Differentiation: Composite, Implicit, and Inverse Functions
Unit 3 of AP Calculus AB, worth 9–13% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Chain rule, implicit differentiation, derivatives of inverse functions and inverse trig functions, higher-order derivatives.
How this unit is tested
Start by identifying the structure of the function you're differentiating: is it a composition (one function stuffed inside another), an equation that mixes x and y so that y cannot be isolated, or a request for a derivative of an inverse function? Each structure signals a different tool. For compositions, work from the outside in — differentiate the outer function first, leaving the inner function alone, then multiply by the derivative of the inner function. If the inner function is itself a composition, repeat the process (chain the chain rule).
For equations that cannot be solved for y, differentiate both sides of the equation term by term with respect to x. Every time you differentiate a term containing y, you must multiply by $\frac{dy}{dx}$ (this is just the chain rule, since y is implicitly a function of x). Terms that mix x and y, like $xy$ or $x^2y^3$, need the product rule before the chain rule is applied to the y-factor. Once every term is differentiated, collect all the $\frac{dy}{dx}$ terms on one side and factor.
For derivatives of inverse functions, remember the relationship is about swapping the roles of input and output: if $g$ is the inverse of $f$, then $g'(a) = \frac{1}{f'(g(a))}$. You almost always need either a small table of values or an explicit formula for $f$. For inverse trig derivatives, memorize the six formulas cold — they show up constantly combined with the chain rule.
For higher-order derivatives, differentiate the first derivative again using whatever rule is appropriate (power, product, quotient, chain). When the first derivative was found implicitly, the second derivative will often still contain $\frac{dy}{dx}$ in it — substitute your earlier expression for $\frac{dy}{dx}$ back in and simplify using the original equation if possible.
What you have to know
Chain Rule
If $y=f(g(x))$, then $\frac{dy}{dx}=f'(g(x))\cdot g'(x)$. Differentiate the outer function and multiply by the derivative of the inner function.
Implicit Differentiation Procedure
Differentiate both sides of an equation with respect to x, applying the chain rule to every term containing y (each contributes a factor of $\frac{dy}{dx}$), then solve algebraically for $\frac{dy}{dx}$.
Derivative of an Inverse Function
If $g=f^{-1}$ and $f$ is differentiable and invertible near the relevant point, then $g'(a)=\dfrac{1}{f'(g(a))}$.
Derivatives of Inverse Trig Functions
$\frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}[\arccos x]=-\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}$, $\frac{d}{dx}[\text{arccot}\,x]=-\frac{1}{1+x^2}$, $\frac{d}{dx}[\text{arcsec}\,x]=\frac{1}{|x|\sqrt{x^2-1}}$, $\frac{d}{dx}[\text{arccsc}\,x]=-\frac{1}{|x|\sqrt{x^2-1}}$.
Higher-Order Derivatives
The second derivative $\frac{d^2y}{dx^2}$ is found by differentiating $\frac{dy}{dx}$ again with respect to x; for implicit relations, any $\frac{dy}{dx}$ appearing in the result should be replaced with its earlier expression in terms of x and y.
14 practice questions
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Find $f'(x)$ if $f(x)=(3x^2-5x)^4$.
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Answer. $f'(x)=4(3x^2-5x)^3(6x-5)$
This is a composition with outer function (something)^4 and inner function $3x^2-5x$. Differentiate the outer power, keeping the inner function unchanged, then multiply by the derivative of the inner function, $6x-5$. -
Find the derivative of $y=\sin(x^3)$.
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Answer. $\frac{dy}{dx}=3x^2\cos(x^3)$
The outer function is sine and the inner function is $x^3$. Differentiate sine to get cosine of the same inner function, then multiply by the derivative of $x^3$, which is $3x^2$. -
For the circle $x^2+y^2=25$, find $\frac{dy}{dx}$ at the point (3,4).
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Answer. $\frac{dy}{dx}=-\frac34$
Differentiating implicitly: $2x+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{x}{y}$. Substituting x=3, y=4 gives $-\frac{3}{4}$. -
If $xy+y^2=6$, find $\frac{dy}{dx}$ in terms of x and y.
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Answer. $\frac{dy}{dx}=\dfrac{-y}{x+2y}$
Differentiate using the product rule on xy: $y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x+2y)=-y$, giving $\frac{dy}{dx}=\frac{-y}{x+2y}$. -
Suppose f(2)=5 and f'(2)=3, and f is one-to-one and differentiable. If g is the inverse of f, find g'(5).
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Answer. $g'(5)=\dfrac13$
Since $f(2)=5$, we have $g(5)=2$. By the inverse function derivative formula, $g'(5)=\frac{1}{f'(g(5))}=\frac{1}{f'(2)}=\frac13$. -
Find the derivative of $y=\arctan(2x)$.
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Answer. $\frac{dy}{dx}=\dfrac{2}{1+4x^2}$
Using $\frac{d}{dx}[\arctan u]=\frac{u'}{1+u^2}$ with $u=2x$, the derivative is $\frac{2}{1+(2x)^2}=\frac{2}{1+4x^2}$. -
Find the derivative of $y=\arcsin(x^2)$.
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Answer. $\frac{dy}{dx}=\dfrac{2x}{\sqrt{1-x^4}}$
Using $\frac{d}{dx}[\arcsin u]=\frac{u'}{\sqrt{1-u^2}}$ with $u=x^2$, the derivative is $\frac{2x}{\sqrt{1-x^4}}$, since $u^2=x^4$. -
Find $f''(x)$ for $f(x)=\sin(x^2)$.
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Answer. $f''(x)=2\cos(x^2)-4x^2\sin(x^2)$
First, $f'(x)=2x\cos(x^2)$ by the chain rule. Differentiate again using the product rule on $2x$ and $\cos(x^2)$: $f''(x)=2\cos(x^2)+2x\cdot(-\sin(x^2))\cdot 2x=2\cos(x^2)-4x^2\sin(x^2)$. -
For the circle $x^2+y^2=1$, show that $\frac{d^2y}{dx^2}$ simplifies to a simple expression in y alone, and give it.
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Answer. $\frac{d^2y}{dx^2}=-\dfrac{1}{y^3}$
Differentiating gives $\frac{dy}{dx}=-\frac{x}{y}$. Differentiating again with the quotient rule and substituting $\frac{dy}{dx}=-\frac{x}{y}$ gives $\frac{d^2y}{dx^2}=-\frac{y^2+x^2}{y^3}$, which equals $-\frac{1}{y^3}$ since $x^2+y^2=1$. -
Find the derivative of $y=\arccos(3x)$.
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Answer. $\frac{dy}{dx}=\dfrac{-3}{\sqrt{1-9x^2}}$
Using $\frac{d}{dx}[\arccos u]=\frac{-u'}{\sqrt{1-u^2}}$ with $u=3x$, the derivative is $\frac{-3}{\sqrt{1-9x^2}}$. -
Find the derivative of $y=e^{\sin(x^2)}$.
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Answer. $\frac{dy}{dx}=2x\cos(x^2)\,e^{\sin(x^2)}$
This is a chain of three functions: exponential, sine, and $x^2$. Differentiate the exponential (keeping the exponent), then multiply by the derivative of $\sin(x^2)$, which is $\cos(x^2)\cdot 2x$, giving the full result. -
Which of the following gives $\frac{dy}{dx}$ for the curve $x^3+y^3=6xy$?
- $\dfrac{2y-x^2}{y^2-2x}$
- $\dfrac{x^2-2y}{y^2-2x}$
- $\dfrac{2y-x^2}{2x-y^2}$
- $\dfrac{x^2+2y}{y^2+2x}$
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Answer. $\dfrac{2y-x^2}{y^2-2x}$
Differentiate both sides: $3x^2+3y^2y'=6y+6xy'$. Collect terms: $y'(3y^2-6x)=6y-3x^2$, so $y'=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}$ after dividing by 3. -
Let $f(x)=x^3+2x+1$ and let g be the inverse of f. Find g'(1).
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Answer. $g'(1)=\dfrac12$
Since $f(0)=0+0+1=1$, we know $g(1)=0$. Also $f'(x)=3x^2+2$, so $f'(0)=2$. By the inverse derivative formula, $g'(1)=\frac{1}{f'(0)}=\frac12$. -
A particle moves along a line with position $s(t)=t^3-6t^2+9t$. Find its acceleration at $t=2$.
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Answer. $0$
Velocity is $s'(t)=3t^2-12t+9$, and acceleration is the second derivative $s''(t)=6t-12$. At $t=2$, $s''(2)=6(2)-12=0$.
What people get wrong
- Forgetting to multiply by the derivative of the inner function in a composition (e.g., writing $\frac{d}{dx}[\sin(x^3)]=\cos(x^3)$ instead of $3x^2\cos(x^3)$). Always ask 'what is the inner function, and what is its derivative?' before finishing.
- Differentiating a mixed term like $xy^2$ implicitly by treating it as a single variable and forgetting the product rule. Write it out as $\frac{d}{dx}[x]\cdot y^2+x\cdot\frac{d}{dx}[y^2]$ first, then apply the chain rule to the $y^2$ piece.
- Mixing up the derivatives of arcsin and arccos, or arctan and arccot — they differ only by a sign. Memorize arcsin/arctan/arcsec as positive and their co-function partners as the negatives of those same expressions.
- Leaving $\frac{dy}{dx}$ unsimplified inside a second-derivative expression instead of substituting the first-derivative formula back in. The AP rubric expects the final answer in terms of x and y (or a number), not in terms of $y'$.
- Using $g'(a)=\frac{1}{f'(a)}$ instead of $g'(a)=\frac{1}{f'(g(a))}$ for inverse function derivatives — the derivative of f must be evaluated at $g(a)$, not at a itself.
- Applying the power rule instead of the chain rule to something like $\arcsin(x^2)$, forgetting that the inner function $x^2$ still needs its own derivative factored in.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.