AP Calculus BC Differentiation: Composite, Implicit, and Inverse Functions — Worked Answer Explanations

Unit 3 · 12 questions explained

1. Question 1 · Easy

   Find $\frac{d}{dx}\left[\sin (x^3)\right]$.

   • A
     $\cos (x^3)$

     Why not A: Applies the outer derivative but forgets the chain-rule factor $3x^2$ from the inner function.
   • B
     $3x^2\cos (x^3)$Correct
   • C
     $3x^2\sin (x^3)$

     Why not C: Applies the chain-rule factor correctly but uses $\sin$ instead of $\cos$ for the outer derivative.
   • D
     $-3x^2\cos (x^3)$

     Why not D: Introduces a spurious negative sign, confusing derivative of $\sin$ with derivative of $\cos$.
   Explanation

   Chain rule: outer function is $\sin (u)$ with $u=x^3$. $\frac{d}{dx}[\sin (x^3)]=\cos (x^3)\cdot \frac{d}{dx}[x^3]=\cos (x^3)\cdot 3x^2=3x^2\cos (x^3)$.

   Key takeaway

   Chain rule: differentiate the outer function, keep the inner unchanged, then multiply by the derivative of the inner function.

2. Question 2 · Easy

   If $y=\arctan (2x)$, find $\frac{dy}{dx}$.

   • A
     $\frac{1}{1+(2x{)}^2}$

     Why not A: Applies the inverse tangent formula but forgets the chain-rule factor of $2$ from the inner function $2x$.
   • B
     $\frac{2}{1+4x^2}$Correct
   • C
     $\frac{1}{1+2x^2}$

     Why not C: Writes $2x^2$ in the denominator instead of $(2x{)}^2=4x^2$; incorrectly squares the coefficient.
   • D
     $\frac{2}{\sqrt{1-4x^2}}$

     Why not D: Uses the derivative formula for $\arcsin$ instead of $\arctan$.
   Explanation

   $\frac{d}{dx}[\arctan (u)]=\frac{1}{1+u^2}\cdot u^{\prime }$. With $u=2x$, $u^{\prime }=2$: $\frac{dy}{dx}=\frac{1}{1+(2x{)}^2}\cdot 2=\frac{2}{1+4x^2}$.

   Key takeaway

   Derivative of $\arctan(u)$: $\frac{1}{1+u^2}\cdot u'$; don't forget the chain-rule factor and expand $(ku)^2 = k^2u^2$.

3. Question 3 · Easy

   Find $\frac{dy}{dx}$ by implicit differentiation if $x^2+y^2=25$.

   • A
     $\frac{x}{y}$

     Why not A: Drops the negative sign when solving $2x+2ydy/dx=0$ for $dy/dx$.
   • B
     $-\frac{y}{x}$

     Why not B: Inverts the ratio; solves for $dy/dx$ by dividing $2y$ by $2x$ instead of $2x$ by $2y$.
   • C
     $-\frac{x}{y}$Correct
   • D
     $\frac{y}{x}$

     Why not D: Inverts ratio and drops negative sign simultaneously.
   Explanation

   Differentiate both sides with respect to $x$: $2x+2y\frac{dy}{dx}=0$. Solve: $2y\frac{dy}{dx}=-2x\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$.

   Key takeaway

   Implicit differentiation: differentiate each term with respect to $x$; terms involving $y$ require the chain rule, giving a $dy/dx$ factor.

4. Question 4 · Easy

   Differentiate $f(x)=e^{{\sin }^{2}x}$.

   • A
     $e^{{\sin }^{2}x}\cdot {\cos }^{2}x$

     Why not A: Differentiates ${\sin }^{2}x$ as ${\cos }^{2}x$ instead of applying the chain rule to get $2\sin x\cos x$.
   • B
     $e^{{\sin }^{2}x}\cdot \sin (2x)$Correct
   • C
     $e^{2\sin x}\cos x$

     Why not C: Differentiates the exponent ${\sin }^{2}x$ but writes $2\sin x$ instead of $2\sin x\cos x$.
   • D
     $2\sin x\cos x$

     Why not D: Differentiates only the exponent and forgets to multiply by $e^{{\sin }^{2}x}$ from the outer derivative.
   Explanation

   Chain rule (twice). Outer: $e^u$ with $u={\sin }^{2}x$. $\frac{d}{dx}[e^{{\sin }^{2}x}]=e^{{\sin }^{2}x}\cdot \frac{d}{dx}[{\sin }^{2}x]=e^{{\sin }^{2}x}\cdot 2\sin x\cos x=e^{{\sin }^{2}x}\sin (2x)$, using the double-angle identity $2\sin x\cos x=\sin (2x)$.

   Key takeaway

   Nested chain rule: $e^{\sin^2 x}$ involves $e^u$, $u = v^2$, $v = \sin x$; differentiate outward in, multiplying each layer's derivative.

5. Question 5 · Easy

   Use implicit differentiation to find $\frac{dy}{dx}$ for $x^3+y^3=6xy$.

   • A
     $\frac{6y-3x^2}{3y^2-6x}$Correct
   • B
     $\frac{3x^2-6y}{6x-3y^2}$

     Why not B: Has the correct terms but inverts the fraction — solves for $dx/dy$ instead of $dy/dx$.
   • C
     $\frac{6y-3x^2}{6x}$

     Why not C: Differentiates the right side as $6y$ only, missing the product rule contribution $6xdy/dx$.
   • D
     $\frac{x^2}{y^2}$

     Why not D: Ignores the right-side $6xy$ term entirely, differentiating only $x^3+y^3=0$.
   Explanation

   Differentiate: $3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}$. Collect $dy/dx$ terms: $3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^2$. Factor: $\frac{dy}{dx}(3y^2-6x)=6y-3x^2$, so $\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}$.

   Key takeaway

   For implicit differentiation of products on the right side, apply the product rule; collect all $dy/dx$ terms on one side and factor.

6. Question 6 · Easy

   Find $\frac{d}{dx}[\arcsin (\sqrt{x})]$.

   • A
     $\frac{1}{\sqrt{x(1-x)}}$

     Why not A: Drops the factor of $2$ from differentiating $\sqrt{x}$; should be $1/(2\sqrt{x})$, not $1/\sqrt{x}$.
   • B
     $\frac{1}{2\sqrt{x}\sqrt{1-x}}$Correct
   • C
     $\frac{1}{\sqrt{1-x}}$

     Why not C: Applies arcsin formula $1/\sqrt{1-u^2}$ with $u=\sqrt{x}$ but forgets to simplify $1-x$ and the chain factor.
   • D
     $\frac{-1}{2\sqrt{x(1-x)}}$

     Why not D: Introduces a spurious negative sign, confusing $\arcsin$ with $\arccos$.
   Explanation

   $\frac{d}{dx}[\arcsin (u)]=\frac{1}{\sqrt{1-u^2}}\cdot u^{\prime }$ with $u=\sqrt{x}$, $u^{\prime }=\frac{1}{2\sqrt{x}}$, $u^2=x$. So: $\frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}\sqrt{1-x}}=\frac{1}{2\sqrt{x(1-x)}}$.

   Key takeaway

   Chain rule on $\arcsin(\sqrt{x})$: apply inverse trig formula and multiply by the derivative of the inner $\sqrt{x}$.

7. Question 7 · Medium

   Find $\frac{d}{dx}[\sinh x]$ using the definition $\sinh x=\frac{e^x-e^{-x}}{2}$.

   • A
     $\sinh x$

     Why not A: Confuses the derivative of $\sinh$ with the derivative of $\cosh$; $d/dx[\cosh x]=\sinh x$.
   • B
     $-\cosh x$

     Why not B: Introduces a spurious negative sign; $d/dx[\cosh x]=\sinh x$, not $-\sinh x$, and $d/dx[\sinh x]=\cosh x$, not $-\cosh x$.
   • C
     $\cosh x$Correct
   • D
     $\tanh x$

     Why not D: Confuses $\tanh x=\sinh x/\cosh x$ with the derivative; $d/dx[\sinh x]\ne \tanh x$.
   Explanation

   $\frac{d}{dx}[\sinh x]=\frac{d}{dx}\left[\frac{e^x-e^{-x}}{2}\right]=\frac{e^x+e^{-x}}{2}=\cosh x$. Note: $d/dx[e^{-x}]=-e^{-x}$, so $d/dx[-e^{-x}/2]=e^{-x}/2$, which is positive.

   Key takeaway

   Hyperbolic derivatives mirror trig: $\frac{d}{dx}\sinh x = \cosh x$ and $\frac{d}{dx}\cosh x = \sinh x$ (both positive, unlike circular trig).

8. Question 8 · Medium

   If $g(x)=f^{-1}(x)$ and $f(3)=7$, $f^{\prime }(3)=4$, find $(g{)}^{\prime }(7)$.

   • A
     $4$

     Why not A: Returns $f^{\prime }(3)$ directly without applying the inverse function derivative formula.
   • B
     $\frac{1}{7}$

     Why not B: Divides by the output value $7$ instead of $f^{\prime }(f^{-1}(7))=f^{\prime }(3)=4$.
   • C
     $\frac{1}{4}$Correct
   • D
     $-4$

     Why not D: Negates the derivative, confusing inverse function with reciprocal function.
   Explanation

   Inverse function derivative formula: $(f^{-1}{)}^{\prime }(y)=\frac{1}{f^{\prime }(f^{-1}(y))}$. Since $f(3)=7$, we have $f^{-1}(7)=3$. Therefore $(g{)}^{\prime }(7)=\frac{1}{f^{\prime }(3)}=\frac{1}{4}$.

   Key takeaway

   $(f^{-1})'(b) = 1/f'(a)$ where $f(a) = b$; find the corresponding input $a$ in the original function first.

9. Question 9 · Medium

   Find $\frac{d^{2}y}{d{x}^2}$ given $y=\sin x$ and then evaluate $\frac{d^{2}y}{d{x}^2}$ for $y=x\sin x$.

   • A
     $2\cos x-x\sin x$Correct
   • B
     $-x\sin x$

     Why not B: Differentiates $\cos x+x(-\sin x)$ and forgets the $+\cos x$ term from the product rule on $x(-\sin x)$.
   • C
     $\cos x-x\sin x$

     Why not C: First derivative is $\sin x+x\cos x$; second derivative requires product rule on $x\cos x$ giving $\cos x-x\sin x$, but misses the additional $\cos x$ from differentiating $\sin x$.
   • D
     $-\sin x$

     Why not D: Computes the second derivative of $\sin x$ alone, ignoring the $x$ factor in $x\sin x$.
   Explanation

   For $y=x\sin x$: $y^{\prime }=\sin x+x\cos x$ (product rule). $y^{\prime \prime }=\cos x+(\cos x+x(-\sin x))=\cos x+\cos x-x\sin x=2\cos x-x\sin x$.

   Key takeaway

   Second derivatives of products require two product rule applications; track every term carefully when differentiating $x\cos x$ in the second step.

10. Question 10 · Hard

    Let $F(x)=f(g(x))$ where $f(u)=u^3+u$ and $g(x)=\cos x$. Find $F^{\prime }(\pi /2)$.

    • A
      $0$

      Why not A: Evaluates $g^{\prime }(\pi /2)=-\sin (\pi /2)=-1$ then multiplies by $f^{\prime }(g(\pi /2))=f^{\prime }(0)=1$, giving $-1$. Choosing $0$ confuses $g(\pi /2)=0$ with $F^{\prime }(\pi /2)$.
    • B
      $-1$Correct
    • C
      $1$

      Why not C: Forgets the negative sign from $g^{\prime }(\pi /2)=-\sin (\pi /2)=-1$.
    • D
      $3$

      Why not D: Evaluates $f^{\prime }(g(\pi /2))$ but uses $g(\pi /2)=1$ instead of $\cos (\pi /2)=0$.
    Explanation

    Chain rule: $F^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$. $g(\pi /2)=\cos (\pi /2)=0$. $g^{\prime }(x)=-\sin x$, so $g^{\prime }(\pi /2)=-\sin (\pi /2)=-1$. $f^{\prime }(u)=3u^2+1$, so $f^{\prime }(0)=1$. Therefore $F^{\prime }(\pi /2)=1\cdot (-1)=-1$.

    Key takeaway

    Chain rule $F'(x) = f'(g(x))\cdot g'(x)$: evaluate $g$ at the point, use that to evaluate $f'$, then multiply by $g'$ at the point.

11. Question 11 · Hard

    Use implicit differentiation to find $\frac{d^{2}y}{d{x}^2}$ for $x^2+y^2=1$ at the point $(0,1)$.

    • A
      $-1$Correct
    • B
      $0$

      Why not B: Evaluates $dy/dx$ at $(0,1)$ as $0$ and stops, confusing $dy/dx$ with $d^{2}y/d{x}^2$.
    • C
      $1$

      Why not C: Drops the negative sign when applying the quotient rule to differentiate $-x/y$.
    • D
      $-\frac{1}{2}$

      Why not D: Mistakenly divides the first derivative formula by $2$ when applying the second implicit differentiation.
    Explanation

    From $x^2+y^2=1$: $\frac{dy}{dx}=-\frac{x}{y}$. Differentiate again: $\frac{d^{2}y}{d{x}^2}=-\frac{y-x(dy/dx)}{y^2}=-\frac{y-x(-x/y)}{y^2}=-\frac{y+x^2/y}{y^2}=-\frac{y^2+x^2}{y^3}=-\frac{1}{y^3}$ (using $x^2+y^2=1$). At $(0,1)$: $-\frac{1}{1^3}=-1$.

    Key takeaway

    For second implicit derivatives, differentiate $dy/dx$ again implicitly, substitute the first derivative formula, then simplify using the original equation.

12. Question 12 · Hard

    Find $\frac{d}{dx}\left[\arctan \left(\frac{y}{x}\right)\right]$ treating $y$ as a differentiable function of $x$.

    • A
      $\frac{1}{1+(y/x{)}^2}$

      Why not A: Applies the outer derivative but ignores the chain rule: needs to multiply by the derivative of $y/x$ with respect to $x$.
    • B
      $\frac{y^{\prime }x-y}{x^2+y^2}$Correct
    • C
      $\frac{y^{\prime }}{x^2+y^2}$

      Why not C: Differentiates $y/x$ as $y^{\prime }/x$ (ignoring quotient rule) and then plugs in, missing the $-y/x^2$ term.
    • D
      $\frac{x{y}^{\prime }-y}{x^2(1+y^2/x^2)}$

      Why not D: Correct intermediate form but fails to simplify denominator: $x^2(1+(y/x{)}^2)=x^2+y^2$.
    Explanation

    Chain rule: $\frac{d}{dx}\arctan (u)\cdot u^{\prime }$ where $u=y/x$. $\frac{d}{dx}[y/x]=\frac{y^{\prime }x-y}{x^2}$ (quotient rule). $\frac{1}{1+u^2}=\frac{1}{1+y^2/x^2}=\frac{x^2}{x^2+y^2}$. Multiply: $\frac{x^2}{x^2+y^2}\cdot \frac{y^{\prime }x-y}{x^2}=\frac{y^{\prime }x-y}{x^2+y^2}$.

    Key takeaway

    Differentiate $\arctan(y/x)$ by chain rule: multiply outer $1/(1+u^2)$ by inner quotient-rule result, then simplify the denominator $x^2(1+(y/x)^2) = x^2+y^2$.