AP Calculus AB Unit 3 — Chain Rule and Implicit Differentiation Worksheet

All 7 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1The Chain Rule

   Differentiate each function. Leave no negative exponents in your final answer.

   1. $y=\sqrt{7-4x^3}$
   2. $g(t)={\sin }^3(4t)$, and then evaluate $g^{\prime }\left({\displaystyle \frac{\pi }{24}}\right)$ exactly.
2. Q2Implicit Differentiation

   The curve $x^3+y^3=9xy$ passes through the point $(2,4)$.

   1. Use implicit differentiation to find ${\displaystyle \frac{dy}{dx}}$ in terms of $x$ and $y$.
   2. Write an equation of the line tangent to the curve at $(2,4)$.
3. Q3Differentiating Inverse Functions

   Let $f(x)=2+\sqrt{x-1}$ on the domain $[1,\infty )$, and let $g$ be the inverse of $f$.

   1. Explain briefly why $f$ has an inverse on $[1,\infty )$.
   2. Find $g^{\prime }(4)$.
4. Q4Differentiating Inverse Trigonometric Functions

   Differentiate.

   1. $y=\arctan (3x)$, and evaluate ${{\displaystyle \frac{dy}{dx}}|}_{x=1/3}$.
   2. $p(x)=\arcsin \left(x^2\right)$, and state the values of $x$ for which your formula applies.
5. Q5Selecting Procedures for Calculating Derivatives

   Let $f(x)={\displaystyle \frac{e^{2x}}{x^2+1}}$.

   1. Name the differentiation rules this derivative needs, and the order in which you apply them.
   2. Find $f^{\prime }(x)$ and write it with the numerator fully factored.
6. Q6Calculating Higher-Order Derivatives

   Find the required derivatives.

   1. For $f(x)=x^4-2x^3+7x$, find $f^{″}(x)$ and $f^{‴}(x)$.
   2. A bead on a vertical spring has displacement $s(t)=\cos (4t)$ centimeters above its rest position, where $t$ is in seconds. Find the acceleration ${\displaystyle \frac{d^{2}s}{d{t}^2}}$, and give its units.
7. Q7Synthesis — drawing on several topics in this unit

   The point $(1,4)$ lies on the curve $2x^2+y^2=18$.

   1. Find ${\displaystyle \frac{dy}{dx}}$ in terms of $x$ and $y$, and write an equation of the tangent line to the curve at $(1,4)$.
   2. Show that ${\displaystyle \frac{d^{2}y}{d{x}^2}}={\displaystyle \frac{-36}{y^3}}$ at every point of the curve with $y\ne 0$.
   3. Evaluate ${\displaystyle \frac{d^{2}y}{d{x}^2}}$ at $(1,4)$ and say what its sign tells you about the curve near that point.

The 7 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The chain rule: the outer derivative is evaluated at the inner output

A composite is a function fed to another function. If h(x) = f(g(x)), then g runs first and hands its output to f, and the derivative follows that same order:

Read the right-hand side literally. It is not f'(x) · g'(x). The first factor is the outer derivative with the whole inner function still sitting inside it, and only the second factor is the inner derivative. In Q1(a) that is why the answer keeps 7 − 4x³ under the root instead of replacing it with x; in Q1(b) it is why cos(4t) appears rather than cos(t).

Taking a nested composite apart

Q1(b) has three layers, not two: something is multiplied by 4, its sine is taken, and the result is cubed. Each layer contributes one factor, and the factors come out in the order you peel them.

A radical is a composite too. Rewriting a root as a power before you start, as Q1(a) invites you to, turns "derivative of a square root" into the power rule plus one chain-rule factor, which is a rule you already trust.

Implicit differentiation: why every y carries a dy/dx

An equation like the one in Q2 does not define y as a single function of x — the curve fails the vertical line test — but near most of its points y still varies with x. That is the whole assumption: y is a function of x, you just do not have a formula for it. So differentiating a term containing y is differentiating a composite, and the chain rule attaches the factor dy/dx:

The second one is the product rule with a chain-rule factor inside it, and it is where most of the lost marks in Q2 live. Any term holding both x and y needs the product rule; any factor of y then needs its dy/dx.

Horizontal and vertical tangents, and the case that is neither

Implicit differentiation almost always leaves dy/dx as a quotient. Once it does, the two special tangents are conditions on the two halves of that quotient — and each condition has two parts:

Finding the points is a system: set the relevant half to zero, substitute that relation back into the equation of the curve, solve for the coordinates, then go back and check the other half at each point you found.

The derivative of an inverse function

Q3 asks for the derivative of an inverse without ever writing a formula for that inverse. The identity that makes it possible is:

The graph of f⁻¹ is the graph of f reflected across y = x, which swaps rise and run — hence the reciprocal. But the reflection also swaps the coordinates, so the point where you measure the slope of f is not at x = a. Writing 1/f'(a) is the same error as the chain-rule one above, in a different costume: the input to f' has to be the inverse's own output.

A tangent line to y = f⁻¹(x) at x = a needs the point (a, b), not (a, a) and not (b, a): the input is a and the output is the b you found in step 2.

Inverse trigonometric derivatives, and the domain that comes with them

The inverse trig derivatives are just the identity above, already worked out for you, and each one arrives with a chain-rule factor when the argument is not a bare x:

Q4 asks you to state where the arcsine formula applies, and that is not a decoration. The denominator must be a positive number under a real square root, so the formula lives only where 1 − u² > 0 — and u is the inner function, not x, so the restriction has to be solved for x. Note also that arcsin is defined at the endpoints of its domain while its derivative is not: a function can be perfectly well defined at a point where the tangent line stands vertical.

Choosing the rule, and the order to apply it

By Q5 the rules are no longer given to you one at a time. The habit that saves time is the same question from the peeling steps: what would you compute last? That operation names the outer rule, and every rule you need inside it goes inside it.

A quotient whose numerator is a composite is a quotient-rule problem with a chain rule living in one slot — not a chain-rule problem, and not two separate exercises. Naming the rules and their order before writing anything, as Q5(a) makes you do, is worth the thirty seconds: it is the step that stops a product rule from being applied to a composite.

Sometimes the best first move is not a rule at all. An expression can often be rewritten — a trigonometric identity applied, a logarithm expanded, a radical turned into a power — into something with fewer layers. Rewriting is legitimate and usually faster, with one caution: check whether the rewrite changed the domain. Cancelling a factor or splitting a logarithm can quietly admit inputs the original expression never accepted, and the derivative you produce then belongs to a different function.

Second derivatives, and the pattern in the nth

Higher-order derivatives are not a new technique — you differentiate, then differentiate the result. What changes is the bookkeeping. Two things are worth being deliberate about:

Q7 combines the two halves of the unit: differentiate implicitly, then differentiate the resulting quotient again. Two things make the second pass work. The first is remembering that dy/dx appears inside it and must itself be differentiated — you already have an expression for it, so substitute that expression in rather than leaving the symbol standing. The second is that the equation of the curve is still available: substituting it back into the second derivative is what collapses a two-variable mess into something clean. Then read the sign. A negative second derivative means concave down at that point, with the tangent line lying above the curve — a conclusion you can check against the shape the curve is supposed to have.

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Getting the most out of it

Write the inner value on its own line

Before you evaluate any composite derivative, write down what the inner function equals at the point in question, as a separate labelled line. Nearly every chain-rule and inverse-function error in this unit is a wrong input to a correct formula, and a number sitting on its own line is hard to substitute in the wrong place.

Do the implicit questions twice

Where a curve can also be solved for y explicitly, do it both ways once. Getting the same slope from implicit differentiation and from the explicit formula is the fastest way to convince yourself that the dy/dx factors are not a ritual — and it is a check you can run on an exam when a slope looks wrong.

Say what rule you are about to use, out loud

"Quotient rule on the outside, chain rule in the numerator." One sentence before you write anything catches the misapplied rule while it is still free to fix. The exam asks for this explicitly on some questions, but the habit pays on all of them.

Common questions