$ ; here evaluated at the base point $a$ with increment $dx=\Delta x$ .

f'(x)

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as $f'(x)$ in $df = f'(x)\,dx$ and then substituted to $f'(x)\,\Delta x$ .
Audio
Observation
The speaker reads it as $f'$ at $x$ .

Symbol

f'(x)

Meaning

Derivative of $f$ evaluated at the variable point $x$ . In the later linearization comparison, the analogous derivative is evaluated at the base point $a$ .

Domain

Requires $f$ to be differentiable at the point where it is evaluated.

dx

Clear evidence

Shown in the video

Evidence

Formula
Observation
Appears in $df = f'(x)\,dx$ and in the header note "Going to set $dx = \Delta x$ ".
Audio
Observation
The speaker says, "if I set $dx$ equal to some change in $x$ or a $\Delta x$ ".

Symbol

Meaning

Differential of the independent variable $x$ . In this lesson it is deliberately identified with the finite increment $\Delta x$ for approximation purposes.

Domain

Used as an input increment paired with the derivative factor.

$\Delta x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written in the header $dx = \Delta x$ , in $= f'(x)\,\Delta x$ , in $(x-a)$ labeled as $\Delta x$ , and in $f'(a)\,\Delta x$ .
Audio
Observation
The speaker calls it "some change in $x$ " and later says $(x-a)$ is "really a $\Delta x$ ".

Symbol

$\Delta x$

Meaning

Finite change in the input variable. In the linearization formula it is represented by $x-a$ when moving from the base point $a$ to the nearby point $x$ .

Domain

Here treated as the same quantity substituted for $dx$ in the differential formula.

$L(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $L(x)=f(a)+f'(a)(x-a)$ and then $L(x)-f(a)=f'(a)\,\Delta x$ .
Audio
Observation
The speaker says, "Our linearization had that $L$ of $x$ was $f$ at $a$ plus $f'$ at $a$ times $x$ minus $a$ ".

Symbol

$L(x)$

Meaning

Linearization (linear approximation) of $f$ at the base point $a$ , viewed as a function of $x$ .

Domain

Defined near the expansion point $a$ ; in this clip it is compared to $f(x)$ through the approximation $L(x)\approx f(x)$ .

Knowledge points · 14

Definition of Independent Differential

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker defines a new independent variable dx.
Formula
Observation
Written text: 'If $y = f(x)$ and f is differentiable, we define a new independent variable dx'

Definition

Explanation

For a differentiable function $y = f(x)$ , dx is defined as a new independent variable representing a differential.

Conditions

f must be a differentiable function

Definition of Dependent Differential

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker defines dy using f'(x) and dx.
Formula
Observation
Written equation: $dy = f$ '(x) dx

Definition

Explanation

The dependent differential dy is defined as the product of the derivative f'(x) and the independent differential dx.

Conditions

dx is defined
f'(x) exists

Prerequisites

Definition of Independent Differential

Definition of Function Differential

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker introduces df as an alternative notation.
Formula
Observation
Written equation: df = f'(x) dx

Definition

Explanation

The differential of a function f can be denoted as df, which is mathematically equivalent to dy when $y = f(x)$ .

Conditions

$y = f(x)$

Prerequisites

Definition of Dependent Differential

Setting up Linear Approximation with Differentials

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker explains setting dx equal to Delta x for linear approximation.
Formula
Observation
Written text: 'Going to set $dx = \Delta x$ '

Method

Explanation

To use differentials for linear approximation, the independent differential dx is set equal to the actual change in x, denoted as $\Delta x$ .

Prerequisites

Definition of Independent Differential

Differential written with $dx=\Delta x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $df = f'(x)\,dx$ and then substitutes to obtain $= f'(x)\,\Delta x$ .
Audio
Observation
The speaker says that $df$ was equal to $f'$ at $x$ times $dx$ , and if $dx$ is set equal to a change in $x$ or a $\Delta x$ , this becomes $f'$ at $x$ times $\Delta x$ .

Formula

Explanation

The clip starts from the differential formula for a one-variable function and explicitly replaces the differential of the independent variable by the finite increment $\Delta x$ . This produces the same algebraic expression but now framed as a linear estimate of a finite change.

Formula

df = f'(x)\,dx = f'(x)\,\Delta x

Conditions

$f$ is differentiable at the point where $f'$ is evaluated.
In this lesson $dx$ is being intentionally identified with $\Delta x$ for approximation purposes.

Linearization of $f$ at $a$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $L(x)=f(a)+f'(a)(x-a)$ in orange.
Audio
Observation
The speaker says, "Our linearization had that $L$ of $x$ was $f$ at $a$ plus $f'$ at $a$ times $x$ minus $a$ ".

Definition

Explanation

The linearization is the tangent-line approximation to $f$ based at the point $a$ . It uses the known value $f(a)$ and the slope $f'(a)$ to build a linear function of $x$ .

Formula

L(x)=f(a)+f'(a)(x-a)

Conditions

$f$ is differentiable at $a$ .
$x$ is taken near $a$ so that the tangent-line approximation is meaningful.

Prerequisites

Differential written with $dx=\Delta x$

Change in the linearization equals slope times input change

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $L(x)-f(a)=f'(a)\,\Delta x$ after rewriting $(x-a)$ as $\Delta x$ .
Audio
Observation
The speaker says, "So linear approximation tells us that $L$ of $x$ minus $f$ of $a$ is equal to $f'$ at $a$ times a $\Delta x$ ".

Formula

Explanation

Subtracting the base value $f(a)$ from the linearization isolates the predicted change produced by moving from $a$ to $x$ . Because $x-a$ is identified with $\Delta x$ , the change in $L$ is exactly $f'(a)\Delta x$ .

Formula

L(x)-f(a)=f'(a)\,\Delta x

Conditions

Uses the linearization $L(x)=f(a)+f'(a)(x-a)$ .
Treats $x-a$ as the finite input increment $\Delta x$ .

Prerequisites

Linearization of $f$ at $a$

Linear approximation to the actual change in $f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $f(x)-f(a)\approx f'(a)\,\Delta x$ .
Audio
Observation
The speaker says, "But linear approximation gives that this $L$ is approximately $f$ at $x$ . So $f$ at $x$ minus $f$ at $a$ is approximately $f'$ at $a$ times $\Delta x$ ".

Formula

Explanation

Since $L(x)$ approximates $f(x)$ near $a$ , the exact change in the linear model is transferred into an approximate statement about the true function change. The result is that the actual difference $f(x)-f(a)$ is approximated by the same slope-times-increment expression.

Formula

f(x)-f(a)\approx f'(a)\,\Delta x

Conditions

$x$ must be sufficiently close to $a$ for the linear approximation to be reasonable.
$f$ is differentiable at $a$ .

Prerequisites

Linearization of $f$ at $a$
Change in the linearization equals slope times input change

Definition of $\Delta f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $\Delta f = f(x)-f(a)$ beside the previous approximation line.
Audio
Observation
The speaker says, "This over here is a change in $f$ ".

Definition

Explanation

The symbol $\Delta f$ denotes the actual finite change in the function value when the input moves from $a$ to $x$ . In this clip it is introduced precisely to connect the linearization discussion with the later differential approximation.

Formula

\Delta f = f(x)-f(a)

Conditions

Base point $a$ and target point $x$ are fixed.
No additional condition beyond evaluating $f$ at those points is stated in the clip.

Prerequisites

Linear approximation to the actual change in $f$

Using differentials to approximate $\Delta f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes the heading "Linearization with differentials" and then $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ .
Audio
Observation
The speaker says the idea is to approximate the change in $f$ by the differential of $f$ , and that the differential is defined as $f'$ at the initial point $a$ times $dx$ , really plugging in $f'(a)\Delta x$ .

Method

Explanation

This is the central synthesis of the clip: the actual change $\Delta f$ is estimated by the differential $df$ . The differential is then expanded using the derivative at the base point and the substitution $dx=\Delta x$ , giving a compact formula for local linear estimation.

Formula

\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x

Conditions

$f$ is differentiable at the base point $a$ .
$x$ is near $a$ so that the linear approximation is appropriate.
The lesson explicitly sets $dx=\Delta x$ for this use.

Prerequisites

Differential written with $dx=\Delta x$
Linear approximation to the actual change in $f$
Definition of $\Delta f$

Linearization with differentials

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board title reads 'Linearization with differentials' and below it: Δf ≈ df = f'(a) $dx = f$ '(a) Δx.
Audio
Observation
'This just gives a new way, a new technique for viewing linearization using differentials...'

Method

Explanation

The video presents a method for approximating the change in a function by replacing the actual change Δf with the differential df. The key relation shown is Δf ≈ df = f'(a) $dx = f$ '(a) Δx, meaning that near the base point a, the function’s change can be estimated by multiplying the derivative at a by the input change.

Formula

\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x

Conditions

f is differentiable at the base point a.
The input change Δx is small enough for the linear approximation to be useful.

Change in function value Δf

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $Δf = f(x) - f(a)$ ≈ f'(a) Δx.
Audio
Observation
'Change of f is then f at 35.8 minus f of 36, which is 6.'

Definition

Explanation

The actual change in the function from the base point a to a nearby point x is defined as $Δf = f(x) - f(a)$ . In the worked example, this becomes $f(35.8) - f(36)$ , and since $f(36)=6$ , it is written as $f(35.8)-6$ .

Formula

\Delta f = f(x) - f(a)

Conditions

x is a nearby input value.
a is the chosen base point.

Claims and conditions · 7

Nature of the Derivative Ratio

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker states that $dy/dx$ looks like a fraction but isn't one because the derivative is not actually a fraction.
Formula
Observation
Written equation: $dy = (dy/dx) dx$

Proposition

Statement

The expression $dy/dx$ represents the derivative f'(x), but it should not be treated as a literal fraction where terms cancel out algebraically.

Hypotheses

dy and dx are differentials
f'(x) is the derivative

Quantifiers

No quantifiers are stated in the source.

Differential identity used at the start

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays $df = f'(x)\,dx$ and then $= f'(x)\,\Delta x$ after the header note $dx=\Delta x$ .
Audio
Observation
The speaker states that $df$ was equal to $f'$ at $x$ times $dx$ , and after setting $dx=\Delta x$ it becomes $f'$ at $x$ times $\Delta x$ .

Proposition

Statement

For the function under discussion, $df = f'(x)\,dx$ , and if $dx$ is set equal to $\Delta x$ , then $df = f'(x)\,\Delta x$ .

Hypotheses

$f$ is differentiable at the point where $f'$ is evaluated.
The lesson adopts the substitution $dx=\Delta x$ .

Quantifiers

Pointwise statement at the chosen evaluation point $x$ .

Exact change in the linearization

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $L(x)-f(a)=f'(a)\,\Delta x$ .
Audio
Observation
The speaker says linear approximation tells us that $L(x)-f(a)$ is equal to $f'(a)$ times $\Delta x$ .

Proposition

Statement

If $L(x)=f(a)+f'(a)(x-a)$ and $\Delta x=x-a$ , then $L(x)-f(a)=f'(a)\,\Delta x$ .

Hypotheses

$L$ is the linearization of $f$ at $a$ .
$\Delta x$ is identified with $x-a$ .

Quantifiers

For the chosen base point $a$ and nearby point $x$ .

Linear approximation of the true function change

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $f(x)-f(a)\approx f'(a)\,\Delta x$ .
Audio
Observation
The speaker says that because $L$ is approximately $f$ at $x$ , $f(x)-f(a)$ is approximately $f'(a)\Delta x$ .

Proposition

Statement

Near the base point $a$ , $f(x)-f(a)\approx f'(a)\,\Delta x$ .

Hypotheses

$f$ is differentiable at $a$ .
$x$ is close enough to $a$ for the linear approximation to be valid.
$\Delta x=x-a$ .

Quantifiers

Local approximation statement for $x$ near $a$ .

Core differential approximation principle

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ under the heading "Linearization with differentials".
Audio
Observation
The speaker says the idea is to approximate the change in $f$ by the differential of $f$ , with the differential given by $f'(a)dx$ and then $f'(a)\Delta x$ .

Proposition

Statement

The actual change $\Delta f$ can be approximated by the differential $df$ , specifically $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ .

Hypotheses

$f$ is differentiable at the initial point $a$ .
The increment satisfies $dx=\Delta x$ in the intended use.
$x$ is near $a$ so that the linear approximation is appropriate.

Quantifiers

Local statement centered at the base point $a$ .

Differential approximation claim

Clear evidence

Shown in the video

Evidence

Formula
Observation
The central displayed relation is Δf ≈ df = f'(a) $dx = f$ '(a) Δx.
Audio
Observation
'So my change in f, I could approximate by $1/12$ the change in x...'

Proposition

Statement

Near a base point a, the actual change in the function satisfies Δf ≈ df = f'(a) Δx.

Hypotheses

f is differentiable at a.
Δx is the input change from a to a nearby point.

Quantifiers

For a differentiable function f and a nearby input change Δx.

Equivalence with prior linearization method

Clear evidence

Shown in the video

Evidence

Audio
Observation
'I add that over and I get the same answer that I had before.'

Uncertainties

The earlier method is not shown in this clip, so the exact previous computation cannot be verified from this segment alone.

Proposition

Statement

Using differentials gives the same final approximation result as the previously used linearization method.

Hypotheses

The same function, base point, and target point are used.
The earlier method is the standard linearization approach referenced by the speaker.

Quantifiers

For the worked example in this lesson.

Derivations and proofs · 6

Relating Differentials to Derivative Notation

Clear evidence

Shown in the video

Evidence

Formula
Observation
Step 1: $dy = f$ '(x) dx
Formula
Observation
Step 2: Substitute f'(x) with $dy/dx$ -> $dy = (dy/dx) dx$

Intuitive argument

Steps

Expression
$dy = f'(x) dx$
Explanation
Start with the definition of the dependent differential.
Justification
Definition of dy
Shown in the video
Expression
$dy = \frac{dy}{dx} dx$
Explanation
Replace f'(x) with the Leibniz notation for the derivative.
Justification
Definition of derivative
Shown in the video

Conclusion

This shows why the ratio of differentials corresponds to the derivative, while warning against treating it as simple fraction cancellation.

Rewriting the differential using $dx=\Delta x$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $df = f'(x)\,dx$ and then directly rewrites it as $= f'(x)\,\Delta x$ under the note "Going to set $dx = \Delta x$ ".
Audio
Observation
The speaker explains that if $dx$ is set equal to a change in $x$ or a $\Delta x$ , the expression becomes $f'(x)\Delta x$ .

Proof

Steps

Expression
$df = f'(x)\,dx$
Explanation
Start from the differential formula displayed on the board.
Justification
Given in the video as the starting identity for $df$ .
Shown in the video
Expression
$dx = \Delta x$
Explanation
Introduce the substitution announced in the header note.
Justification
Explicitly stated by the speaker and written on the board.
Shown in the video
Expression
$df = f'(x)\,\Delta x$
Explanation
Replace $dx$ by $\Delta x$ in the differential formula.
Justification
Direct substitution into the previous equality.
Shown in the video

Conclusion

The differential can be written as $df = f'(x)\,\Delta x$ once $dx$ is identified with the finite input change $\Delta x$ .

From linearization to the change in $L(x)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $L(x)=f(a)+f'(a)(x-a)$ , marks $(x-a)$ as $\Delta x$ , and then writes $L(x)-f(a)=f'(a)\,\Delta x$ .
Audio
Observation
The speaker first states the linearization formula, then says $(x-a)$ is really a $\Delta x$ , and concludes that $L(x)-f(a)=f'(a)\Delta x$ .

Proof

Steps

Expression
$L(x)=f(a)+f'(a)(x-a)$
Explanation
Write the linearization of $f$ at the base point $a$ .
Justification
Formula explicitly presented in the video.
Shown in the video
Expression
$\Delta x = x-a$
Explanation
Identify the horizontal displacement from $a$ to $x$ with the finite increment $\Delta x$ .
Justification
Stated by the speaker and indicated by the arrow label on the board.
Shown in the video
Expression
$L(x)-f(a)=f'(a)(x-a)$
Explanation
Subtract $f(a)$ from both sides of the linearization formula.
Justification
Algebraic rearrangement of the displayed equation.
Derived from the video
Expression
$L(x)-f(a)=f'(a)\,\Delta x$
Explanation
Replace $(x-a)$ by $\Delta x$ .
Justification
Substitution using the previous identification.
Shown in the video

Conclusion

The exact change predicted by the linearization is $L(x)-f(a)=f'(a)\,\Delta x$ .

Connecting linear approximation to the differential formula

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board successively shows $f(x)-f(a)\approx f'(a)\,\Delta x$ , then $\Delta f = f(x)-f(a)$ , and finally $\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$ .
Audio
Observation
The speaker says $L$ is approximately $f$ at $x$ , identifies $f(x)-f(a)$ as a change in $f$ , and then summarizes that the change in $f$ can be approximated by the differential of $f$ .

Proof

Steps

Expression
$L(x)\approx f(x)$
Explanation
Use the defining purpose of linearization near the base point.
Justification
Stated verbally in the video as the approximation provided by linear approximation.
Shown in the video
Expression
$f(x)-f(a)\approx f'(a)\,\Delta x$
Explanation
Transfer the exact change formula for $L$ into an approximate statement about the true function values.
Justification
Follows from replacing $L(x)$ by its approximation $f(x)$ in $L(x)-f(a)=f'(a)\Delta x$ .
Shown in the video
Expression
$\Delta f = f(x)-f(a)$
Explanation
Introduce notation for the actual finite change in the function value.
Justification
Explicitly written and named on the board.
Shown in the video
Expression
$\Delta f \approx f'(a)\,\Delta x$
Explanation
Rewrite the previous approximation using the new symbol $\Delta f$ .
Justification
Substitution of the definition of $\Delta f$ into the approximate equality.
Derived from the video
Expression
$\Delta f \approx df = f'(a)\,dx = f'(a)\,\Delta x$
Explanation
Recognize the right-hand side as the differential evaluated at the base point with $dx=\Delta x$ .
Justification
Synthesis stated under the heading "Linearization with differentials" and supported by the earlier differential formula.
Shown in the video

Conclusion

The clip derives the practical rule $\Delta f \approx df$ , with $df$ computed as $f'(a)\,dx$ and then as $f'(a)\,\Delta x$ .

Computing the differential approximation in the example

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board successively writes f'(36)= $1/12$ , then Δf ≈ $1/12(-0.2)$ , then = -1/60.
Audio
Observation
'So my change in f, I could approximate by $1/12$ the change in x, which was minus 0.2... This becomes my negative 1 over 60.'

Numerical verification

Steps

Expression
$\Delta f \approx df = f'(36)\,\Delta x$
Explanation
Substitute the example base point $a=36$ into the general differential approximation formula.
Justification
Use the displayed method Δf ≈ df = f'(a)Δx.
Shown in the video
Expression
$= \frac{1}{12}(-0.2)$
Explanation
Insert the given derivative value f'(36)= $1/12$ and the input change $Δx=-0.2$ .
Justification
These values are explicitly stated on the board and in the audio.
Shown in the video
Expression
$= -\frac{1}{60}$
Explanation
Multiply $1/12$ by -0.2 = -1/5 to obtain -1/60.
Justification
Arithmetic simplification.
Shown in the video

Conclusion

The differential approximation for the change in f in this example is Δf ≈ -1/60.

Using Δf to recover the nearby function value

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $Δf = f(35.8) - 6$ and an arrow points back toward the computed value -1/60.
Audio
Observation
'Change of f is then f at 35.8 minus f of 36, which is 6. I add that over and I get the same answer that I had before.'

Uncertainties

The final numerical value for $f(35.8)$ is not written out in this clip, although the setup is shown.

Numerical verification

Steps

Expression
$\Delta f = f(35.8) - f(36)$
Explanation
Write the definition of the actual change in f between 36 and 35.8.
Justification
Definition $Δf = f(x)-f(a)$ with $x=35.8$ and $a=36$ .
Shown in the video
Expression
$= f(35.8) - 6$
Explanation
Substitute the known value $f(36)=6$ .
Justification
The speaker explicitly states f of 36 is 6.
Shown in the video
Expression
$f(35.8) \approx 6 + \left(-\frac{1}{60}\right)$
Explanation
Replace Δf by its differential approximation -1/60 and solve for $f(35.8)$ .
Justification
From Δf ≈ -1/60 and the equation $Δf = f(35.8)-6$ .
Derived from the video

Conclusion

The clip sets up $f(35.8)-6$ ≈ -1/60, so $f(35.8)$ is obtained by adding the approximated change to 6.

Worked examples · 2

Reference to the earlier $\sqrt{35.8}$ example

Approximate timing

Shown in the video

Evidence

Audio
Observation
The speaker says, "And the stuff we did with the square root of 35.8 would work in the same way. It's not as direct because you have to go through this whole change of $f$ step first."

Uncertainties

No worked numerical example is shown in this clip.
Only the verbal reference to a prior example is available, so the full setup and computation cannot be extracted from the provided segment.

Problem

The speaker mentions that a previously discussed approximation problem involving $\sqrt{35.8}$ can be handled by the same differential method.

Given

A prior example involving $\sqrt{35.8}$ is referenced verbally.
The current framework is $\Delta f \approx df = f'(a)\,\Delta x$ .

Goal

Explain how the earlier numerical approximation fits the differential viewpoint developed in this clip.

Steps

Expression
Explanation
The clip does not reproduce the earlier calculation; it only states that the same method applies.
Justification
Verbal reference only, with no displayed setup or arithmetic in this segment.
Shown in the video
Expression
Explanation
The speaker notes that using differentials is less direct because one must first pass through the change-in- $f$ formulation.
Justification
Explicitly stated in the closing audio.
Shown in the video

Answer

The video asserts that the $\sqrt{35.8}$ example works in the same way under the differential framework, but the actual computation is not shown here.

Verification

Cannot be verified within this clip because no numerical data, intermediate steps, or final value are displayed.

Approximating $f(35.8)$ using differentials

Clear evidence

Shown in the video

Evidence

Formula
Observation
The lower board shows f'(36)= $1/12$ , Δf ≈ $1/12(-0.2) = -1/60$ , and $Δf = f(35.8)-6$ .
Audio
Observation
The speaker narrates the whole substitution and interpretation step by step.

Uncertainties

The original function f is not named in this clip.
The final decimal or fractional value of $f(35.8)$ is not explicitly written before the clip ends.

Problem

Given f'(36)= $1/12$ , $f(36)=6$ , and moving from $x=36$ to $x=35.8$ , approximate the change in f and use it to estimate $f(35.8)$ .

Given

Base point $a = 36$ .
Target point $x = 35.8$ .
f'(36) = $1/12$ .
$f(36) = 6$ .
$Δx = 35.8 - 36 = -0.2$ .

Goal

Compute Δf ≈ df and relate it to $f(35.8)-f(36)$ .

Steps

Expression
$\Delta x = 35.8 - 36 = -0.2$
Explanation
Find the input change from the base point to the target point.
Justification
Direct subtraction of the two x-values.
Shown in the video
Expression
$\Delta f \approx df = f'(36)\,\Delta x$
Explanation
Apply the differential approximation formula at the base point 36.
Justification
Use the method Δf ≈ df = f'(a)Δx shown on the board.
Shown in the video
Expression
$= \frac{1}{12}(-0.2) = -\frac{1}{60}$
Explanation
Substitute the given derivative and input change, then simplify.
Justification
Arithmetic substitution and multiplication.
Shown in the video
Expression
$\Delta f = f(35.8) - f(36) = f(35.8) - 6$
Explanation
Rewrite the actual change in f using the definition of Δf and the known value $f(36)=6$ .
Justification
Definition $Δf=f(x)-f(a)$ .
Shown in the video
Expression
$f(35.8) \approx 6 - \frac{1}{60}$
Explanation
Add the approximated change to the known base value to estimate $f(35.8)$ .
Justification
Solve $f(35.8)-6$ ≈ -1/60 for $f(35.8)$ .
Derived from the video

Answer

Δf ≈ -1/60, so $f(35.8)$ is estimated by $6 + (-1/60) = 6 - 1/60$ .

Verification

The speaker states that after adding the approximated change back, this gives the same answer as the earlier linearization method.

Visual events · 6

Writing Definitions

Clear evidence

Shown in the video

Evidence

Animation
Observation
Handwritten text appears on screen defining differentials.

Objects

Text: Definition If $y = f(x)$ ...
Equation: $dy = f$ '(x) dx

Changes

Text is written sequentially.

Invariants

Background slide remains static.

Interpretation

Visual reinforcement of the spoken definitions.

Highlighting Fraction Structure

Clear evidence

Shown in the video

Evidence

Animation
Observation
Purple brackets appear under dy and dx in the fraction $dy/dx$ .

Objects

Equation: $dy = (dy/dx) dx$
Purple brackets

Changes

Brackets are drawn to emphasize the numerator and denominator.

Invariants

Equation content does not change.

Interpretation

Visual aid to discuss the 'fraction-like' appearance of the derivative notation.

Sequential color-coded derivation on the whiteboard

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Blue writing presents the differential line $df=f'(x)dx=f'(x)\Delta x$ ; orange writing develops the linearization chain $L(x)=f(a)+f'(a)(x-a)$ to $\Delta f=f(x)-f(a)\approx f'(a)\Delta x$ ; red arrows label $(x-a)$ as $\Delta x$ ; the final summary returns to blue under the heading "Linearization with differentials".
Animation
Observation
Equations are added sequentially from top to bottom as the speaker talks through the derivation.

Objects

Blue differential formulas
Orange linearization formulas
Red arrow labels identifying $\Delta x$
Heading text "Linearization with differentials"

Changes

The board first establishes $df=f'(x)dx$ and then rewrites it with $\Delta x$ .
It next introduces $L(x)$ and transforms the linearization into a change formula.
It then converts the exact change in $L$ into an approximation for the true change in $f$ .
Finally it condenses the result into $\Delta f\approx df=f'(a)dx=f'(a)\Delta x$ .

Invariants

The base-point structure of the linearization remains centered at $a$ .
The derivative factor remains the slope controlling the predicted change.
The identification $dx=\Delta x$ persists through the final summary.

Interpretation

The visual organization contrasts the differential viewpoint (blue) with the linearization viewpoint (orange), then merges them in the concluding formula showing that the differential is the linearized change in the function.

Annotation identifying $(x-a)$ as $\Delta x$

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A red arrow points from the factor $(x-a)$ in the linearization formula to a handwritten label $\Delta x$ .
Audio
Observation
The speaker says, "this here is really a $\Delta x$ . It's a change in $x$ values from $a$ to $x$ ".

Objects

The term $(x-a)$ in $L(x)=f(a)+f'(a)(x-a)$
Red arrow
Handwritten label $\Delta x$

Changes

The symbolic displacement $x-a$ is visually relabeled as the finite increment $\Delta x$ .

Invariants

The algebraic meaning of the factor remains the distance from the base point $a$ to the target point $x$ .

Interpretation

This annotation is the bridge that lets the linearization formula be rewritten in the same increment notation used by the differential formula.

Scroll reveals the differential-based example area

Clear evidence

Shown in the video

Evidence

Animation
Observation
At about 2 seconds the page scrolls upward, moving the top formulas partly out of view and centering the section titled 'Linearization with differentials'.
Diagram
Observation
The right sidebar remains fixed with 'Calculus 1', 'Linearization and Differentials', and 'Video #2: Differentials'.

Objects

Whiteboard formulas
Title 'Linearization with differentials'
Right sidebar text

Changes

The upper formulas shift upward and partly out of frame.
The lower working area becomes centered for the example.

Invariants

The sidebar course information stays fixed.
The main differential formula remains visible.

Interpretation

The visual transition shifts attention from the general setup to the concrete example using differentials.

Step-by-step writing of the worked example

Clear evidence

Shown in the video

Evidence

Animation
Observation
Blue handwriting appears sequentially: first f'(36)= $1/12$ , then Δf ≈ $1/12(-0.2)$ , then = -1/60, then $f(35.8)-6$ .
Diagram
Observation
An orange underline emphasizes parts of the formula, and a pink/magenta arrow points from $f(35.8)-6$ back toward -1/60.

Objects

f'(36)= $1/12$
Δf ≈ $1/12(-0.2)$
-1/60
$f(35.8)-6$
pink arrow

Changes

Each new line is written in sequence as the speaker explains the substitution.
The arrow visually links the expression for Δf to its numerical approximation.

Invariants

The general formula Δf ≈ df = f'(a) $dx = f$ '(a)Δx stays on screen above the example.

Interpretation

The animation shows the logical flow from the general differential formula to a specific numerical approximation and then to recovering $f(35.8)$ .

Misconceptions · 5

Treating Derivatives as Fractions

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker explicitly warns: 'it's kind of like you're canceling out fractions but you're not because this thing in here really isn't a fraction'.

Misconception

Students might think they can algebraically cancel dx in the expression ( $dy/dx$ ) * dx as if it were a standard fraction.

Clarification

While the notation resembles a fraction and the ratio holds, the derivative itself is a limit, not a quotient of infinitesimal numbers, so formal cancellation rules don't strictly apply in the same way.

Thinking the differential route bypasses the change-in- $f$ idea

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the earlier $\sqrt{35.8}$ method "would work in the same way. It's not as direct because you have to go through this whole change of $f$ step first."

Misconception

One might think using differentials immediately replaces the linearization procedure without any intermediate conceptual step.

Clarification

The video stresses that the differential approach still relies on first understanding the actual change $\Delta f=f(x)-f(a)$ ; the differential formula is a compact way to approximate that change, not a separate unrelated trick.

Confusing the differential $dx$ with the finite increment $\Delta x$

Clear evidence

Derived from the video

Evidence

Formula
Observation
The board explicitly writes "Going to set $dx = \Delta x$ " before substituting into $df=f'(x)dx$ .
Audio
Observation
The speaker says, "if I set $dx$ equal to some change in $x$ or a $\Delta x$ ".

Uncertainties

This clarification is inferred from the video's explicit substitution step rather than stated as a warning in those words.

Misconception

Students may treat $dx$ and $\Delta x$ as always identical in every context.

Clarification

In this clip the equality $dx=\Delta x$ is introduced as a deliberate substitution for applying the differential to approximate a finite change. The video does not claim a general metaphysical identity beyond this approximation setup.

Confusing actual change with differential

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board distinguishes $Δf = f(x)-f(a)$ from df = f'(a)dx and writes Δf ≈ df.
Audio
Observation
The speaker says 'my change in f, I could approximate by...' rather than equating them exactly.

Misconception

One might think Δf and df are exactly the same quantity.

Clarification

The video treats Δf as the actual change in the function and df as the linear approximation to that change, related by Δf ≈ df = f'(a)Δx.

Ignoring direction of the input change

Clear evidence

Shown in the video

Evidence

Audio
Observation
'the change in x, which was minus 0.2, because we're going from 36 to 35.8.'
Formula
Observation
The board uses (-0.2) explicitly in the product.

Misconception

One might use 0.2 instead of -0.2 when moving from 36 down to 35.8.

Clarification

Because the input decreases, $Δx = dx = -0.2$ , and this sign must be included in the approximation.

Concept relations · 11

Definition of Dependent Differential → Definition of Function Differential

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says 'Different notation but the same idea'.

Equivalent

Explanation

dy and df represent the same mathematical object when $y = f(x)$ .

Definition of Independent Differential → Setting up Linear Approximation with Differentials

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker transitions to 'How is this useful for linearization'.

Application

Explanation

The concept of differentials is applied to set up linear approximation problems.

Linearization of $f$ at $a$ → Using differentials to approximate $\Delta f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The orange linearization chain leads to $f(x)-f(a)\approx f'(a)\Delta x$ , and the final blue line rewrites this as $\Delta f\approx df=f'(a)dx=f'(a)\Delta x$ .
Audio
Observation
The speaker explicitly compares the two by saying the differential expression "looks a lot like our linearization" and then summarizes "Linearization with differentials".

Application

Explanation

The differential approximation method is applied to the same local situation analyzed by linearization, converting the tangent-line estimate into the compact rule $\Delta f\approx df$ .

Definition of $\Delta f$ → Linear approximation to the actual change in $f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board first writes $f(x)-f(a)\approx f'(a)\Delta x$ and then names the left-hand side as $\Delta f=f(x)-f(a)$ .
Audio
Observation
The speaker says, "This over here is a change in $f$ " while pointing to $f(x)-f(a)$ .

Contains

Explanation

The definition $\Delta f=f(x)-f(a)$ packages the left-hand side of the approximation formula into a single symbol, making the later statement $\Delta f\approx df$ possible.

Differential written with $dx=\Delta x$ → Using differentials to approximate $\Delta f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The opening line $df=f'(x)dx=f'(x)\Delta x$ reappears in the closing summary specialized to the base point $a$ as $df=f'(a)dx=f'(a)\Delta x$ .
Audio
Observation
The speaker begins with the general differential formula and ends by saying the differential is defined using $f'$ at the initial point $a$ .

Special case

Explanation

The final approximation rule specializes the differential formula to the base point $a$ and uses it as the linear estimate of the actual change $\Delta f$ .

Change in the linearization equals slope times input change → Linear approximation to the actual change in $f$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board moves from $L(x)-f(a)=f'(a)\Delta x$ to $f(x)-f(a)\approx f'(a)\Delta x$ .
Audio
Observation
The speaker says that because $L$ is approximately $f$ at $x$ , the same right-hand side approximates the true change in $f$ .

Generalizes

Explanation

The exact change formula for the linear model is turned into an approximate statement about the original nonlinear function by invoking $L(x)\approx f(x)$ .

Change in function value Δf → Linearization with differentials

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board first defines $Δf = f(x)-f(a)$ and then applies Δf ≈ df = f'(a)Δx.

Prerequisite

Explanation

The differential method depends on knowing what Δf means before approximating it by df.

Linearization with differentials → Approximating $f(35.8)$ using differentials

Clear evidence

Shown in the video

Evidence

Formula
Observation
The example substitutes $a=36$ , f'(36)= $1/12$ , and $Δx=-0.2$ into Δf ≈ f'(a)Δx.

Application

Explanation

The worked example is a direct application of the differential approximation method.

Derivative value at the base point in the example → Approximating $f(35.8)$ using differentials

Clear evidence

Shown in the video

Evidence

Formula
Observation
f'(36)= $1/12$ is written separately and then used in Δf ≈ $1/12(-0.2)$ .

Prerequisite

Explanation

The example requires the derivative value at the base point before the approximation can be computed.

Input change in the example → Approximating $f(35.8)$ using differentials

Clear evidence

Shown in the video

Evidence

Formula
Observation
The factor (-0.2) in the example comes from moving from 36 to 35.8.

Prerequisite

Explanation

The numerical approximation uses the input change $Δx = -0.2$ .

Linearization with differentials → Equivalence with prior linearization method

Clear evidence

Shown in the video

Evidence

Audio
Observation
The closing statement says this is a new technique for viewing linearization using differentials.

Proof dependency

Explanation

The claim that this method gives the same answer relies on applying the differential formulation to the same linearization problem.

Find an answer · 16

What is the definition of the independent differential dx?

Clear evidence

Shown in the video

Evidence

Audio
Observation
Definition of dx.

Knowledge points

Definition of Independent Differential

How do you calculate the dependent differential dy?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Formula for dy.

Knowledge points

Definition of Dependent Differential

Why can't we treat $dy/dx$ as a simple fraction?

Clear evidence

Shown in the video

Evidence

Audio
Observation
Warning about fractions.

Knowledge points

Nature of the Derivative Ratio
Treating Derivatives as Fractions

What is the differential $df$ in terms of $f'(x)$ and $dx$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Opening board line $df=f'(x)dx$ .
Audio
Observation
Speaker defines $df$ as $f'$ at $x$ times $dx$ .

Knowledge points

Differential written with $dx=\Delta x$

Why does the video replace $dx$ with $\Delta x$ in the differential formula?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Header note "Going to set $dx = \Delta x$ " and substitution to $f'(x)\Delta x$ .
Audio
Observation
Speaker says he is setting $dx$ equal to a change in $x$ or a $\Delta x$ .

Knowledge points

Differential written with $dx=\Delta x$
Confusing the differential $dx$ with the finite increment $\Delta x$

What is the formula for the linearization $L(x)$ of $f$ at $a$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Board writes $L(x)=f(a)+f'(a)(x-a)$ .
Audio
Observation
Speaker recites the linearization formula at the base point $a$ .

Knowledge points

Linearization of $f$ at $a$

How is the factor $(x-a)$ in the linearization related to $\Delta x$ ?

Clear evidence

Shown in the video

Evidence

Diagram
Observation
Red arrow labels $(x-a)$ as $\Delta x$ .
Audio
Observation
Speaker says $(x-a)$ is really a $\Delta x$ , a change in $x$ values from $a$ to $x$ .

Knowledge points

Linearization of $f$ at $a$
Change in the linearization equals slope times input change

Why can we say $f(x)-f(a)\approx f'(a)\Delta x$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Board writes $f(x)-f(a)\approx f'(a)\Delta x$ .
Audio
Observation
Speaker explains this follows because $L(x)$ is approximately $f(x)$ .

Knowledge points

Linear approximation to the actual change in $f$
s120-rel-change-in-L-to-approx-change-in-f

What does $\Delta f$ mean in this lesson?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Board writes $\Delta f=f(x)-f(a)$ .
Audio
Observation
Speaker calls this expression a change in $f$ .

Knowledge points

Definition of $\Delta f$

How do differentials approximate the actual change in a function value?

Clear evidence

Shown in the video

Evidence

Formula
Observation
Final summary line $\Delta f\approx df=f'(a)dx=f'(a)\Delta x$ .
Audio
Observation
Speaker says the idea is to approximate the change in $f$ by the differential of $f$ .

Knowledge points

Using differentials to approximate $\Delta f$
Core differential approximation principle

What is the relationship between linearization and differentials in this clip?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board moves from $L(x)=f(a)+f'(a)(x-a)$ to $\Delta f\approx df$ .
Audio
Observation
Speaker explicitly compares the differential expression to the linearization and titles the conclusion "Linearization with differentials".

Knowledge points

Linearization of $f$ at $a$
Using differentials to approximate $\Delta f$
s120-rel-diff-from-linearization

How would the earlier $\sqrt{35.8}$ approximation fit the differential method?

Approximate timing

Shown in the video

Evidence

Audio
Observation
Speaker references the earlier $\sqrt{35.8}$ example and says it would work in the same way but is less direct.

Uncertainties

The actual numerical example is not shown in this segment.

Knowledge points

Reference to the earlier $\sqrt{35.8}$ example
Using differentials to approximate $\Delta f$

Coverage and review notes

Covered · Introduction to the topic of differentials and linear approximation.

Covered · Definition of independent differential dx.

Covered · Definition of dependent differential dy.

Covered · Discussion on the relationship between differentials and derivatives, and the misconception of fraction cancellation.

Covered · Introduction of df notation.

Covered · Application of differentials to linearization by setting dx = Delta x.

Covered · Opening differential formula and the substitution $dx=\Delta x$ are fully captured.

Covered · Transitional spoken comparison from the differential expression to the upcoming linearization formula.

Covered · Linearization formula is written and $(x-a)$ is identified as $\Delta x$ .

Covered · Exact change in $L(x)$ is derived and then converted toward an approximation for $f(x)-f(a)$ .

Covered · Approximation $f(x)-f(a)\approx f'(a)\Delta x$ is written and named as $\Delta f$ .

Covered · Final synthesis $\Delta f\approx df=f'(a)dx=f'(a)\Delta x$ is presented, followed by the verbal reference to the earlier $\sqrt{35.8}$ example.

Covered · The entire 44-second clip consists of a single worked explanation of linearization with differentials, including the general formula, the numerical substitution, and the interpretation for recovering $f(35.8)$ .

Explore the knowledge in this video

Reviewed subject paths

Linearization and Differentials - Video 2 - Differentials

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Independent Differential (dx)

Dependent Differential (dy)

Differential vs. Fraction Misconception

Function Differential (df)

Linearization Setup

Differential formula with dx=Δxdx=\Delta xdx=Δx

Linearization at a base point

(x−a)(x-a)(x−a) is the input increment Δx\Delta xΔx

Exact change predicted by the linearization

Approximating the true change in fff

Definition of Δf\Delta fΔf

Linearization with differentials

Connection to the earlier 35.8\sqrt{35.8}35.8​ example

Linearization with differentials

Meaning of Δf

Worked example setup

Computing the approximated change

Recovering f(35.8)f(35.8)f(35.8) from the approximation

Why differentials are useful here

Detailed learning notes

y

f(x)f(x)f(x)

dx

dy

f'(x)

df

Δx\Delta xΔx

df

f'(x)

dx

Δx\Delta xΔx

L(x)L(x)L(x)

Definition of Independent Differential

Definition of Dependent Differential

Definition of Function Differential

Setting up Linear Approximation with Differentials

Differential written with dx=Δxdx=\Delta xdx=Δx

Linearization of fff at aaa

Change in the linearization equals slope times input change

Linear approximation to the actual change in fff

Definition of Δf\Delta fΔf

Using differentials to approximate Δf\Delta fΔf

Linearization with differentials

Change in function value Δf

Nature of the Derivative Ratio

Differential identity used at the start

Exact change in the linearization

Linear approximation of the true function change

Core differential approximation principle

Differential approximation claim

Equivalence with prior linearization method

Relating Differentials to Derivative Notation

Rewriting the differential using dx=Δxdx=\Delta xdx=Δx

From linearization to the change in L(x)L(x)L(x)

Connecting linear approximation to the differential formula

Computing the differential approximation in the example

Using Δf to recover the nearby function value

Reference to the earlier 35.8\sqrt{35.8}35.8​ example

Approximating f(35.8)f(35.8)f(35.8) using differentials

Writing Definitions

Highlighting Fraction Structure

Sequential color-coded derivation on the whiteboard

Annotation identifying (x−a)(x-a)(x−a) as Δx\Delta xΔx

Scroll reveals the differential-based example area

Step-by-step writing of the worked example

Treating Derivatives as Fractions

Thinking the differential route bypasses the change-in-fff idea

Confusing the differential dxdxdx with the finite increment Δx\Delta xΔx

Confusing actual change with differential

Ignoring direction of the input change

Definition of Dependent Differential → Definition of Function Differential

Definition of Independent Differential → Setting up Linear Approximation with Differentials

Linearization of fff at aaa → Using differentials to approximate Δf\Delta fΔf

Definition of Δf\Delta fΔf → Linear approximation to the actual change in fff

Differential written with dx=Δxdx=\Delta xdx=Δx → Using differentials to approximate Δf\Delta fΔf

Differential formula with $dx=\Delta x$

$(x-a)$ is the input increment $\Delta x$

Approximating the true change in $f$

Definition of $\Delta f$

Connection to the earlier $\sqrt{35.8}$ example

Recovering $f(35.8)$ from the approximation

$f(x)$

$\Delta x$

$\Delta x$

$L(x)$

Differential written with $dx=\Delta x$

Linearization of $f$ at $a$

Linear approximation to the actual change in $f$

Definition of $\Delta f$

Using differentials to approximate $\Delta f$

Rewriting the differential using $dx=\Delta x$

From linearization to the change in $L(x)$

Reference to the earlier $\sqrt{35.8}$ example

Approximating $f(35.8)$ using differentials

Annotation identifying $(x-a)$ as $\Delta x$

Thinking the differential route bypasses the change-in- $f$ idea

Confusing the differential $dx$ with the finite increment $\Delta x$

Linearization of $f$ at $a$ → Using differentials to approximate $\Delta f$

Definition of $\Delta f$ → Linear approximation to the actual change in $f$

Differential written with $dx=\Delta x$ → Using differentials to approximate $\Delta f$

Change in the linearization equals slope times input change → Linear approximation to the actual change in $f$

Linearization with differentials → Approximating $f(35.8)$ using differentials

Derivative value at the base point in the example → Approximating $f(35.8)$ using differentials

Input change in the example → Approximating $f(35.8)$ using differentials

Why can't we treat $dy/dx$ as a simple fraction?

What is the differential $df$ in terms of $f'(x)$ and $dx$ ?

Why does the video replace $dx$ with $\Delta x$ in the differential formula?

What is the formula for the linearization $L(x)$ of $f$ at $a$ ?

How is the factor $(x-a)$ in the linearization related to $\Delta x$ ?

Why can we say $f(x)-f(a)\approx f'(a)\Delta x$ ?

What does $\Delta f$ mean in this lesson?

How would the earlier $\sqrt{35.8}$ approximation fit the differential method?