MIT OpenCourseWare · YouTube · 11:33
This 180-second excerpt from an MIT multivariable calculus recitation introduces a worked example on total differentials and the chain rule. After a title card, the instructor presents the setup , , , and states two tasks: (a) write in terms of , and (b) compute in two ways. He then fully solves part (a) by using the formula , computing and , and concluding . The clip then shifts to part (b), where he introduces a dependency graph with at the top, and in the middle, and and at the bottom as a way to organize the chain rule. The actual chain-rule expansion and final value of are not reached before the excerpt ends. This clip from MIT 18.02SC Multivariable Calculus demonstrates computing ∂z/∂u using the chain rule and a dependency graph for ² + y², ² − v², y = uv. The instructor traces paths and , obtains ∂z/∂ux + 2vy, and explains that leaving the answer in mixed variables is acceptable. He contrasts the chain rule as a quick prescription with total differentials as a clearer conceptual tool for understanding variable dependence. This 180-second whiteboard segment works a multivariable calculus example connecting total differentials and the chain rule. The left side of the board shows , the chain-rule formula , its simplified value , and a dependency graph with above and below. On the right, the instructor rewrites and , substitutes them into , expands to , and states that the coefficient of is . This clip completes a multivariable chain-rule example by showing that the total differential dz can be written in standard form as ()du + ()dv, so each partial derivative is read from the corresponding coefficient. The instructor extracts xu+2vy from the expanded differential, corrects an algebraic slip in the coefficient, and checks that the same value comes from the dependency-graph chain rule on the middle board. He then compares the two methods: the dependency graph is faster for a single requested derivative, while the total-differential method is more algebraic after the initial calculus and less error-prone, though it also produces automatically.
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
The clip opens on a course title slide for multivariable calculus, naming the topic as total differentials and the chain rule.
The scene cuts to a blackboard with the full example already written: , , and . The instructor explains that is a function of the intermediate variables and , while those in turn depend on auxiliary variables and .
He states the two tasks. In part (a), the goal is only to write the total differential in terms of and , so and are deliberately left out of the final expression at this stage. In part (b), the goal is to compute by two methods: first the chain rule, then total differentials.
Returning after inviting the viewer to pause, he begins part (a) by recalling the defining formula for the total differential of a two-variable function: .
He then applies this to the specific function . Differentiating with respect to gives , and differentiating with respect to gives . Substituting these into the formula yields the completed answer for part (a): .
The instructor turns to part (b) and says he will first use the chain rule. Before writing any derivative formula, he introduces a dependency graph as an organizing device for the variable relationships.
On the board he places at the top, draws arrows down to and , and then draws arrows from both and down to and . As he explains, this picture records that depends on and , while each of and depends jointly on and .
By the end of the clip, the graph makes clear that there are two routes from to , one through and one through , but the actual chain-rule expansion for has not yet been written.
The instructor begins part b) of the problem: given ² + y², ² − v², y = uv, compute ∂z/∂u using the chain rule. A dependency graph on the board shows z at the top with arrows to x and y, and arrows from x and y down to u and v.
He explains the chain rule procedure: identify every path from z to u in the dependency graph. There are two such paths: and . Each path contributes a product of partial derivatives to the sum.
He writes the chain rule formula: ∂ = (∂z/∂x)(∂x/∂u) + (∂z/∂y)(∂y/∂u). This directly mirrors the two paths traced on the graph.
Next, he computes each factor. From ² + y², he gets ∂z/∂ and ∂z/∂. From ² − v², he gets ∂x/∂. From y = uv, treating v as constant, he gets ∂y/∂.
Substituting these into the chain rule formula yields ∂z/∂ux + 2vy. He writes this as the final answer on the board.
He then addresses a common student instinct: since x and y are themselves functions of u and v, one might feel compelled to substitute their expressions back in. He explicitly states this is unnecessary and that an answer with mixed variables (containing both x, y and u, v) is perfectly acceptable.
He argues that keeping mixed variables actually reveals something important: it shows how the differentials depend on one another, which is the conceptual heart of the problem.
Finally, he transitions toward the alternative method of total differentials. He candidly compares the two approaches: the chain rule is the fastest computational prescription, but total differentials provide a clearer explanation of the underlying dependence structure. He expresses his personal preference for total differentials when time allows deeper exploration.
The clip opens on a two-part board setup. On the left, part (a) already states the total differential and specializes it to ; part (b) shows the chain-rule formula for together with a dependency graph. On the right, the instructor begins a fresh derivation by repeating the same starting differential, making clear that the goal is to transform from the variables to the variables.
The next step is to express the intermediate differential in the new independent variables. The instructor says that is itself a function of and , then writes . This supplies the first ingredient needed for substitution into the formula for .
He then handles the second intermediate variable. Recalling that , he differentiates the product and writes . At this point all three required differential identities are visible on the right side: the original , the new , and the new .
With those pieces in place, the instructor emphasizes that the remaining work is substitution rather than a new conceptual step. He replaces and inside , producing . The purpose is to answer how depends on and by rewriting its differential entirely in the basis.
The expression is then expanded and regrouped. Collecting the coefficients of gives , and collecting the coefficients of gives , so the board ends with . This is the algebraic form from which partial derivatives can be read directly.
Finally, the instructor points to the coefficient of and states that one definition of is exactly that coefficient. Thus the right-side total-differential computation reproduces the left-side chain-rule result, where the prewritten formula had already simplified to .
The instructor begins by rewriting the total differential of z in the independent variables u and v. On the right side of the board he writes , establishing the standard form that will be used to read off partial derivatives from coefficients.
He then compares this new line with the earlier expanded expression already on the board, , and states that the two sides represent the same differential. From that equality, the coefficient multiplying must be .
While extracting that coefficient, he notices an algebraic slip in the second term, corrects it verbally and on the board, and settles on . Thus the total-differential method gives .
As a check, he points back to the middle board, where the chain rule has already been applied through the dependency graph: . The two methods agree.
The remainder of the clip is a method comparison. The instructor says that if speed matters, for example on an exam, the dependency-graph method is quickest: identify the variable dependencies, trace all paths from to the chosen independent variable, and multiply the partial derivatives along each edge.
With more time available, he prefers the total-differential method because after the initial differentiation the rest is mostly algebra, which he finds less error-prone. He also notes a tradeoff: this method naturally produces the full differential, so it yields as well even when only was requested.
The example starts with an outer function and then re-expresses the intermediate variables as and . This creates a composition in which depends on and indirectly through and . The instructor uses this setup to motivate both a total-differential calculation and a later chain-rule calculation.
For a differentiable function , the total differential is the sum of the partial derivative with respect to times and the partial derivative with respect to times . In this clip the instructor states this formula before applying it to the specific polynomial example.
Applying the total differential formula to , the instructor computes and , then substitutes them to obtain the answer requested in part (a). The important point emphasized in the audio is that part (a) is expressed only in terms of and , not yet in terms of and .
To begin part (b), the instructor draws a graph with at the top, and in the middle, and and at the bottom. Arrows encode direct dependence. This visual tool is introduced specifically to keep track of which intermediate variables contribute to before writing the chain-rule formula.
The clip fully completes part (a), giving . It only begins part (b): the instructor announces that will be computed first by the chain rule and then by differentials, and he draws the dependency graph, but the actual expansion and final value are not reached within this excerpt.
When z depends on x and y, and x, y depend on u and v, the partial derivative ∂z/∂u is found by summing over all paths from z to u in the dependency graph. Each path contributes the product of partial derivatives along its edges. For the graph and , this gives ∂z/∂u = (∂)(∂) + (∂z/∂y)(∂y/∂u).
Given ² + y², ² − v², y = uv: ∂z/∂, ∂x/∂, ∂z/∂, ∂y/∂. Substituting: ∂z/∂ux + 2vy.
After applying the chain rule, the result may contain both intermediate variables (x, y) and independent variables (u, v). Substituting ² − v² and y = uv back into the answer is optional. The instructor emphasizes that mixed-variable forms are useful because they display how differentials depend on one another.
The instructor characterizes the chain rule as a quick computational prescription but notes it lacks explanatory depth. Total differentials, by contrast, make the dependence relationships between variables more transparent. He prefers total differentials when the goal is understanding rather than speed.
The video starts from the standard total differential formula for a function of two variables and instantiates it with the specific partial derivatives shown on the board. This is the object that will later be rewritten in terms of and .
The left side of the board displays the multivariable chain rule for the case where depends on intermediate variables and , each of which depends on and . The formula sums the contributions from the two paths leading from to .
Because is treated as a function of the new independent variables, its differential is rewritten before substitution into . The board gives the linear combination of and that will replace .
The instructor explicitly recalls that equals the product and differentiates it using the product rule. This provides the second substitution needed to convert into the basis.
Once the intermediate differentials are known, the method becomes algebraic substitution. The formulas for and are inserted directly into the original expression for .
After substitution, the expression is expanded and grouped by the independent differentials. This step isolates the coefficient that corresponds to the partial derivative with respect to and the one corresponding to .
The final conceptual point is that once is written as a linear combination of and , the multiplier of is the partial derivative of with respect to . This links the differential calculation back to the chain rule.
The static graph on the left encodes the functional dependence structure: sits above and , and both and sit above and . It visually explains why the chain-rule formula for has two summed terms.
When z is viewed as depending on the independent variables u and v, its total differential can be written as a linear combination of du and dv. This form is the basis for identifying partial derivatives as coefficients.
If the same differential dz has already been expanded into terms involving du and dv, then the coefficient of du is exactly . In the worked example, the expanded form gives the coefficient 4xu+2vy.
From the board computation, after substituting dx and dy into dz and collecting terms, the derivative with respect to u is found to be 4xu+2vy. The instructor corrects an intermediate algebraic slip before finalizing this expression.
The same value is obtained by applying the multivariable chain rule through the dependency graph: multiply the partial derivatives along each path from z to u and add the path contributions. This reproduces 4xu+2vy.
The clip contrasts two strategies. The dependency-graph method is presented as fastest when only one derivative is needed, especially under time pressure. The total-differential method is presented as more algebraic after the initial calculus and less error-prone, but longer because it computes extra information such as .
Because the total differential is rewritten fully in terms of du and dv, the coefficient of dv is also obtained automatically. In this example that means appears even though the requested quantity was only .
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
The instructor says, "So we have a function z which is x squared plus y squared."
The board shows .
z
Dependent variable and function value in the example.
Function of and ; ultimately depends on and through and .
The instructor says that depends on the two variables and , and that itself depends on and .
The board shows .
x
Intermediate variable in and dependent variable in .
Function of and .
The instructor says that depends on the two variables and , and that also depends on and .
The board shows .
y
Intermediate variable in and dependent variable in .
Function of and .
The instructor says, "Now the variables x and y themselves depend on two auxiliary variables u and v."
The board shows and .
u
Auxiliary independent variable underlying and .
Independent variable in the setup.
The instructor says, "Now the variables x and y themselves depend on two auxiliary variables u and v."
The board shows and .
v
Auxiliary independent variable underlying and .
Independent variable in the setup.
The instructor says, "we just want to compute the total differential dz in terms of dx and dy."
The board shows and then .
dz
Total differential of .
Linear combination of and in part (a).
The instructor says the total differential should be written in terms of and .
The board shows and then .
dx
Differential of the intermediate variable .
Appears as a basis differential in the expression for .
The instructor says the total differential should be written in terms of and .
The board shows and then .
dy
Differential of the intermediate variable .
Appears as a basis differential in the expression for .
The instructor says, "the partial derivative of z in the x direction is 2x."
The board shows in and substitutes it as . The subscript is clear.
Partial derivative of with respect to .
Coefficient of in the total differential formula.
The instructor says, "the partial derivative of z in the y is 2y."
The board shows in and substitutes it as . The subscript is clear.
Partial derivative of with respect to .
Coefficient of in the total differential formula.
The instructor says, "we're going to compute the partial derivative partial z partial u in two different ways."
The board shows in part (b).
Partial derivative of with respect to after composing .
Target quantity in part (b); computation is not completed within this clip.
Board shows .
z
Dependent variable defined as a function of x and y.
Function of x and y
The instructor says, "the total differential dz is just the partial derivative of z in the x direction dx plus z in the y direction dy."
The board shows .
For a function , the total differential is the linear combination of the partial derivatives with respect to each independent variable multiplied by the corresponding differentials. In this clip the instructor applies it directly to .
is treated as a differentiable function of two variables and .
The expression is written in terms of and only.
The instructor says, "So we have a function z which is x squared plus y squared ... Now the variables x and y themselves depend on two auxiliary variables u and v."
The board shows , , and .
The problem defines an outer function and then expresses the intermediate variables and in terms of auxiliary variables and . This creates a composition that motivates both the total differential and the chain rule discussion.
and are intermediate variables.
and are auxiliary independent variables.
The instructor says, "Whenever I think about the chain rule, I like to draw this dependency graph ... this is just a way for me to organize how the different variables depend on one another."
The board shows a three-level graph with at the top, and in the middle, and and at the bottom, with arrows from to and , and from and to and .
The instructor introduces a visual bookkeeping device for the chain rule. Nodes represent variables and directed edges represent dependence, so one can read off which paths connect the final target variable to the chosen independent variable before writing derivative formulas.
Useful when several intermediate variables depend on the same underlying variables.
In this clip the graph is drawn but the full chain-rule formula is not yet written.
Board shows .
The total differential of z is written as , where and are the partial derivatives of z with respect to x and y.
z is a differentiable function of x and y.
Instructor says that if we take a partial derivative ∂z/∂u, we have to go through our dependency graph every way that we can get from z to u, and each path gives a term in the summation.
Instructor writes ∂ = (∂z/∂x)(∂x/∂u) + (∂z/∂y)(∂y/∂u).
To compute ∂ when z depends on x and y, and x and y depend on u and v, sum over all paths from z to u in the dependency graph. Each path contributes the product of partial derivatives along that path.
, , .
Instructor says he is perfectly happy with an answer that has mixed variables like this, and that substituting x's formula for u and v is not really necessary.
When using the chain rule or total differentials, the resulting expression may contain both intermediate variables (like x, y) and independent variables (like u, v). Substituting the intermediate variables back into the final answer is optional unless specifically required.
Left side part (a) displays and then .
At the start the instructor repeats, “We already saw that is .”
The clip uses the standard first-order total differential formula for a function of two variables. Here is treated as depending on and , and its differential is written as the sum of the partial derivative with respect to times plus the partial derivative with respect to times . The board then substitutes the specific values and .
is viewed as a differentiable function of and .
The displayed example uses the specific partial derivatives and .
Left side part (b) displays .
A dependency graph with at the top, and in the middle, and and at the bottom visually encodes the same composition.
The board presents the multivariable chain rule for the case where depends on and , and each of those depends on and . For the partial with respect to , the contribution comes from both paths and , so the formula is a sum of two products.
and , are differentiable enough for the displayed chain-rule computation.
The formula shown is specifically for the -partial; the analogous -partial is not separately written on the left side in this clip.
The instructor says, “Now we want to use the fact that is itself a function of and .”
He writes on the right side of the board.
The explicit formula for is not shown in the clip; only its differential is written.
To change variables in the total differential, the instructor first rewrites the differential of the intermediate variable using the independent variables and . The displayed result is a linear combination of and with coefficients and .
is a differentiable function of and .
This step is used before substituting into .
The instructor says, “So remember was . So taking of , we get .”
He writes on the right side of the board.
The second intermediate differential is obtained by differentiating the product . The product rule gives one term from differentiating while holding fixed and one term from differentiating while holding fixed.
.
Standard product differentiation is applied to obtain the total differential.
The instructor says, “once you've done these computations, now it's just substitution,” and then, “all we need to do is substitute in our formulas for here.”
He writes .
After listing the needed total differentials, the method becomes purely algebraic: replace and in by their expressions in and . This converts the differential from the -basis to the -basis.
The earlier formulas , , and have already been computed.
Substitution is valid because all three are identities among differentials.
The instructor says, “now we just expand everything out,” then “let's collect all the things involving ,” and finally “if we collect the terms in .”
The board ends with .
The substituted expression is expanded and regrouped according to the independent differentials and . The coefficient multiplying becomes , and the coefficient multiplying becomes .
Algebraic distributivity and commutativity are used to expand products and group like differential terms.
The result is still written with and appearing symbolically rather than being replaced by their formulas.
The instructor says, "the partial derivative of z in the x direction is 2x" and "the partial derivative of z in the y is 2y."
The board shows after substituting into .
If , then and .
is given explicitly by .
Differentiation is with respect to and while treating the other variable as constant.
For the function shown in this example.
The instructor concludes, "Okay, and that's all we have to do for a."
The board shows .
For , the total differential expressed in terms of and is .
Use the total differential formula .
Use and .
For the specific function in part (a) of the worked example.
Left side part (b) explicitly displays the equality .
The dependency graph shows depending on and , and both and depending on and .
If depends on and , and and depend on and , then .
is differentiable in the variables used.
and are differentiable in the variables used.
The displayed formula is for the partial with respect to .
For the composed function shown in the dependency graph, the partial derivative with respect to is the sum over the two intermediate paths through and .
The instructor says, “one definition of the partial derivative is this coefficient.”
The coefficient being indicated is the multiplier of in .
The clip ends before a separate written equation identifying the coefficient is added.
When is written as , the coefficient is .
has already been rewritten in the independent differentials and .
The expression is organized as a linear combination of and .
For any such rewritten total differential, the coefficient attached to gives the -partial of .
"what we have here on these two sides is essentially the same expression."
Instructor points back and forth between the expanded dz expression and the newly written standard-form dz expression.
The expanded differential expression obtained by substitution and the standard-form expression du + ()dv represent the same differential.
Both expressions are algebraic rewritings of dz for the same dependent variable z in terms of u and v.
For the worked example on the board.
"if we want to compute partial z partial u, then that's just equal to this coefficient here."
Instructor circles the coefficient (4xu+2vy) multiplying du.
In the expression dz = (4xu+2vy)du + (2yu-4xv)dv, the coefficient of du equals .
The displayed expression is the total differential of z written in the independent differentials du and dv.
For the specific worked expression on the board.
"when we computed total differentials, we got an expression for partial z partial v at the end of the day even though we weren't asked to do that."
Right board contains the dv coefficient (2yu-4xv), corresponding to .
Using the total-differential method in this example produces the derivative with respect to v as well, even when only the derivative with respect to u was requested.
The computation rewrites dz fully in terms of du and dv.
For the example discussed in the clip.
The instructor states the definition of , computes the two partial derivatives, and says that completes part (a).
The board shows and then .
Start from the general formula for the total differential of a function of two variables.
Definition/formula stated by the instructor.
Differentiate with respect to , treating as constant.
Direct partial differentiation of the displayed function.
Differentiate with respect to , treating as constant.
Direct partial differentiation of the displayed function.
Substitute the computed partial derivatives into the total differential formula.
Algebraic substitution into .
Part (a) is completed with .
The instructor says they will compute first using the chain rule and begins by drawing a dependency graph.
The board shows above and , with arrows down to and ; no final chain-rule formula is written before the clip ends.
The actual chain-rule expression for is not reached within this clip.
The second method using total differentials is announced but not executed here.
Identify the target derivative and note that depends on indirectly through and .
Stated problem setup in part (b).
Draw a dependency graph with at the top, and in the middle, and and at the bottom.
The instructor explicitly uses the graph to organize variable dependence before applying the chain rule.
Read off that there are two routes from to : and .
Visible arrow structure in the diagram.
The clip establishes the structural setup for applying the chain rule to , but stops before writing the derivative formula or computing the result.
Instructor explains the chain rule using the dependency graph and computes each partial derivative step by step.
Instructor writes the full derivation on the board.
Apply the chain rule by summing over the two paths from z to u in the dependency graph: and .
Chain rule for multivariable functions.
Compute the partial derivative of with respect to x.
Power rule for partial differentiation.
Compute the partial derivative of with respect to u.
Power rule for partial differentiation.
Compute the partial derivative of with respect to y.
Power rule for partial differentiation.
Compute the partial derivative of y = uv with respect to u.
Product rule / treating v as constant.
Substitute the computed partial derivatives into the chain rule formula.
Algebraic substitution.
Simplify the expression.
Algebraic simplification.
The partial derivative ∂ is 4ux + 2vy.
The right column successively shows , , , then the substituted and collected expression.
The narration explicitly frames the process as substitution followed by expansion and collection.
Start from the total differential of in the original variables and .
This is the displayed formula from part (a) on the left side and repeated on the right side.
Rewrite the differential of the intermediate variable in terms of and .
The instructor states that is a function of and and writes this differential on the board.
Rewrite the differential of the intermediate variable in terms of and .
The instructor reminds the viewer that and applies the product differential.
Substitute the expressions for and into the formula for .
This is direct substitution of previously established differential identities.
Expand the products and collect the coefficients of and .
Algebraic distribution and grouping of like differential terms produce the displayed final line.
Identify the coefficient of as the partial derivative of with respect to .
The instructor explicitly says that one definition of is this coefficient.
By rewriting in the basis, the coefficient of is obtained as , which the video identifies with .
Left side part (b) shows .
The clip does not separately narrate every symbol in this prewritten left-side computation; it is visible on the board throughout.
Write the chain rule for the composed dependence of on through and .
This is the formula displayed under part (b) beside the dependency graph.
Substitute the specific partial derivatives visible from the example.
The board uses , , , and .
Multiply and simplify each term.
Elementary algebra gives the displayed simplified result.
The left-side worked example computes directly from the chain rule, matching the coefficient found from the total-differential substitution on the right.
Instructor explains that the two sides are the same expression and identifies the coefficient of du.
Board shows (2u du - 2v dv) + 2y(v du + u dv) = (4xu+2vy)du + (2yu-4xv)dv and du + dv.
Instructor circles (4xu+2vy) and later corrects the second term to 2vy.
The exact original function is not stated in this clip, although the displayed partials imply and .
Start from the total differential of z in the intermediate variables x and y, as already written on the board.
Given board work from the earlier part of the lesson.
Substitute the differentials of x and y in terms of the independent variables u and v.
Given board work from the earlier part of the lesson.
Insert the expressions for dx and dy into the formula for dz.
Algebraic substitution.
Collect the coefficients of du and dv.
Expansion and collection of like differential terms.
Write the same differential in standard total-differential form.
Definition of the total differential with respect to u and v.
Compare the coefficient of du in the two expressions for dz.
Equality of the two representations of the same differential implies equality of corresponding coefficients.
The partial derivative is read off as the coefficient of du in the expanded total differential, namely 4xu+2vy.
"why don't we go back to the middle of the board and we'll see that we got the same thing. So 4xu plus 2vy ... and then going back to the middle of the board, that's what we found again."
Middle board shows xu + 2vy.
Use the chain rule for z as a function of x and y, each depending on u.
Multivariable chain rule shown on the middle board.
Substitute the displayed values of the partial derivatives from the board.
Values already written on the board.
Simplify the products.
Basic algebra.
Compare with the coefficient obtained from the total-differential method.
Direct comparison of the two results.
The dependency-graph/chain-rule computation gives the same value for as the coefficient extracted from the total differential.
The instructor introduces the problem, solves part (a), and begins part (b) with a dependency graph.
The board shows , , , part (a) , and part (b) target .
Part (b) is not completed in this clip.
The requested second method using total differentials is not shown here.
Given , , and , (a) write the total differential in terms of ; (b) compute in two ways, using the chain rule and differentials.
Part (a) asks for only in terms of and .
Part (b) asks for by two methods.
Complete part (a) and begin the setup for part (b).
Write the general total differential formula for a function of two variables.
Formula stated by the instructor at the start of part (a).
Compute the partial derivatives of .
Direct differentiation with respect to and .
Substitute the partial derivatives into the total differential formula to finish part (a).
Algebraic substitution.
For part (b), introduce a dependency graph showing depending on and , and both and depending on and .
The instructor says this graph organizes how the variables depend on one another before applying the chain rule.
Part (a): . Part (b): only the dependency-graph setup is shown in this clip; the final value of is not reached.
Part (a) can be checked by comparing the substituted expression with the general formula . For part (b), the visible graph confirms two indirect paths from to , but no numerical or symbolic verification of the final derivative is present in this clip.
Board shows the problem setup: Suppose , , y = uv.
Instructor works through part b) computing ∂z/∂u using the chain rule.
Given , , y = uv, compute ∂z/∂u using the chain rule and a dependency graph.
y = uv
Compute ∂z/∂u.
Identify all dependencies among the variables.
Setup for chain rule.
Write the chain rule formula based on paths from z to u.
Chain rule.
Compute each individual partial derivative from the given formulas.
Differentiation rules.
Substitute and simplify.
Algebra.
ux + 2vy
Instructor states this is the partial derivative and notes that mixed variables are acceptable.
The board contains a complete worked example with parts (a) and (b) on the left and a parallel derivation on the right.
The instructor narrates the right-side derivation step by step from repeating to identifying the coefficient of .
The original explicit formula for is not written in the clip, though its partials and are used.
Given , with depending on and , rewrite in terms of and read off .
is displayed.
is derived on the board.
is derived from .
A dependency graph shows above , which are above .
Express in the variables and , then identify the coefficient of as .
Begin with the known total differential of in the original variables.
This is the starting formula repeated by the instructor and written on the right side.
Compute the differential of the intermediate variable .
The instructor says is a function of and and writes this result.
Compute the differential of .
The instructor explicitly recalls and differentiates the product.
Substitute the formulas for and into .
The instructor describes this stage as “just substitution.”
Expand and collect terms by and .
The instructor says to expand everything out and collect the terms involving and then .
Read off the coefficient of .
The instructor states that one definition of is this coefficient.
, so .
The result agrees with the prewritten left-side chain-rule computation .
Board displays , du - 2v dv, du + u dv, and the chain-rule line xu + 2vy.
Instructor compares the coefficient method with the middle-board chain-rule result.
The original functions , , are not explicitly stated in this clip; only their differential data are visible.
For the displayed composite setup with z depending on x and y, and x,y depending on u and v, compute .
\,\,dy
\,du - 2v\,dv
\,du + u\,dv
,\ ,\ ,\
Find .
Substitute dx and dy into the total differential for dz.
Direct substitution from the given differential relations.
Expand and collect terms in du and dv.
Algebraic simplification.
Read off the coefficient of du from the total differential.
Standard form du + ()dv.
Verify the same result using the chain rule via the dependency graph.
Multivariable chain rule shown on the board.
xu + 2vy
The instructor checks the coefficient-of-du result against the middle-board chain-rule computation and states that they agree.
A title slide reads "Calculus", "Multivariable Calculus", "David Jordan", and "Total differentials and the chain rule" with MIT branding.
Title text
MIT logo
Course branding
Static slide remains on screen until the cut to the blackboard.
No mathematical derivation is shown on the slide.
The slide identifies the lecture topic as total differentials and the chain rule in multivariable calculus.
The blackboard shows the full problem statement: , , , part (a) asking for in terms of , and part (b) asking for in two ways.
The instructor verbally explains the same setup and invites the viewer to pause and try the problem.
Given equations
Part (a) prompt
Part (b) prompt
Instructor pointing at the board
The instructor points to different lines while explaining the meaning of the setup.
The written problem statement stays unchanged during this interval.
This visual establishes the composite dependence structure that motivates both the total differential and the chain rule.
The instructor writes and then beneath it .
He narrates the meaning of the total differential and the values of the partial derivatives.
Formula line for
Substituted line
First the general formula is written, then the specific partial derivatives are substituted below it.
Only and appear in the final expression for in part (a).
The visual sequence shows the direct application of the total differential formula to the given function.
The instructor draws a graph with at the top, arrows to and , and arrows from and down to and .
He says this is a dependency graph used to organize how variables depend on one another for the chain rule.
The graph is complete, but the chain-rule formula derived from it is not written before the clip ends.
Node
Nodes and
Nodes and
Directed arrows
The graph is built level by level from top to bottom.
The dependence pattern is fixed: depends on ; depend on .
The picture encodes the paths that will contribute to under the chain rule.
Board shows a dependency graph with z at the top, arrows pointing down to x and y, and arrows from x and y pointing down to u and v.
z
x
y
u
v
arrows
Instructor points to paths and while explaining the chain rule.
Graph structure remains unchanged throughout the clip.
The graph visually encodes the functional dependencies: z depends on x and y; x and y each depend on u and v. Paths from z to u correspond to terms in the chain rule sum.
The blackboard is split into a left worked summary and a right live derivation; the left side contains parts (a), (b), and a dependency graph, while the right side is filled in during the clip.
Left column part (a):
Left column part (b): chain-rule formula and simplified result
Dependency graph with nodes
Right column: sequentially written derivation of in terms of
The right column grows from to , then , then the substituted and collected expression.
The instructor repeatedly points between the right-column differentials and the left-column chain-rule formula.
The left-side formulas and dependency graph remain visible throughout the clip.
The overall goal stays the same: express the dependence of on and .
The layout visually separates the general rule and summary computation on the left from the step-by-step differential substitution on the right, making the equivalence between chain-rule differentiation and total-differential manipulation explicit.
A labeled graph shows at the top, and below it, and and at the bottom, with arrows indicating dependence.
Node
Nodes and
Nodes and
Directed edges from to and from to
None during the clip; the graph is static reference material.
The graph consistently represents as depending on and , and as depending on .
The graph encodes the composition structure behind the chain rule: each path from up to contributes a product term to .
Instructor writes du + dv on the right side of the board while speaking.
right side of chalkboard
chalk
instructor's hand
A new line for dz is written.
The terms du and dv appear sequentially.
The previously written expanded dz expression remains visible on the board.
The visual act introduces the canonical form used to identify partial derivatives as coefficients of du and dv.
Instructor circles the coefficient (4xu+2vy) and points between it and the newly written .
He says the desired partial derivative is equal to this coefficient.
expanded dz expression
coefficient (4xu+2vy)
standard-form dz expression
The coefficient (4xu+2vy) is circled.
Attention is directed from the circled term to .
The rest of the differential expression stays unchanged during the circling.
The gesture visually encodes the rule that the partial derivative with respect to u is the coefficient multiplying du.
"that should be v, one of those is an x ... Is it y? 2vy."
Instructor pauses, looks back at the board, and corrects the second term in the circled coefficient to 2vy.
The exact pre-correction handwritten character is not fully legible from the available view, but the spoken correction establishes the intended final term as 2vy.
circled coefficient (4xu+2vy)
The instructor revises the second term in the coefficient.
The final accepted expression is 4xu+2vy.
The first term 4xu remains unchanged.
The overall method of reading the coefficient of du remains the same.
This event shows error checking within the algebraic manipulation before the final identification of .
Instructor points from the right-side coefficient to the middle-board chain-rule computation.
He says they get the same thing, 4xu+2vy.
right-side coefficient
middle-board chain-rule formula
Gaze and pointing shift from the right board to the middle board.
The same algebraic expression is highlighted in two places.
Both computations concern .
The visual comparison demonstrates that the total-differential coefficient method and the dependency-graph chain rule yield the same derivative.
Left board shows a dependency graph with z at the top, x and y in the middle, and u and v at the bottom, connected by directed edges.
Instructor describes tracing all paths from z to u and multiplying the partial derivatives along each edge.
nodes z, x, y, u, v
directed edges
The instructor references the graph while explaining the faster method.
The graph structure itself does not change during this segment.
The diagram encodes the chain-rule decomposition: each path from z to u corresponds to one additive term in .
The instructor says, "So u and v aren't going to enter into the picture" when describing part (a).
One might think the auxiliary variables and should already appear in the answer to part (a).
Part (a) asks only for in terms of and , so the correct intermediate-stage answer is without substituting and yet.
The instructor says he likes to draw a dependency graph whenever he thinks about the chain rule.
The graph separates the roles of , , , , and before any formula is written.
One might try to differentiate directly without first identifying all intermediate paths from to .
The video models the safer procedure: first organize the dependence structure with a graph, then use it to determine which terms belong in the chain-rule expansion.
Instructor says: 'I could substitute for x its formula for u and v, but that's not really necessary... I'm perfectly happy with an answer that has mixed variables like this.'
Students may think that after applying the chain rule, they must always substitute x and y back in terms of u and v to get a 'final' answer.
The instructor explicitly states that leaving the answer in mixed variables (e.g., 4ux + 2vy) is acceptable and often preferable, as it shows how the differentials depend on one another.
Instructor says: 'somehow this chain rule, it's just to me it's just a prescription. It's not an explanation.'
Students may treat the chain rule formula as a self-explanatory concept rather than a computational recipe.
The instructor distinguishes the chain rule as a quick prescription from total differentials, which he considers clearer as an explanation of how variables depend on one another.
The instructor says, “the nice thing about this is once you've done these computations, now it's just substitution,” and later, “You see, it's just substitution.”
After computing and , one might think another conceptual step is needed to find how depends on and .
The video stresses that once the relevant total differentials are listed, the remaining work is algebraic substitution into .
The instructor says, “one definition of the partial derivative is this coefficient.”
He points to the multiplier of in the final expression for .
One may view the expanded expression for only as an algebraic simplification and miss its meaning.
The clip explicitly identifies the coefficient of in the rewritten total differential as .
"it involves computing a lot more derivatives that we didn't actually use in the final answer. For instance, when we computed total differentials, we got an expression for partial z partial v at the end of the day even though we weren't asked to do that."
One might expect the total-differential approach to produce exactly the single derivative being asked for.
In this example, rewriting dz fully in terms of du and dv automatically produces both coefficients, so appears even when only was requested.
"that should be v, one of those is an x ... Is it y? 2vy."
Instructor stops and corrects the term in the circled coefficient.
During expansion and collection of terms, a variable can be accidentally written as the wrong symbol (for example x instead of y).
The instructor catches the slip and corrects the coefficient to 4xu+2vy before using it as .
After stating , the instructor immediately applies the total differential formula in part (a).
The board moves from the given function to .
The explicit function is the object to which the total differential formula is applied in part (a).
The general formula is specialized to using the computed partials.
The result of part (a) is obtained by instantiating the general total differential formula with the specific partial derivatives of the example.
The instructor introduces the dependency graph specifically to organize the variable dependence already present in the problem setup.
The graph nodes are exactly the variables from the setup: , , , , .
The dependency graph is a visual reorganization of the same composite dependence encoded by , , and .
The instructor says the graph is used when thinking about the chain rule for .
The drawn arrows show two routes from down to through and through .
The actual chain-rule formula is not written in this clip.
The dependency graph is introduced as the organizational tool needed to compute the target derivative by the chain rule.
Instructor says we have to go through our dependency graph every way that we can get from z to u.
The chain rule method is applied by reading paths off the dependency graph.
Instructor says the chain rule is the quickest way, but he prefers total differentials because they are clearer as an explanation.
Total differentials and the chain rule are contrasted: the chain rule is faster computationally, while total differentials offer more conceptual clarity about variable dependence.
The right-side derivation starts from and ends by reading off from the coefficient of .
The left side separately displays the chain-rule formula for .
The total differential formula is the starting object that gets transformed; the chain-rule formula is the derivative identity recovered from the transformed differential.
and are written before the substituted expression for .
The instructor says the next step is to substitute in the formulas for and .
The differential of in terms of must be available before it can be substituted into .
is written immediately before the substituted expression for .
The instructor derives from and then uses it in the next line.
The differential of in terms of is also required before the full substitution into can be carried out.
The board moves from to .
The instructor says to expand everything out and collect the terms involving and .
Substitution produces an unsimplified expression; collecting like differential terms turns it into the coefficient form needed to read off partial derivatives.
The coefficient of in the collected expression is .
The instructor identifies that coefficient as .
The collected expression contains the information from which the partial derivative with respect to is extracted as the coefficient of .
Left side part (b) shows .
Right side derivation yields the coefficient of as .
This equivalence is inferred by comparing the two displayed results; the clip does not contain an explicit spoken sentence saying the two sides match.
The total-differential substitution method and the direct chain-rule computation produce the same expression for .
The instructor defines the total differential and writes the formula on the board.
appears explicitly.
The board shows the completed answer for part (a).
The instructor says and are not going to enter into the picture in part (a).
The instructor calls the drawing a dependency graph and says it organizes how variables depend on one another.
The graph shows over over with directed arrows.
The instructor announces that part (b) will compute using the chain rule.
The graph reveals two paths from to through and .
The final chain-rule expression is not shown in this clip.
Instructor explains how to compute ∂ using the chain rule and dependency graph.
Instructor says mixed-variable answers are fine and substitution is not necessary.
Instructor compares the chain rule (quickest, prescription) with total differentials (clearer explanation).
The board displays .
Part (b) shows .
The instructor says they want to know how depends on and , then substitutes the formulas for and .
“So remember was . So taking of , we get .”
Covered · Title slide identifies the topic as total differentials and the chain rule; no worked mathematics yet.
Covered · The instructor states the full problem setup and the two tasks, clarifying that part (a) uses only and part (b) will use two methods.
Covered · Part (a) is solved explicitly: the total differential formula is written, the partial derivatives are computed, and the result is obtained.
Covered · Part (b) begins by introducing a dependency graph for the chain rule; the clip ends before the derivative formula or final value is written.
Covered · Full 180-second clip covers the chain rule computation of ∂ via dependency graph, the acceptability of mixed-variable answers, and the instructor's comparison of the chain rule with total differentials.
Covered · Continuous whiteboard lecture segment: the instructor repeats , derives and in terms of , substitutes them into , collects coefficients, and identifies the coefficient of as ; the left side simultaneously shows the chain-rule formula, a simplified computation, and a dependency graph.
Covered · Instructor writes the standard total-differential form du + ()dv.
Covered · Instructor identifies as the coefficient of du, corrects the term to 2vy, and obtains 4xu+2vy.
Covered · The result is checked against the middle-board chain-rule computation, which gives the same expression.
Covered · Instructor compares the dependency-graph method and the total-differential method, discussing speed, algebraic simplicity, and extra derivatives produced.
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