Gradient

1. Formula
   Observation

   The board shows $f(x,y)=x^2\sin(y)$.

2. Audio
   Observation

   The speaker says, "let's say it's f of x y equals x squared sine of y."

$f(x,y)$

A two-variable scalar function used as the example for computing a gradient.

Two-variable real function; the video does not state an explicit domain.

1. Formula
   Observation

   x appears in $f(x,y)=x^2\sin(y)$, $\frac{\partial f}{\partial x}=2x\sin(y)$, and $\nabla f(x,y)$.

2. Audio
   Observation

   The speaker treats x as the variable when differentiating with respect to x and as a constant when differentiating with respect to y.

x

First input coordinate/variable of the two-variable function.

Real variable in the displayed multivariable example; no explicit domain is stated.

1. Formula
   Observation

   y appears in $f(x,y)=x^2\sin(y)$, $\frac{\partial f}{\partial y}=x^2\cos(y)$, and $\nabla f(x,y)$.

2. Audio
   Observation

   The speaker treats y as the variable when differentiating with respect to y and as a constant when differentiating with respect to x.

y

Second input coordinate/variable of the two-variable function.

Real variable in the displayed multivariable example; no explicit domain is stated.

1. Formula
   Observation

   The board writes $\frac{\partial f}{\partial x}=2x\sin(y)$.

2. Audio
   Observation

   The speaker says, "partial of f with respect to x" and explains that x is the variable and y is the constant.

$\frac{\partial f}{\partial x}$

Partial derivative of f with respect to x, holding y fixed.

Defined for the displayed function $f(x,y)=x^2\sin(y)$; result shown as $2x\sin(y)$.

1. Formula
   Observation

   The board writes $\frac{\partial f}{\partial y}=x^2\cos(y)$.

2. Audio
   Observation

   The speaker says, "the partial derivative with respect to y" and explains that x is considered a constant.

$\frac{\partial f}{\partial y}$

Partial derivative of f with respect to y, holding x fixed.

Defined for the displayed function $f(x,y)=x^2\sin(y)$; result shown as $x^2\cos(y)$.

1. Formula
   Observation

   The board writes $\nabla f$ and later $\nabla f(x,y)$.

2. Audio
   Observation

   The speaker says, "You denote it with a little upside down triangle. The name of that symbol is nabla, but you often just pronounce it del."

$\nabla$

Nabla/del operator symbol used to denote the gradient.

Used here as gradient notation for scalar functions.

1. Formula
   Observation

   The board writes $\nabla f$ = [$2x\sin(y)$; $x^2\cos(y)$].

2. Audio
   Observation

   The speaker says, "what the gradient does is it just puts both of these together in a vector."

$\nabla f$

Gradient of f, written as a vector whose components are the partial derivatives of f.

For the displayed two-variable example, it outputs a two-dimensional vector.

1. Formula
   Observation

   The speaker adds (x,y) after $\nabla f$, yielding $\nabla f(x,y)$.

2. Audio
   Observation

   The speaker says, "This is actually a vector valued function... This is a function that takes in a point in two dimensional space and outputs a two dimensional vector."

$\nabla f(x,y)$

The gradient viewed explicitly as a function of the input point (x,y).

Takes a point in two-dimensional space and outputs a two-dimensional vector in this example.

1. Formula
   Observation

   The gradient is written inside large square brackets with one entry above another.

2. Audio
   Observation

   The speaker says the gradient "puts both of these together in a vector" and refers to "the first one" and "the bottom one."

[a;b]

Column-vector notation used on the board to stack the partial derivatives vertically.

Used for two-component vectors in this clip; the speaker also mentions a three-dimensional output for three variables.

1. Formula
   Observation

   The lower part of the board begins $\nabla f$ = [$\frac{\partial f}{\partial x}$; ...].

2. Audio
   Observation

   The speaker says, "the gradient of any function is equal to a vector with its partial derivatives, partial of f with respect to x."

1. The final general formula is not completed before the clip ends; only the top component $\frac{\partial f}{\partial x}$ is visible by the end.

$\nabla f$ = [$\frac{\partial f}{\partial x}$; $\ldots$]

General gradient notation introduced at the end as a vector containing the function’s partial derivatives.

Intended for a general function; the full number of components is not completed within this clip.

1. Formula
   Observation

   Displayed at the top center as $f(x,y) = x^2 \sin (y)$.

$f(x,y)$

A scalar-valued function of two variables.

Two-dimensional input space (x,y).

1. Formula
   Observation

   Used in $f(x,y)$ and partial derivative with respect to x.

x

First independent variable of the function.

Real numbers.

1. Audio
   Observation

   The speaker says, "So here I'm going to talk about the gradient. And in this video, I'm only going to describe how you compute the gradient, and in the next couple ones, I'm going to give the geometric interpretation."

2. Diagram
   Observation

   The title "Gradient" is written at the upper left.

This segment announces that the current lesson will focus on computing the gradient rather than explaining its geometric meaning. The speaker explicitly separates computation from the later geometric interpretation.

1. Applies to the present video segment only.

2. Geometric interpretation is deferred to later videos.

1. Formula
   Observation

   The board writes $f(x,y)=x^2\sin(y)$.

2. Audio
   Observation

   The speaker says, "let's say you have some sort of function, and I'm just going to make it a two variable function, and let's say it's f of x y equals x squared sine of y."

The clip introduces a concrete two-variable function as the worked example for gradient computation. The function depends on variables x and y and combines a polynomial factor $x^2$ with the trigonometric factor $\sin(y)$.

1. The function has two input variables, x and y.

2. No explicit domain is stated in the video.

1. Scope of this clip: computational introduction to the gradient

1. Audio
   Observation

   The speaker says, "The gradient is a way of packing together all the partial derivative information of a function."

2. Formula
   Observation

   Immediately afterward the board begins listing partial derivatives of the example function.

The gradient is defined operationally as the object that gathers all partial derivative information of a function into one vector. In this clip, that means collecting the partial derivatives with respect to each input variable.

1. Presented for a multivariable function.

2. In the example, the function has two variables.

1. Example scalar function $f(x,y)=x^2\sin(y)$

1. Formula
   Observation

   The board writes $\frac{\partial f}{\partial x}=2x\sin(y)$.

2. Audio
   Observation

   The speaker says, "we consider x the variable and y the constant... the derivative of x is 2x, so we see that this will be 2x times that constant sine of y."

When differentiating $f(x,y)=x^2\sin(y)$ with respect to x, y is treated as constant, so $\sin(y)$ is also constant. The derivative of $x^2$ is 2x, giving the displayed result.

1. Differentiate with respect to x.

2. Hold y constant.

3. Applied to $f(x,y)=x^2\sin(y)$.

1. Example scalar function $f(x,y)=x^2\sin(y)$
2. Gradient as a collection of partial derivative information

1. Formula
   Observation

   The board writes $\frac{\partial f}{\partial y}=x^2\cos(y)$.

2. Audio
   Observation

   The speaker says, "x is considered a constant, so x squared is also considered a constant... that same constant times the cosine of y, which is the derivative of sine."

When differentiating $f(x,y)=x^2\sin(y)$ with respect to y, x is treated as constant, so $x^2$ is constant. The derivative of $\sin(y)$ is $\cos(y)$, giving the displayed result.

1. Differentiate with respect to y.

2. Hold x constant.

3. Applied to $f(x,y)=x^2\sin(y)$.

1. Example scalar function $f(x,y)=x^2\sin(y)$
2. Gradient as a collection of partial derivative information

1. Formula
   Observation

   The board writes $\nabla f$ = [$2x\sin(y)$; $x^2\cos(y)$].

2. Audio
   Observation

   The speaker says, "what the gradient does is it just puts both of these together in a vector... the first one is the partial derivative with respect to x... and the bottom one, partial derivative with respect to y."

The gradient is formed by stacking the computed partial derivatives into a vector. For this two-variable example, the top component is $\frac{\partial f}{\partial x}$ and the bottom component is $\frac{\partial f}{\partial y}$.

1. The function has two variables in this example.

2. Components are ordered as x-partial then y-partial.

1. Partial derivative of f with respect to x
2. Partial derivative of f with respect to y
3. Gradient as a collection of partial derivative information

1. Formula
   Observation

   The symbol $\nabla$ is written before f.

2. Audio
   Observation

   The speaker says, "You denote it with a little upside down triangle. The name of that symbol is nabla, but you often just pronounce it del. You'd say del f or gradient of f."

The gradient is denoted by $\nabla f$. The symbol $\nabla$ is called nabla, though the speaker notes it is often pronounced "del," so $\nabla f$ may be read as "del f" or "gradient of f."

1. Used to denote the gradient of a function.

1. Gradient of the example function as a column vector

1. Formula
   Observation

   The speaker rewrites the left side as $\nabla f(x,y)$.

2. Audio
   Observation

   The speaker says, "this is actually a vector valued function... This is a function that takes in a point in two dimensional space and outputs a two dimensional vector."

The clip emphasizes that $\nabla f$ is not merely a static vector expression but a function of the input point. In the two-variable example, it maps a point (x,y) in two-dimensional space to a two-dimensional vector.

1. The input is a point (x,y) in two-dimensional space.

2. The output is a two-dimensional vector in this example.

1. Gradient of the example function as a column vector
2. Notation $\nabla f$ for the gradient

1. Audio
   Observation

   The speaker says, "you could also imagine doing this with three different variables, then you would have three partial derivatives and a three dimensional output."

The speaker states that the same construction extends beyond two variables: with three input variables, the gradient would contain three partial derivatives and produce a three-dimensional vector.

1. Applies when the function has three variables.

2. Only stated verbally; no three-variable formula is written in this clip.

1. Gradient as a vector-valued function of the input point

1. Formula
   Observation

   The lower board begins writing $\nabla f$ = [$\frac{\partial f}{\partial x}$; ...].

2. Audio
   Observation

   The speaker says, "the gradient of any function is equal to a vector with its partial derivatives, partial of f with respect to x."

1. The general formula is incomplete by the end of the clip; only the first component is clearly established before cutoff.

At the end of the clip, the speaker starts writing a more general expression for the gradient of any function as a vector made from its partial derivatives. The visible and audible portion establishes the top component $\frac{\partial f}{\partial x}$, but the full formula is not completed within this segment.

1. Intended as a general statement for a function with multiple variables.

2. Full component list is not completed in this clip.

1. Gradient as a collection of partial derivative information
2. Notation $\nabla f$ for the gradient

1. Formula
   Observation

   Shows ∇$f(x,y)$ constructed from ∂$f/x$ and ∂$f/y$.

2. Audio
   Observation

   Speaker says 'we call these partial derivatives... I like to think of the gradient as the full derivative because it kind of captures all of the information that you need.'

The gradient of a scalar function $f(x,y)$ is a vector whose components are the partial derivatives of f with respect to each independent variable. It is denoted by ∇f.

1. f must be a differentiable scalar function of multiple variables.

1. Computing Partial Derivatives

1. Formula
   Observation

   Displays ∂f/∂$x = 2x \sin (y)$ and ∂f/∂$y = x^2 \cos (y)$.

To find the partial derivative with respect to one variable, treat all other variables as constants and differentiate normally.

1. The function must be differentiable with respect to the variable being differentiated.

1. Audio
   Observation

   The speaker says, "The gradient is a way of packing together all the partial derivative information of a function."

For a multivariable function, the gradient is the vector obtained by collecting the function’s partial derivative information.

1. The function has partial derivatives with respect to its input variables.

2. The clip presents this in the context of a two-variable example.

Stated generally for a function, but demonstrated only on $f(x,y)=x^2\sin(y)$ in this clip.

1. Audio
   Observation

   The speaker says, "This is a function that takes in a point in two dimensional space and outputs a two dimensional vector."

2. Formula
   Observation

   The board shows $\nabla f(x,y)=\begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$.

For the displayed two-variable function, $\nabla f(x,y)$ is a vector-valued function whose input is a point in two-dimensional space and whose output is a two-dimensional vector.

1. The function has two input variables x and y.

2. The gradient is formed from the two partial derivatives shown on the board.

Explicitly stated for the two-variable example; the speaker separately mentions a three-variable analogue.

1. Audio
   Observation

   The speaker says, "you could also imagine doing this with three different variables, then you would have three partial derivatives and a three dimensional output."

1. No three-variable formula is written on screen in this clip.

If the same construction is applied to a function of three variables, the gradient has three partial derivatives and a three-dimensional output.

1. The function has three input variables.

2. The gradient is built by collecting one partial derivative per variable.

Stated verbally as a general extension, without a written example in this clip.

1. Formula
   Observation

   The board writes $\frac{\partial f}{\partial x}=2x\sin(y)$.

2. Audio
   Observation

   The speaker explains that x is the variable and y is the constant, and that the derivative of $x^2$ is 2x.

1. Expression

   $$f(x,y)=x^2\sin(y)$$
   Explanation

   Start from the given two-variable function.

   Justification

   Given example function written on the board.

   Shown in the video
2. Expression

   $$\text{Treat } y \text{ as constant, so } \sin(y) \text{ is constant with respect to } x.$$
   Explanation

   For the partial derivative with respect to x, only x varies.

   Justification

   Definition of partial differentiation with respect to x, as stated by the speaker.

   Shown in the video
3. Expression

   $$\frac{d}{dx}(x^2)=2x$$
   Explanation

   Differentiate the x-dependent factor.

   Justification

   Power rule for ordinary single-variable differentiation, invoked verbally by the speaker.

   Shown in the video
4. Expression

   $$\frac{\partial f}{\partial x}=2x\sin(y)$$
   Explanation

   Multiply the derivative of $x^2$ by the constant factor $\sin(y)$.

   Justification

   Constant-multiple rule under partial differentiation with respect to x.

   Shown in the video

The partial derivative of the example function with respect to x is $\frac{\partial f}{\partial x}=2x\sin(y)$.

1. Formula
   Observation

   The board writes $\frac{\partial f}{\partial y}=x^2\cos(y)$.

2. Audio
   Observation

   The speaker explains that x is considered a constant and that the derivative of sine is cosine.

1. Expression

   $$f(x,y)=x^2\sin(y)$$
   Explanation

   Start again from the same function.

   Justification

   Given example function written on the board.

   Shown in the video
2. Expression

   $$\text{Treat } x \text{ as constant, so } x^2 \text{ is constant with respect to } y.$$
   Explanation

   For the partial derivative with respect to y, only y varies.

   Justification

   Definition of partial differentiation with respect to y, as stated by the speaker.

   Shown in the video
3. Expression

   $$\frac{d}{dy}(\sin(y))=\cos(y)$$
   Explanation

   Differentiate the y-dependent factor.

   Justification

   Standard derivative of sine, explicitly mentioned by the speaker.

   Shown in the video
4. Expression

   $$\frac{\partial f}{\partial y}=x^2\cos(y)$$
   Explanation

   Multiply the constant factor $x^2$ by the derivative $\cos(y)$.

   Justification

   Constant-multiple rule under partial differentiation with respect to y.

   Shown in the video

The partial derivative of the example function with respect to y is $\frac{\partial f}{\partial y}=x^2\cos(y)$.

1. Formula
   Observation

   The board writes $\nabla f = \begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$.

2. Audio
   Observation

   The speaker says the gradient "puts both of these together in a vector," with the x-partial on top and the y-partial on the bottom.

1. Expression

   $$\frac{\partial f}{\partial x}=2x\sin(y)$$
   Explanation

   Use the previously computed x-partial as the first component.

   Justification

   Already derived on the board from the example function.

   Shown in the video
2. Expression

   $$\frac{\partial f}{\partial y}=x^2\cos(y)$$
   Explanation

   Use the previously computed y-partial as the second component.

   Justification

   Already derived on the board from the example function.

   Shown in the video
3. Expression

   $$\nabla f = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix}$$
   Explanation

   Place the partial derivatives into a column vector in the order x then y.

   Justification

   Definition of the gradient as the vector collecting partial derivative information, stated by the speaker.

   Shown in the video
4. Expression

   $$\nabla f = \begin{bmatrix} 2x\sin(y) \\ x^2\cos(y) \end{bmatrix}$$
   Explanation

   Substitute the computed component formulas into the vector.

   Justification

   Direct substitution of the two derived partial derivatives.

   Shown in the video

For $f(x,y)=x^2\sin(y)$, the gradient is $\nabla f = \begin{bmatrix} 2x\sin(y) \\ x^2\cos(y) \end{bmatrix}$.

1. Formula
   Observation

   Step-by-step calculation shown on screen for $f(x,y) = x^2 \sin (y)$.

1. Expression

   $$f(x,y) = x^2 \sin(y)$$
   Explanation

   Start with the given scalar function.

   Justification

   Given in the problem statement.

   Shown in the video
2. Expression

   $$\frac{\partial f}{\partial x} = 2x \sin(y)$$
   Explanation

   Differentiate f with respect to x, treating y as a constant.

   Justification

   Power rule and constant multiple rule for partial differentiation.

   Shown in the video
3. Expression

   $$\frac{\partial f}{\partial y} = x^2 \cos(y)$$
   Explanation

   Differentiate f with respect to y, treating x as a constant.

   Justification

   Derivative of sine is cosine; $x^2$ is treated as a constant coefficient.

   Shown in the video
4. Expression

   $$\nabla f(x,y) = \begin{bmatrix} 2x \sin(y) \\ x^2 \cos(y) \end{bmatrix}$$
   Explanation

   Assemble the partial derivatives into the gradient vector.

   Justification

   Definition of the gradient vector.

   Shown in the video

The gradient of $f(x,y) = x^2 \sin (y)$ is the vector [$2x \sin (y)$, $x^2 \cos (y)$]^T.

1. Formula
   Observation

   The board successively shows $f(x,y)=x^2\sin(y)$, $\frac{\partial f}{\partial x}=2x\sin(y)$, $\frac{\partial f}{\partial y}=x^2\cos(y)$, and $\nabla f(x,y)=\begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$.

2. Audio
   Observation

   The speaker narrates each step of computing the partial derivatives and then assembling them into the gradient.

Given the two-variable function $f(x,y)=x^2\sin(y)$, compute its gradient.

1. $f(x,y)=x^2\sin(y)$

2. The function has two variables, x and y.

3. The gradient is formed by collecting the partial derivatives into a vector.

Find $\nabla f(x,y)$ explicitly.

1. Expression

   $$\frac{\partial f}{\partial x}$$
   Explanation

   Begin by computing the partial derivative with respect to x.

   Justification

   The speaker says to start by computing the partial derivatives of the function.

   Shown in the video
2. Expression

   $$\frac{\partial f}{\partial x}=2x\sin(y)$$
   Explanation

   Treat y as constant, differentiate $x^2$ to get 2x, and keep $\sin(y)$ as the constant multiplier.

   Justification

   Partial differentiation rule plus the power rule, as explained in the audio.

   Shown in the video
3. Expression

   $$\frac{\partial f}{\partial y}$$
   Explanation

   Next compute the partial derivative with respect to y.

   Justification

   The speaker moves from the x-partial to the y-partial.

   Shown in the video
4. Expression

   $$\frac{\partial f}{\partial y}=x^2\cos(y)$$
   Explanation

   Treat x as constant, so $x^2$ remains constant, and differentiate $\sin(y)$ to get $\cos(y)$.

   Justification

   Partial differentiation rule plus the standard derivative of sine, as explained in the audio.

   Shown in the video
5. Expression

   $$\nabla f = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix}$$
   Explanation

   Combine the two partial derivatives into a column vector.

   Justification

   Definition of the gradient as packing together all partial derivative information.

   Shown in the video
6. Expression

   $$\nabla f(x,y)=\begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$$
   Explanation

   Substitute the computed components and emphasize that the gradient depends on the input point (x,y).

   Justification

   Direct substitution from the previous steps; the speaker explicitly adds (x,y) to stress vector-valued dependence.

   Shown in the video

$\nabla f(x,y)=\begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$

The answer matches the final board expression and follows directly from the two displayed partial derivatives.

1. Formula
   Observation

   Complete worked example displayed on the whiteboard.

Compute the gradient vector ∇f for the function $f(x,y) = x^2 \sin (y)$.

1. $f(x,y) = x^2 \sin(y)$

Find $\nabla f(x,y)$.

1. Expression

   $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 \sin(y)) = 2x \sin(y)$$
   Explanation

   Calculate the partial derivative with respect to x.

   Justification

   Standard rules of partial differentiation.

   Shown in the video
2. Expression

   $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 \sin(y)) = x^2 \cos(y)$$
   Explanation

   Calculate the partial derivative with respect to y.

   Justification

   Standard rules of partial differentiation.

   Shown in the video
3. Expression

   $$\nabla f(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} 2x \sin(y) \\ x^2 \cos(y) \end{bmatrix}$$
   Explanation

   Construct the gradient vector using the computed partial derivatives.

   Justification

   Definition of the gradient.

   Shown in the video

$\nabla f(x,y) = \begin{bmatrix} 2x \sin(y) \\ x^2 \cos(y) \end{bmatrix}$

The result matches the final expression written on the board.

1. Diagram
   Observation

   The title "Gradient" appears at the upper left, then the formula $f(x,y)=x^2\sin(y)$ is written near the top center.

2. Audio
   Observation

   The speaker introduces the topic as computing the gradient and chooses a two-variable example function.

1. Title text "Gradient"

2. Formula $f(x,y)=x^2\sin(y)$

1. The board begins mostly blank.

2. The title is written first.

3. The example function is added below/right of the title.

1. The topic remains the gradient throughout this interval.

2. The example function stays fixed once written.

The visual setup establishes that the clip will work from one concrete scalar function of two variables toward a gradient computation.

1. Diagram
   Observation

   Two lines are written beneath the example function: $\frac{\partial f}{\partial x}=2x\sin(y)$ and $\frac{\partial f}{\partial y}=x^2\cos(y)$.

2. Audio
   Observation

   The speaker explains one partial derivative at a time, first with respect to x and then with respect to y.

1. Expression $\frac{\partial f}{\partial x}=2x\sin(y)$

2. Expression $\frac{\partial f}{\partial y}=x^2\cos(y)$

1. The x-partial is written first.

2. The y-partial is written second below it.

1. Both expressions refer to the same original function $f(x,y)=x^2\sin(y)$.

2. The order of variables remains x then y.

The board visually separates the two ingredient computations that will later be assembled into the gradient vector.

1. Diagram
   Observation

   To the right, the board writes $\nabla f$ and then a large column vector containing $2x\sin(y)$ above $x^2\cos(y)$.

2. Audio
   Observation

   The speaker says the gradient puts the two partial derivatives together in a vector and later adds (x,y) to emphasize vector-valued dependence.

1. Symbol $\nabla f$

2. Column vector \begin{bmatrix}$2x\sin(y)$\\ This formula needs review. Please report it using the page feedback control.{bmatrix)

3. Added argument (x,y)

1. The gradient symbol is introduced.

2. The vector brackets are drawn.

3. The top component $2x\sin(y)$ is inserted.

4. The bottom component $x^2\cos(y)$ is inserted.

5. The notation is expanded to $\nabla f(x,y)$.

1. The top component corresponds to the x-partial.

2. The bottom component corresponds to the y-partial.

3. The original function and its partials remain visible on the left.

The animation of writing makes clear that the gradient is not a new unrelated quantity but a structured packaging of the previously computed partial derivatives.

1. Diagram
   Observation

   A new lower line begins with $\nabla f$ = and a column vector whose top entry is $\frac{\partial f}{\partial x}$.

2. Audio
   Observation

   The speaker says, "the gradient of any function is equal to a vector with its partial derivatives, partial of f with respect to x."

1. The rest of the general vector is not completed before the clip ends.

1. New line beginning $\nabla f$ =

2. Top component $\frac{\partial f}{\partial x}$

1. Attention shifts from the specific example to a general formula.

2. Only the first component of the general vector is written before cutoff.

1. The notation still uses $\nabla f$.

2. The structure remains a vector of partial derivatives.

The ending visually transitions from the worked example to the general rule that the gradient is a vector built from partial derivatives.

1. Diagram
   Observation

   Static layout showing function definition, partial derivatives, gradient formula, and nabla operator explanation.

1. Function $f(x,y)$

2. Partial derivatives ∂$f/x$ and ∂$f/y$

3. Gradient vector ∇f

4. Nabla operator ∇

1. Initial state shows pre-calculated partials and gradient.

2. Speaker writes out the general definition of ∇f.

3. Speaker introduces the nabla symbol as a vector of operators.

4. Speaker expands the nabla concept to higher dimensions.

1. The specific example $f(x,y) = x^2 \sin (y)$ remains visible throughout.

2. The calculated gradient for the example remains unchanged.

The visual progression moves from a concrete numerical example to the abstract operator definition of the gradient, reinforcing the connection between the two.

1. Audio
   Observation

   The speaker says, "I hate showing the computation before the geometric intuition since usually it should go the other way around, but the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition."

One might expect the formula for the gradient to make its geometric interpretation obvious right away.

The speaker explicitly warns that in this case the computation and the geometric intuition seem unrelated at first, and that the connection will be explained in later videos.

1. Audio
   Observation

   The speaker says, "maybe I should emphasize this is actually a vector valued function... This is a function that takes in a point in two dimensional space and outputs a two dimensional vector."

2. Formula
   Observation

   The notation is revised from $\nabla f$ to $\nabla f(x,y)$.

A learner may read $\nabla f$ as just one vector expression rather than as a function of the input point.

The clip emphasizes that $\nabla f(x,y)$ is vector-valued: each input point (x,y) produces a corresponding output vector.

1. Audio
   Observation

   Speaker addresses the question 'what's its dimension?' for the nabla symbol.

Students might assume the nabla symbol has a fixed dimension regardless of the context.

The dimension of the nabla vector corresponds to the number of independent variables in the function it acts upon. For a 2D function, it has 2 components; for a 3D function, it has 3 components.

1. Audio
   Observation

   The speaker defines the gradient as packing together partial derivative information and then computes the partials before forming $\nabla f$.

2. Formula
   Observation

   The board moves from $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ to $\nabla f = \begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$.

The example gradient is built directly from the previously computed partial derivatives.

1. Formula
   Observation

   The bottom component of the gradient vector is exactly the previously written $\frac{\partial f}{\partial y}$.

The y-partial supplies the second component of the gradient vector in the worked example.

1. Audio
   Observation

   After describing the two-variable gradient as a map to a two-dimensional vector, the speaker says that with three variables there would be three partial derivatives and a three-dimensional output.

The three-variable statement extends the same gradient construction from two inputs and two output components to three inputs and three output components.

1. Formula
   Observation

   The board first shows the specific gradient $\nabla f(x,y)=\begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$, then begins a lower general line $\nabla f$ = [$\frac{\partial f}{\partial x}$; ...].

2. Audio
   Observation

   The speaker transitions from the worked example to "the gradient of any function."

1. The general formula is incomplete by the end of the clip.

The worked two-variable example motivates the more general rule that the gradient is a vector of partial derivatives.

1. Formula
   Observation

   The notation changes from $\nabla f$ to $\nabla f(x,y)$.

2. Audio
   Observation

   The speaker explains that adding (x,y) emphasizes that the gradient is a vector-valued function.

The notation $\nabla f$ is refined into $\nabla f(x,y)$ to express that the gradient depends on the input point.

1. Formula
   Observation

   Gradient is explicitly built from partial derivatives.

Understanding how to compute partial derivatives is necessary before constructing the gradient vector.

1. Audio
   Observation

   Speaker links the nabla symbol directly to the operation of finding the gradient.

The nabla operator, when applied to a scalar function, produces its gradient.

1. Audio
   Observation

   The speaker defines the gradient as packing together partial derivative information and then constructs it from the two partials.

1. Gradient as a collection of partial derivative information
2. Gradient of the example function as a column vector

1. Formula
   Observation

   The board shows $\frac{\partial f}{\partial x}=2x\sin(y)$.

2. Audio
   Observation

   The speaker explains treating y as constant.

1. Partial derivative of f with respect to x
2. Derivation of $\frac{\partial f}{\partial x}$ for $f(x,y)=x^2\sin(y)$

1. Formula
   Observation

   The board shows $\frac{\partial f}{\partial y}=x^2\cos(y)$.

2. Audio
   Observation

   The speaker explains treating x as constant and using the derivative of sine.

1. Partial derivative of f with respect to y
2. Derivation of $\frac{\partial f}{\partial y}$ for $f(x,y)=x^2\sin(y)$

1. Formula
   Observation

   The gradient is written as $\begin{bmatrix}2x\sin(y)\\ x^2\cos(y)\end{bmatrix}$.

2. Audio
   Observation

   The speaker says the gradient puts the partial derivatives together in a vector.

1. Gradient of the example function as a column vector
2. Assembling the gradient vector from the two partial derivatives

1. Audio
   Observation

   The speaker names the upside-down triangle symbol as nabla and says it is often pronounced del.

1. Notation $\nabla f$ for the gradient

1. Audio
   Observation

   The speaker says the gradient is a vector-valued function taking in a point and outputting a vector.

1. Gradient as a vector-valued function of the input point
2. Treating $\nabla f$ as only a single fixed vector

1. Audio
   Observation

   The speaker says that with three variables there would be three partial derivatives and a three-dimensional output.

1. Extension of the gradient idea to three variables
2. Three-variable analogue of the gradient

1. Formula
   Observation

   Only the beginning of the general vector is written, with top entry $\frac{\partial f}{\partial x}$.

1. The full general formula is not completed within the clip.

1. Beginning of the general gradient formula

1. Formula
   Observation

   Worked example on screen.

1. Definition of the Gradient Vector
2. Computing Partial Derivatives
3. Example: Gradient of $f(x,y) = x^2 \sin (y)$

1. Audio
   Observation

   Explanation of the triangle symbol.

1. The Nabla Symbol as an Operator Vector

1. Audio
   Observation

   Discussion about 2D vs 3D functions.

1. The Nabla Symbol as an Operator Vector
2. Dimension of the Nabla Operator