This deck covers partial derivatives - hold the other variable constant, and read the result as a slope - then higher-order and mixed partials with Clairaut's theorem, the total differential and linear approximation, the multivariable chain rule with tree diagrams, and implicit differentiation. It targets the classic traps: forgetting to hold a variable constant, confusing the order of a mixed partial, and dropping a term by using the single-variable chain rule where there are several paths.
Subject: Calculus III · 122 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Compute a partial derivative by holding every other variable constant.
2. Find higher-order and mixed partials, and use Clairaut's theorem.
3. Write the total differential and use it for linear approximation and error estimates.
4. Apply the multivariable chain rule with a tree diagram, summing over every path.
5. Differentiate implicitly using the chain rule.
Warm-up
Discussion prompt
Before we open Week 6 - Partial Derivatives & Chain Rules: without looking back, what was the main idea of Week 5 - Functions of Several Variables, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers functions of two and three variables: domain, range, graphs as surfaces, level curves and level surfaces, and the limits story. It targets three misconceptions: that two agreeing paths prove a limit exists, that a boundary point can be substituted freely, and that a 0/0 form automatically means there is no limit.
Concept
In Calculus I a function had one input. Now a function can take several inputs at once.
\[ z = f(x, y) \]
The output depends on two things. So there is no single derivative. There is one rate of change for each input direction.
Counterexample
Discussion prompt
In Calculus I a function had one input. Now a function can take several inputs at once.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The output depends on two things. So there is no single derivative. There is one rate of change for each input direction.
Intuition
Picture the surface as a hilly landscape. Your position is set by two map coordinates, east and north.
\[ \text{height} = f(x, y) \]
How steep the hill is depends on which way you walk. Walking east gives one slope, walking north gives another. Each one is a partial derivative.
Analogy
Discussion prompt
Explain Think of a landscape by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Picture the surface as a hilly landscape. Your position is set by two map coordinates, east and north.
Concept
To measure the slope in the x direction, freeze y at a fixed value and let only x change.
partial derivative — The derivative of a multivariable function with respect to one variable, taken while treating every other variable as a constant.
That is the whole trick. Hold the others still, then differentiate as usual.
Explain it
Discussion prompt
Explain The partial derivative idea to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
To measure the slope in the x direction, freeze y at a fixed value and let only x change.
Intuition
Imagine a machine with two knobs. The partial with respect to x asks: if I turn only the x knob and leave the y knob alone, how fast does the output move?
Because y is not moving, every part of the formula that contains only y behaves like a fixed number.
Concept
The partial of z with respect to x can be written several equivalent ways:
\[ \frac{\partial z}{\partial x} \;=\; \frac{\partial f}{\partial x} \;=\; f_x \]
The rounded symbol signals a partial. The subscript form names the variable you differentiated with respect to.
\[ \frac{\partial z}{\partial y} \;=\; f_y \]
Concept
Formally, the partial is a one-variable limit in disguise. Only x changes; y is pinned.
\[ f_x(x,y) = \lim_{h \to 0} \frac{f(x+h,\, y) - f(x,\, y)}{h} \]
\[ f_y(x,y) = \lim_{h \to 0} \frac{f(x,\, y+h) - f(x,\, y)}{h} \]
Intuition
Once you pin the other variable, all your Calculus I rules come straight back: power rule, product rule, quotient rule, chain rule.
The only new habit is deciding, for the current variable, which letters are frozen numbers.
Ranking
Put in order
Put the moves of Compute both partials from the limit definition into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Add h to the x slot and subtract the original.
Worked example
Find the partial of the function below with respect to x, using the definition.
\[ f(x,y) = x^2 y \]
Form the difference quotient, changing only x
Why: Add h to the x slot and subtract the original. The y stays exactly where it was.
\[ \frac{(x+h)^2 y - x^2 y}{h} \]
Expand the numerator
Why: Multiply out the square, then factor the common y.
\[ \frac{(x^2 + 2xh + h^2)y - x^2 y}{h} = \frac{2xhy + h^2 y}{h} \]
Cancel h and take the limit
Why: Divide each term by h, then let h go to zero. The leftover h term vanishes.
\[ \lim_{h\to 0}\,(2xy + hy) = 2xy \]
Verify against the shortcut rule
Why: Treating y as a constant and using the power rule on x squared gives the same 2xy. The definition and the shortcut agree.
\[ f_x = 2xy \]
Picture it
Animation
Shows: Each line of the worked example "Compute both partials from the limit definition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Treating y as a constant and using the power rule on x squared gives the same 2xy. The definition and the shortcut agree.
Step zero
Discussion prompt
Both partials of a two-term function — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate with respect to x, holding y fixed
Answer:
Worked example
Find both first partials of this function.
\[ f(x,y) = x^2 y + 3y \]
Differentiate with respect to x, holding y fixed
Why: The factor y is a constant multiplier on x squared, so its derivative is 2xy. The term 3y has no x, so it is a constant and dies.
\[ f_x = 2xy + 0 = 2xy \]
Differentiate with respect to y, holding x fixed
Why: Now x squared is a constant multiplier on y, giving x squared. The term 3y differentiates to 3.
\[ f_y = x^2 + 3 \]
Verify numerically near a point
Why: At the point (1,2), the x-partial predicts a slope of 2. Nudging x from 1 to 1.001 raises f by about 0.002, matching a slope near 2.
\[ f_x(1,2) = 2(1)(2) = 4,\qquad f_y(1,2) = 1 + 3 = 4 \]
Picture it
Animation
Shows: Each line of the worked example "Both partials of a two-term function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the point (1,2), the x-partial predicts a slope of 2. Nudging x from 1 to 1.001 raises f by about 0.002, matching a slope near 2.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Finding the x-partial of the function below by mistakenly differentiating the y as well, as if a product rule in x were needed.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This treats y as if it changed with x.
When differentiating with respect to x, the variable y is a frozen constant. A constant factor stays put.
Why: This treats y as if it changed with x. It adds a bogus second term.
Trap
Finding the x-partial of the function below by mistakenly differentiating the y as well, as if a product rule in x were needed.
\[ f(x,y) = x^2 y \]
Applies the product rule to x squared and y both
Why: This treats y as if it changed with x. It adds a bogus second term.
\[ f_x \;\overset{?}{=}\; 2xy + x^2 \quad \text{(wrong)} \]
When differentiating with respect to x, the variable y is a frozen constant. A constant factor stays put.
\[ f(x,y) = x^2 y \]
Treat y as a constant multiplier
Why: Only x squared is differentiated; y rides along untouched. There is no extra term.
\[ f_x = 2xy \quad \text{(right)} \]
Notation
Annotate
From Trap: forgetting to hold the other variable constant — read this one piece at a time. What is each part doing?
On: \( f_x \;\overset{?}{=}\; 2xy + x^2 \quad \text{(wrong)} \)
Concept
Freezing y at a value cuts the surface with a vertical plane. That cut is a curve, called a trace.
The x-partial is the ordinary slope of that trace curve. It is the steepness of the surface as you walk in the x direction.
\[ f_x(a,b) = \text{slope of the trace } y=b \text{ at } x=a \]
Intuition
Slice the hill with a wall running east-west. Where the wall meets the surface you see a single curve, like a ridge line drawn on the wall.
Its slope is the x-partial. Slice with a north-south wall instead and its slope is the y-partial.
Picture it
Figure (svg): A surface patch with two tangent lines at a point, one along the x direction and one along the y direction.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
At one point on the surface there are two tangent directions and two slopes. The two partials record them.
Concept
Figure (svg): A surface patch with two tangent lines at a point, one along the x direction and one along the y direction.
At one point on the surface there are two tangent directions and two slopes. The two partials record them.
Fill the middle
Fill in the blanks
From Partials of an exponential product — finish the line. Write what belongs on the right of the equals sign before you look.
f(x,y) = e^{xy}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The outer function is the exponential; the inner exponent xy differentiates to y when x is the variable.
Worked example
Find both first partials.
\[ f(x,y) = e^{xy} \]
Differentiate with respect to x
Why: The outer function is the exponential; the inner exponent xy differentiates to y when x is the variable. Chain rule in one variable still applies.
\[ f_x = y\,e^{xy} \]
Differentiate with respect to y
Why: Now the exponent xy differentiates to x. The exponential factor is unchanged.
\[ f_y = x\,e^{xy} \]
Verify by symmetry
Why: Swapping the roles of x and y in the formula swaps the two partials, exactly as it should because f is symmetric in x and y.
\[ f_x = y\,e^{xy},\qquad f_y = x\,e^{xy} \]
Picture it
Animation
Shows: Each line of the worked example "Partials of an exponential product", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Swapping the roles of x and y in the formula swaps the two partials, exactly as it should because f is symmetric in x and y.
Fill the middle
Fill in the blanks
From Partials with an inner chain rule — finish the line. Write what belongs on the right of the equals sign before you look.
f(x,y) = \sin(xy)
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Derivative of sine is cosine, then multiply by the derivative of the inside xy with respect to x, which is y.
Worked example
Find both first partials.
\[ f(x,y) = \sin(xy) \]
Differentiate with respect to x
Why: Derivative of sine is cosine, then multiply by the derivative of the inside xy with respect to x, which is y.
\[ f_x = y\cos(xy) \]
Differentiate with respect to y
Why: Same outer cosine, but the inside xy now differentiates to x.
\[ f_y = x\cos(xy) \]
Verify at a convenient point
Why: At the point where xy equals zero, cosine equals 1, so the partials reduce to y and x. That is easy to confirm by hand.
\[ f_x(0,0) = 0,\qquad f_y(0,0) = 0 \]
Picture it
Animation
Shows: Each line of the worked example "Partials with an inner chain rule", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the point where xy equals zero, cosine equals 1, so the partials reduce to y and x. That is easy to confirm by hand.
Concept
The same rule extends to any number of inputs. Differentiate with respect to one variable and freeze all the rest.
\[ w = f(x,y,z) \;\Rightarrow\; f_x,\; f_y,\; f_z \]
There is one partial for each input variable.
Hypothesis
Predict first
Partials of a three-variable function is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Partial with respect to x
Why: Only the first term has an x. Treat y and z as constants; the second term has no x, so it drops.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find all three first partials.
\[ f(x,y,z) = x^2 y + y z^3 \]
Partial with respect to x
Why: Only the first term has an x. Treat y and z as constants; the second term has no x, so it drops.
\[ f_x = 2xy \]
Partial with respect to y
Why: Both terms contain y, and y appears to the first power in each, so each becomes its constant coefficient.
\[ f_y = x^2 + z^3 \]
Partial with respect to z
Why: Only the second term has a z. Its power rule gives three z squared, times the constant y.
\[ f_z = 3yz^2 \]
Verify each term was used exactly once
Why: The x-term fed f_x and f_y; the z-term fed f_y and f_z. Every variable that appears in a term shows up in that term's partials.
\[ f_x = 2xy,\quad f_y = x^2 + z^3,\quad f_z = 3yz^2 \]
Picture it
Animation
Shows: Each line of the worked example "Partials of a three-variable function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The x-term fed f_x and f_y; the z-term fed f_y and f_z. Every variable that appears in a term shows up in that term's partials.
Pattern
1. Pick the variable you are differentiating with respect to
Why: That is the only letter allowed to change.
2. Treat every other variable as a fixed number
Why: Constant factors stay; terms with no target variable become zero.
3. Differentiate using ordinary one-variable rules
Why: Power, product, quotient, and chain rules all still apply to the target variable.
4. Repeat for each variable in turn
Why: One partial per input direction.
Elimination
Eliminate the wrong options
What is the partial derivative of f with respect to x?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Holding y constant, x squared times y differentiates to 2xy, and the term 3y has no x so it is a constant that differentiates to zero. The result is 2xy.
Check
Consider the function below and differentiate with respect to x.
\[ f(x,y) = x^2 y + 3y \]
Check your understanding
What is the partial derivative of f with respect to x?
Answer: A
Why: Holding y constant, x squared times y differentiates to 2xy, and the term 3y has no x so it is a constant that differentiates to zero. The result is 2xy.
Concept
A first partial is itself a function of x and y. So you can differentiate it again.
Differentiating twice gives a second-order partial derivative. There are four of them for a function of two variables.
\[ f_{xx},\quad f_{yy},\quad f_{xy},\quad f_{yx} \]
Concept
The subscripts read left to right in the order you differentiated.
\[ f_{xy} = (f_x)_y = \frac{\partial}{\partial y}\!\left(\frac{\partial f}{\partial x}\right) \]
So the subscript form differentiates with respect to the first listed letter first. Watch out: the rounded notation stacks the other way.
\[ f_{xy} = \frac{\partial^2 f}{\partial y\,\partial x} \]
Intuition
A pure second partial like the x-x one measures how the x-slope changes as you move further in the x direction. That is the bending, or concavity, along that direction.
A mixed partial measures how the slope in one direction changes as you move in the other direction, a kind of twist.
Step zero
Discussion prompt
All four second partials — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: First partials
Answer:
Worked example
Find every second-order partial of this function.
\[ f(x,y) = x^3 y^2 \]
First partials
Why: Differentiate once with respect to each variable, holding the other constant.
\[ f_x = 3x^2 y^2,\qquad f_y = 2x^3 y \]
Differentiate f_x again with respect to x, then y
Why: The x-x one lowers the x power; the x-y one differentiates the y-squared factor.
\[ f_{xx} = 6x y^2,\qquad f_{xy} = 6x^2 y \]
Differentiate f_y again with respect to y, then x
Why: The y-y one lowers the y power; the y-x one differentiates the x-cubed factor.
\[ f_{yy} = 2x^3,\qquad f_{yx} = 6x^2 y \]
Verify the mixed partials match
Why: Both mixed partials came out to the same expression, a preview of Clairaut's theorem in the next slides.
\[ f_{xy} = f_{yx} = 6x^2 y \]
Picture it
Animation
Shows: Each line of the worked example "All four second partials", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both mixed partials came out to the same expression, a preview of Clairaut's theorem in the next slides.
Concept
The two mixed partials differentiate with respect to both variables but in the opposite order.
\[ f_{xy} = (f_x)_y \qquad \text{versus} \qquad f_{yx} = (f_y)_x \]
You might expect the order to matter. Remarkably, for the functions you meet in this course, it does not.
Concept
If the mixed partials are continuous near a point, the order of differentiation does not change the answer.
Clairaut's theorem — If f and its mixed partials are continuous on an open region, then the two mixed second partials are equal on that region.
\[ f_{xy} = f_{yx} \]
Intuition
Both mixed partials measure the same twist of the surface, just approached from two directions.
For a smooth surface those two measurements land on the same value. The theorem lets you differentiate in whichever order is easier.
Estimation
Predict first
Confirm the mixed partials are equal for this function.
Commit before you compute: what does Verify Clairaut on a two-term function come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the two results are identical
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Both mixed partials equal the same expression, so Clairaut's theorem is confirmed for this function.
Worked example
Confirm the mixed partials are equal for this function.
\[ f(x,y) = x^3 y - x y^2 \]
First partials
Why: Differentiate once with respect to each variable.
\[ f_x = 3x^2 y - y^2,\qquad f_y = x^3 - 2xy \]
Differentiate f_x with respect to y
Why: The three-x-squared-y term gives three-x-squared; the minus-y-squared term gives minus two-y.
\[ f_{xy} = 3x^2 - 2y \]
Differentiate f_y with respect to x
Why: The x-cubed term gives three-x-squared; the minus-two-x-y term gives minus two-y.
\[ f_{yx} = 3x^2 - 2y \]
Verify the two results are identical
Why: Both mixed partials equal the same expression, so Clairaut's theorem is confirmed for this function.
\[ f_{xy} = f_{yx} = 3x^2 - 2y \]
Picture it
Animation
Shows: Each line of the worked example "Verify Clairaut on a two-term function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both mixed partials equal the same expression, so Clairaut's theorem is confirmed for this function.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Asked for the mixed partial f-x-y, a student differentiates twice with respect to x instead, confusing the subscripts.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Differentiating f_x again with respect to x gives a different function.
The mixed partial f-x-y means differentiate with respect to x first, then with respect to y.
Why: Differentiating f_x again with respect to x gives a different function. This is the x-x partial, not the mixed one.
Trap
Asked for the mixed partial f-x-y, a student differentiates twice with respect to x instead, confusing the subscripts.
\[ f(x,y) = x^3 y^2 \]
Computes f_xx and calls it f_xy
Why: Differentiating f_x again with respect to x gives a different function. This is the x-x partial, not the mixed one.
\[ f_{xx} = 6x y^2 \quad \text{(this is NOT } f_{xy}\text{)} \]
The mixed partial f-x-y means differentiate with respect to x first, then with respect to y.
\[ f(x,y) = x^3 y^2 \]
Differentiate f_x with respect to y
Why: Starting from three-x-squared-y-squared, differentiating the y-squared factor gives the correct mixed partial.
\[ f_{xy} = 6x^2 y \quad \text{(right)} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Starting from three-x-squared-y-squared, differentiating the y-squared factor gives the correct mixed partial.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Differentiating f_x again with respect to x gives a different function. This is the x-x partial, not the mixed one.
Check
Use the function below and compute the mixed partial that differentiates with respect to x first, then y.
\[ f(x,y) = x^3 y^2 \]
Check your understanding
What is the mixed partial derivative f_xy?
Answer: A
Why: First f_x equals three x squared times y squared. Differentiating that with respect to y lowers the y power and doubles it, giving six x squared times y.
Concept
A function of two variables is differentiable at a point when it is well approximated near that point by a flat plane, the tangent plane.
Having both partials exist is not quite enough on its own, but for the smooth functions in this course, existence and continuity of the partials guarantee differentiability.
Concept
The total differential collects both partials to predict how the output changes when both inputs move a little.
total differential — The quantity dz that estimates the change in z from small changes dx and dy in the inputs, using both partial derivatives.
\[ dz = f_x\,dx + f_y\,dy \]
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of partial derivative, Clairaut's theorem, total differential as Week 6 - Partial Derivatives & Chain Rules uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
The true change in height is the difference in the actual function values. The differential dz is the tangent-plane estimate of that change.
\[ \Delta z = f(x+dx,\, y+dy) - f(x,y) \;\approx\; dz \]
For small steps the two are very close. The differential is the linear part of the true change.
Fill the middle
Fill in the blanks
From Compute a total differential — finish the line. Write what belongs on the right of the equals sign before you look.
dz(2,3) = 12\,dx + 4\,dy
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The x-partial holds y constant; the y-partial holds x constant.
Worked example
Write the total differential for this function.
\[ z = x^2 y \]
Find both partials
Why: The x-partial holds y constant; the y-partial holds x constant.
\[ z_x = 2xy,\qquad z_y = x^2 \]
Assemble the differential
Why: Multiply each partial by the differential of its variable and add.
\[ dz = 2xy\,dx + x^2\,dy \]
Verify with a small step at (2,3)
Why: The differential predicts a change of 12 dx plus 4 dy. For dx equal to dy equal to 0.01 that is 0.16, and the actual change in x squared times y is about 0.1606, a close match.
\[ dz(2,3) = 12\,dx + 4\,dy \]
Picture it
Animation
Shows: Each line of the worked example "Compute a total differential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The differential predicts a change of 12 dx plus 4 dy. For dx equal to dy equal to 0.01 that is 0.16, and the actual change in x squared times y is about 0.1606, a close match.
Concept
Rearranging the differential gives a formula that estimates a nearby function value from a known one.
\[ f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) \]
This is the linearization of f at the base point. It replaces the curved surface with its tangent plane.
Intuition
Near the base point the tangent plane hugs the surface. Reading a height off the plane is much easier than off the true surface.
The farther you move from the base point, the more the plane drifts from the surface, so keep the steps small.
Ranking
Put in order
Put the moves of Approximate a value with the tangent plane into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Three squared plus four squared is 25, whose square root is 5.
Worked example
Estimate the function below near the base point (3,4), where it is easy to evaluate.
\[ f(x,y) = \sqrt{x^2 + y^2} \]
Evaluate at the base point
Why: Three squared plus four squared is 25, whose square root is 5.
\[ f(3,4) = \sqrt{9+16} = 5 \]
Find the partials at the base point
Why: Each partial is a variable over the square root; at (3,4) the denominator is 5.
\[ f_x(3,4) = \tfrac{3}{5} = 0.6,\qquad f_y(3,4) = \tfrac{4}{5} = 0.8 \]
Apply the linearization to estimate f(3.02, 3.97)
Why: Use dx equal to 0.02 and dy equal to minus 0.03, then add the two partial contributions to the base value.
\[ f \approx 5 + 0.6(0.02) + 0.8(-0.03) = 4.988 \]
Verify against the exact value
Why: The exact square root of 3.02 squared plus 3.97 squared is about 4.9881, so the estimate 4.988 is accurate to three decimals.
\[ \sqrt{3.02^2 + 3.97^2} \approx 4.9881 \]
Picture it
Animation
Shows: Each line of the worked example "Approximate a value with the tangent plane", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The exact square root of 3.02 squared plus 3.97 squared is about 4.9881, so the estimate 4.988 is accurate to three decimals.
Concept
If measured inputs carry small errors, the differential estimates the resulting error in the computed output.
\[ dz = f_x\,dx + f_y\,dy \]
Here dx and dy are the measurement errors, and dz is the propagated error in z.
Missing information
Discussion prompt
A rectangle is measured as 30 by 24, with each side possibly off by up to 0.1. Estimate the error in the computed area.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The x-partial of x-times-y is y; the y-partial is x.
Worked example
A rectangle is measured as 30 by 24, with each side possibly off by up to 0.1. Estimate the error in the computed area.
\[ A = x y \]
Write the differential of the area
Why: The x-partial of x-times-y is y; the y-partial is x.
\[ dA = y\,dx + x\,dy \]
Insert the measurements and the errors
Why: Use x equals 30, y equals 24, and the maximum errors dx equal dy equal 0.1.
\[ dA = 24(0.1) + 30(0.1) = 2.4 + 3.0 = 5.4 \]
State the result
Why: The area is 720 with an estimated maximum error of about 5.4 square units.
\[ A = 720 \pm 5.4 \]
Verify via relative error
Why: The relative error 5.4 over 720 equals 0.0075, which matches the sum of the relative errors, 0.1 over 30 plus 0.1 over 24.
\[ \frac{dA}{A} = \frac{0.1}{30} + \frac{0.1}{24} = 0.0075 \]
Picture it
Animation
Shows: Each line of the worked example "Error in a computed area", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The relative error 5.4 over 720 equals 0.0075, which matches the sum of the relative errors, 0.1 over 30 plus 0.1 over 24.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Assuming the differential dz gives the exact change in the function, even for a large step.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The differential is only the linear estimate.
The differential estimates the change; the exact change is the actual difference of function values.
Why: The differential is only the linear estimate. For a step of size 1 it misses the curvature and is off.
Trap
Assuming the differential dz gives the exact change in the function, even for a large step.
\[ z = x^2 y, \quad \text{from } (2,3) \text{ with } dx = dy = 1 \]
Claims the true change equals dz for a big step
Why: The differential is only the linear estimate. For a step of size 1 it misses the curvature and is off.
\[ dz = 12(1) + 4(1) = 16 \quad (\text{not the exact } \Delta z) \]
The differential estimates the change; the exact change is the actual difference of function values.
\[ \Delta z = f(3,4) - f(2,3) \]
Compute the true change directly
Why: Three squared times four is 36; two squared times three is 12; the exact change is 24, well above the estimate 16 because the step is large.
\[ \Delta z = 36 - 12 = 24 \quad (\text{use } dz \text{ only for small steps}) \]
Translation
\( z = x^2 y, \quad \text{from } (2,3) \text{ with } dx = dy = 1 \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Prediction
Predict first
Which expression is the total differential dz?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: dz = 2xy dx + x^2 dy
Why: The x-partial of x squared times y is 2xy, and the y-partial is x squared. The differential pairs each partial with the matching input differential, giving 2xy dx plus x squared dy.
Check
Find the total differential of the function below.
\[ z = x^2 y \]
Check your understanding
Which expression is the total differential dz?
Answer: A
Why: The x-partial of x squared times y is 2xy, and the y-partial is x squared. The differential pairs each partial with the matching input differential, giving 2xy dx plus x squared dy.
Concept
In Calculus I, if y depends on u and u depends on x, you multiply the two rates along the single chain.
\[ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} \]
The multivariable version keeps this multiply-along-a-path idea, but there is now more than one path.
Concept
Suppose z depends on x and y, and both x and y depend on a single variable t. Then z is ultimately a function of t alone.
\[ z = f(x,y),\quad x = x(t),\quad y = y(t) \]
There are two routes from t up to z: through x, and through y. Add the contribution of each route.
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \]
Picture it
Figure (svg): A tree diagram with z at the top branching to x and y, which both branch down to t.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Trace every path from t up to z. For each path, multiply the derivatives along its branches. Then add the paths together.
Intuition
Figure (svg): A tree diagram with z at the top branching to x and y, which both branch down to t.
Trace every path from t up to z. For each path, multiply the derivatives along its branches. Then add the paths together.
Two branches reach z, so the sum has two terms. That is the heart of the rule.
Step zero
Discussion prompt
A chain-rule dz/dt — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the inner partials and the t-derivatives
Answer:
Worked example
Find dz/dt for the setup below.
\[ z = x^2 y,\quad x = t^2,\quad y = t^3 \]
Find the inner partials and the t-derivatives
Why: Differentiate z with respect to x and y, and differentiate x and y with respect to t.
\[ z_x = 2xy,\; z_y = x^2,\quad \tfrac{dx}{dt} = 2t,\; \tfrac{dy}{dt} = 3t^2 \]
Assemble the two-term sum
Why: Add the x-path and the y-path contributions.
\[ \frac{dz}{dt} = 2xy\,(2t) + x^2\,(3t^2) \]
Substitute x and y in terms of t
Why: Replace x with t squared and y with t cubed, then simplify each term.
\[ = 2(t^2)(t^3)(2t) + (t^2)^2(3t^2) = 4t^6 + 3t^6 = 7t^6 \]
Verify by direct substitution
Why: Substituting first gives z equal to t to the seventh, whose ordinary derivative is seven t to the sixth. The chain rule agrees.
\[ z = t^4\cdot t^3 = t^7 \;\Rightarrow\; \frac{dz}{dt} = 7t^6 \]
Picture it
Animation
Shows: Each line of the worked example "A chain-rule dz/dt", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting first gives z equal to t to the seventh, whose ordinary derivative is seven t to the sixth. The chain rule agrees.
Estimation
Predict first
Find dz/dt for a point moving on a circle.
Commit before you compute: what does A chain-rule dz/dt with trig come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the identity
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The point stays on the unit circle, so z equals cosine squared plus sine squared equals 1, a constant, whose derivative is zero.
Worked example
Find dz/dt for a point moving on a circle.
\[ z = x^2 + y^2,\quad x = \cos t,\quad y = \sin t \]
Find the pieces
Why: Both inner partials are simple, and the t-derivatives are the standard trig ones.
\[ z_x = 2x,\; z_y = 2y,\quad \tfrac{dx}{dt} = -\sin t,\; \tfrac{dy}{dt} = \cos t \]
Sum the two paths
Why: Substitute x and y and combine the terms.
\[ \frac{dz}{dt} = 2\cos t(-\sin t) + 2\sin t(\cos t) = 0 \]
Verify with the identity
Why: The point stays on the unit circle, so z equals cosine squared plus sine squared equals 1, a constant, whose derivative is zero. The result matches.
\[ z = \cos^2 t + \sin^2 t = 1 \;\Rightarrow\; \frac{dz}{dt} = 0 \]
Picture it
Animation
Shows: Each line of the worked example "A chain-rule dz/dt with trig", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The point stays on the unit circle, so z equals cosine squared plus sine squared equals 1, a constant, whose derivative is zero. The result matches.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Using the single-variable chain rule and following only the x route, forgetting that t also flows through y.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This ignores the second branch.
Both x and y depend on t, so both paths must be summed.
Why: This ignores the second branch. The y route contributes another term that is missing here.
Trap
Using the single-variable chain rule and following only the x route, forgetting that t also flows through y.
\[ z = x^2 y,\quad x = t^2,\quad y = t^3 \]
Keeps only the x-path term
Why: This ignores the second branch. The y route contributes another term that is missing here.
\[ \frac{dz}{dt} \;\overset{?}{=}\; z_x\frac{dx}{dt} = 4t^6 \quad (\text{wrong}) \]
Both x and y depend on t, so both paths must be summed.
\[ z = x^2 y,\quad x = t^2,\quad y = t^3 \]
Add both path contributions
Why: The x-path gives four t to the sixth and the y-path gives three t to the sixth. The correct total is seven t to the sixth, confirmed by direct substitution z equals t to the seventh.
\[ \frac{dz}{dt} = 4t^6 + 3t^6 = 7t^6 \quad (\text{right}) \]
Notation
Annotate
From Trap: dropping a path in the chain rule — read this one piece at a time. What is each part doing?
On: \( \frac{dz}{dt} = 4t^6 + 3t^6 = 7t^6 \quad (\text{right}) \)
Concept
Now let x and y each depend on two variables, s and t. Then z depends on both s and t, so there are two partials to find.
\[ z = f(x,y),\quad x = x(s,t),\quad y = y(s,t) \]
\[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \]
\[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} \]
Picture it
Figure (svg): A tree with z branching to x and y, and each of x and y branching to both s and t.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
To get the s-partial, follow only the branches that end at s: one through x and one through y. Add those two paths.
Intuition
Figure (svg): A tree with z branching to x and y, and each of x and y branching to both s and t.
To get the s-partial, follow only the branches that end at s: one through x and one through y. Add those two paths.
For the t-partial, follow the branches that end at t instead. Each partial is its own two-term sum.
Pattern
1. Draw the tree from the top function down to the independent variables
Why: Each level shows what depends on what.
2. To differentiate with respect to one independent variable, find every path that reaches it
Why: There is one term per path.
3. Multiply the derivatives along each path
Why: Each branch contributes a partial or ordinary derivative.
4. Add the paths together
Why: Never keep just one path; summing over all of them is what makes it the multivariable chain rule.
Worked example
Find the s-partial and the t-partial of z.
\[ z = x^2 y,\quad x = s + t,\quad y = s - t \]
Gather the pieces
Why: Compute the inner partials of z and the partials of x and y with respect to s and t.
\[ z_x = 2xy,\; z_y = x^2,\quad x_s = 1,\, x_t = 1,\, y_s = 1,\, y_t = -1 \]
Build the s-partial
Why: Add the two paths that reach s.
\[ \frac{\partial z}{\partial s} = 2xy(1) + x^2(1) = 2xy + x^2 \]
Substitute and simplify the s-partial
Why: Replace x with s plus t and y with s minus t, then expand.
\[ = 2(s+t)(s-t) + (s+t)^2 = 3s^2 + 2st - t^2 \]
Verify by substituting first
Why: Writing z as a function of s and t and differentiating with respect to s gives the same three-s-squared plus two-s-t minus t-squared.
\[ z = (s+t)^2(s-t) \;\Rightarrow\; \frac{\partial z}{\partial s} = 3s^2 + 2st - t^2 \]
Picture it
Animation
Shows: Each line of the worked example "Two independent variables", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Writing z as a function of s and t and differentiating with respect to s gives the same three-s-squared plus two-s-t minus t-squared.
Concept
The pattern scales to any number of intermediate and independent variables.
For a function of intermediate variables, each themselves functions of the independent variables, the partial with respect to one independent variable sums one term per intermediate variable.
\[ \frac{\partial w}{\partial t_j} = \sum_{i} \frac{\partial w}{\partial x_i}\,\frac{\partial x_i}{\partial t_j} \]
Estimation
Predict first
Find dw/dt when w depends on three intermediate variables, each a function of t.
Commit before you compute: what does Three intermediate variables come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify directly
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The product x-y-z equals t times t squared times t cubed, which is t to the sixth, whose derivative is six t to the fifth.
Worked example
Find dw/dt when w depends on three intermediate variables, each a function of t.
\[ w = xyz,\quad x = t,\; y = t^2,\; z = t^3 \]
Write the three-term sum
Why: One path per intermediate variable, so three terms this time.
\[ \frac{dw}{dt} = yz\frac{dx}{dt} + xz\frac{dy}{dt} + xy\frac{dz}{dt} \]
Substitute the pieces
Why: Use the t-derivatives 1, two-t, and three-t-squared, and replace x, y, z by their t-expressions.
\[ = (t^2\cdot t^3)(1) + (t\cdot t^3)(2t) + (t\cdot t^2)(3t^2) \]
Simplify
Why: Each term becomes t to the fifth times a coefficient.
\[ = t^5 + 2t^5 + 3t^5 = 6t^5 \]
Verify directly
Why: The product x-y-z equals t times t squared times t cubed, which is t to the sixth, whose derivative is six t to the fifth. The chain rule matches.
\[ w = t^{6} \;\Rightarrow\; \frac{dw}{dt} = 6t^5 \]
Picture it
Animation
Shows: Each line of the worked example "Three intermediate variables", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The product x-y-z equals t times t squared times t cubed, which is t to the sixth, whose derivative is six t to the fifth. The chain rule matches.
Check
Use the chain rule on the setup below.
\[ z = x^2 y,\quad x = t^2,\quad y = t^3 \]
Check your understanding
What is dz/dt?
Answer: A
Why: The x-path gives 2xy times 2t equal to 4t^6 and the y-path gives x^2 times 3t^2 equal to 3t^6. Adding both paths gives 7t^6, which matches differentiating t^7 directly.
Elimination
Eliminate the wrong options
Which expression correctly gives the partial of z with respect to s?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: There are two paths from s up to z, one through x and one through y. Multiply the derivatives along each path and add, giving f_x times x_s plus f_y times y_s.
Check
Let z depend on x and y, and let x and y each depend on s and t. Consider the partial of z with respect to s.
\[ z = f(x,y),\quad x = g(s,t),\quad y = h(s,t) \]
Check your understanding
Which expression correctly gives the partial of z with respect to s?
Answer: A
Why: There are two paths from s up to z, one through x and one through y. Multiply the derivatives along each path and add, giving f_x times x_s plus f_y times y_s.
Concept
Sometimes a curve is given by an equation that mixes x and y together, not solved for y.
\[ F(x,y) = 0 \]
Think of the left side as a function whose value is held at zero. Differentiating with respect to x, treating y as a function of x, uses the chain rule.
Concept
Applying the chain rule to the constant zero and solving for the slope gives a clean formula.
\[ F_x + F_y\,\frac{dy}{dx} = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{F_x}{F_y} \]
The minus sign is essential, and the y-partial goes in the denominator.
Explain it
Discussion prompt
Explain The implicit-differentiation formula to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Applying the chain rule to the constant zero and solving for the slope gives a clean formula.
Step zero
Discussion prompt
Implicit slope of a circle — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Move everything to one side as F
Answer:
Worked example
Find the slope dy/dx on the circle below.
\[ x^2 + y^2 = 25 \]
Move everything to one side as F
Why: Write the equation as F equals zero so the formula applies.
\[ F(x,y) = x^2 + y^2 - 25 \]
Compute the two partials
Why: Each is a simple power-rule derivative.
\[ F_x = 2x,\qquad F_y = 2y \]
Apply the formula
Why: Minus the x-partial over the y-partial, and the twos cancel.
\[ \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \]
Verify by direct implicit differentiation
Why: Differentiating both sides directly gives two-x plus two-y times y-prime equals zero, which solves to the same minus x over y.
\[ 2x + 2y\,y' = 0 \;\Rightarrow\; y' = -\frac{x}{y} \]
Picture it
Animation
Shows: Each line of the worked example "Implicit slope of a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating both sides directly gives two-x plus two-y times y-prime equals zero, which solves to the same minus x over y.
Concept
If z is defined implicitly by an equation in x, y, and z, the same idea gives its partials.
\[ F(x,y,z) = 0 \]
\[ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z},\qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} \]
Analogy
Discussion prompt
Explain Implicit partials for a surface by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
If z is defined implicitly by an equation in x, y, and z, the same idea gives its partials.
Ranking
Put in order
Put the moves of Implicit partials on a sphere into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Move the 1 across and differentiate with respect to each variable.
Worked example
Find the partials of z for the sphere below.
\[ x^2 + y^2 + z^2 = 1 \]
Write F and its partials
Why: Move the 1 across and differentiate with respect to each variable.
\[ F = x^2 + y^2 + z^2 - 1,\quad F_x = 2x,\, F_y = 2y,\, F_z = 2z \]
Apply the formulas
Why: Minus the relevant partial over the z-partial; the twos cancel each time.
\[ \frac{\partial z}{\partial x} = -\frac{x}{z},\qquad \frac{\partial z}{\partial y} = -\frac{y}{z} \]
Verify by differentiating directly
Why: Differentiating the sphere with respect to x, treating z as a function of x and y, gives two-x plus two-z times the z-partial equals zero, matching minus x over z.
\[ 2x + 2z\,\frac{\partial z}{\partial x} = 0 \;\Rightarrow\; \frac{\partial z}{\partial x} = -\frac{x}{z} \]
Picture it
Animation
Shows: Each line of the worked example "Implicit partials on a sphere", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating the sphere with respect to x, treating z as a function of x and y, gives two-x plus two-z times the z-partial equals zero, matching minus x over z.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Writing the implicit partial as the ratio of partials, but dropping the negative sign.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The chain rule leaves the F-x term on the opposite side, so a negative is required when solving.
Solving the chain-rule equation moves the F-x term across, introducing the negative sign.
Why: The chain rule leaves the F-x term on the opposite side, so a negative is required when solving.
Trap
Writing the implicit partial as the ratio of partials, but dropping the negative sign.
\[ F(x,y,z) = 0 \]
Omits the minus sign
Why: The chain rule leaves the F-x term on the opposite side, so a negative is required when solving.
\[ \frac{\partial z}{\partial x} \;\overset{?}{=}\; \frac{F_x}{F_z} \quad (\text{wrong}) \]
Solving the chain-rule equation moves the F-x term across, introducing the negative sign.
\[ F_x + F_z\,\frac{\partial z}{\partial x} = 0 \]
Solve for the partial with the correct sign
Why: Subtract F-x and divide by F-z; the result carries the minus sign.
\[ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \quad (\text{right}) \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Pattern
1. Move everything to one side to form F equal to zero
Why: The formula needs the equation in this form.
2. Compute the partials of F treating all variables as independent
Why: No chain rule here; just ordinary partials of F.
3. Divide and attach the minus sign
Why: The dependent variable's partial goes in the denominator, with a leading negative.
4. Simplify and, if you can, verify by direct differentiation
Why: A second method that agrees is your safety check.
Real world
Discussion prompt
Outside this lesson: where does Week 6 - Partial Derivatives & Chain Rules actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe: implicit differentiation is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers partial derivatives - hold the other variable constant, and read the result as a slope - then higher-order and mixed partials with Clairaut's theorem, the total differential and linear approximation, the multivariable chain rule with tree diagrams, and implicit differentiation. It targets the classic traps: forgetting to hold a variable constant, confusing the order of a mixed partial, and dropping a term by using the single-variable chain rule where there are several paths.
Check
Find the slope dy/dx for the curve defined implicitly below.
\[ x^2 + y^2 = 25 \]
Check your understanding
What is dy/dx?
Answer: A
Why: With F equal to x squared plus y squared minus 25, the partials are 2x and 2y. The formula gives minus 2x over 2y, and the twos cancel to leave minus x over y.
Concept
You now have one derivative for each direction, a way to combine them into a differential, and a rule for chaining through intermediate variables.
Every tool rests on the same move: change one thing at a time, and add up the separate contributions.
Counterexample
Discussion prompt
You now have one derivative for each direction, a way to combine them into a differential, and a rule for chaining through intermediate variables.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Every tool rests on the same move: change one thing at a time, and add up the separate contributions.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: taking a partial derivative · Recipe: chain rule via a tree diagram · Recipe: implicit differentiation · From one variable to many · Think of a landscape. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A partial derivative differentiates with respect to one variable while every other variable is held constant.
Mixed second partials are equal for smooth functions, which is Clairaut's theorem, so you may differentiate in whichever order is easier.
The total differential combines both partials to estimate small changes, powering linear approximation and error propagation.
The multivariable chain rule sums over every path in the tree diagram. Never keep just one path, and never drop a term.
| Tool | Key formula in words |
|---|---|
| Partial derivative | hold the others constant, then differentiate |
| Total differential | x-partial times dx plus y-partial times dy |
| Chain rule | sum the product of derivatives along each path |
| Implicit derivative | minus F-x over F-y, with the dependent variable underneath |
Want this taught 1-on-1? Alexander tutors Calculus III — $55/session, free consultation.