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

Unit 3 · 9% of the AP exam · 12 questions explained

1. Question 1 · Easy

   Find $\frac{dy}{dx}$ if $y=(3x^2+1{)}^5$.

   • A
     $5(3x^2+1{)}^4$

     Why not A: Forgot to multiply by the derivative of the inner function.
   • B
     $30x(3x^2+1{)}^4$Correct
   • C
     $5(6x{)}^4$

     Why not C: Differentiated only the inner function.
   • D
     $5x(3x^2+1{)}^4$

     Why not D: Used $x$ instead of $6x$ from the inner derivative.
   Explanation

   Chain rule: $\frac{dy}{dx}=5(3x^2+1{)}^4\cdot \frac{d}{dx}(3x^2+1)=5(3x^2+1{)}^4\cdot 6x=30x(3x^2+1{)}^4$.

   Key takeaway

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

2. Question 2 · Easy

   If $f(x)=e^{\sin x}$, find $f^{\prime }(x)$.

   • A
     $e^{\cos x}$

     Why not A: Differentiated the exponent instead of applying the chain rule to the base.
   • B
     $e^{\sin x}$

     Why not B: Forgot to multiply by the derivative of the exponent.
   • C
     $\cos x\cdot e^{\sin x}$Correct
   • D
     $\sin x\cdot e^{\cos x}$

     Why not D: Mixed up the exponent and its derivative.
   Explanation

   Chain rule with $y=e^u$ where $u=\sin x$: $\frac{dy}{dx}=e^u\cdot \frac{du}{dx}=e^{\sin x}\cdot \cos x$.

   Key takeaway

   $\dfrac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot u'(x)$ — the exponential survives; multiply by the inner derivative.

3. Question 3 · Easy

   Differentiate $y=\ln (x^4+7)$ with respect to $x$.

   • A
     $\frac{1}{x^4+7}$

     Why not A: Forgot to multiply by the derivative of the inner function.
   • B
     $\frac{4}{x^4+7}$

     Why not B: Dropped the $x^3$ factor from the inner derivative.
   • C
     $\frac{4x^3}{x^4+7}$Correct
   • D
     $4\ln (x^3)$

     Why not D: Used a log-power rule incorrectly.
   Explanation

   $\frac{d}{dx}[\ln (u)]=\frac{u^{\prime }}{u}$. Here $u=x^4+7$ and $u^{\prime }=4x^3$, so $\frac{dy}{dx}=\frac{4x^3}{x^4+7}$.

   Key takeaway

   Chain rule for $\ln(u)$: result is $u'/u$. Don't forget to differentiate the argument.

4. Question 4 · Easy

   If $y=\arctan (2x)$, what is $\frac{dy}{dx}$?

   • A
     $\frac{1}{1+4x^2}$

     Why not A: Forgot to multiply by the derivative of $2x$.
   • B
     $\frac{2}{1+4x^2}$Correct
   • C
     $\frac{2}{1+2x^2}$

     Why not C: Did not square the $2x$ in the denominator.
   • D
     $\frac{1}{\sqrt{1-4x^2}}$

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

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

   Key takeaway

   $\dfrac{d}{dx}[\arctan(u)] = \dfrac{u'}{1+u^2}$ — always square the full inner function.

5. Question 5 · Medium

   Use implicit differentiation to find $\frac{dy}{dx}$ for $x^2+y^2=25$.

   • A
     $\frac{x}{y}$

     Why not A: Sign error: should be negative.
   • B
     $-\frac{y}{x}$

     Why not B: Inverted $x$ and $y$.
   • C
     $-\frac{x}{y}$Correct
   • D
     $\frac{2x}{2y}$

     Why not D: Left coefficients unsimplified and omitted the negative sign.
   Explanation

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

   Key takeaway

   Implicit differentiation: apply chain rule to $y$-terms, writing $\frac{dy}{dx}$, then isolate $\frac{dy}{dx}$.

6. Question 6 · Medium

   Find $\frac{dy}{dx}$ if $y=x^2\sin (3x)$.

   • A
     $2x\cos (3x)$

     Why not A: Forgot the product rule; only differentiated $x^2$.
   • B
     $x^2\cos (3x)+2x\sin (3x)$

     Why not B: Applied chain rule to $\sin (3x)$ but forgot the factor of $3$.
   • C
     $3x^2\cos (3x)+2x\sin (3x)$Correct
   • D
     $3x^2\cos (3x)$

     Why not D: Only differentiated the trig factor, ignoring $x^2$.
   Explanation

   Product rule: $\frac{dy}{dx}=(x^2{)}^{\prime }\sin (3x)+x^2(\sin (3x){)}^{\prime }$. The derivative of $\sin (3x)$ by chain rule is $3\cos (3x)$. So $\frac{dy}{dx}=2x\sin (3x)+x^2\cdot 3\cos (3x)=3x^2\cos (3x)+2x\sin (3x)$.

   Key takeaway

   Combine product rule with chain rule: differentiate each factor, then apply chain rule to any composite piece.

7. Question 7 · Medium

   Differentiate $y={\tan }^{-1}\left(\frac{x}{3}\right)$ with respect to $x$.

   • A
     $\frac{3}{9+x^2}$Correct
   • B
     $\frac{1}{9+x^2}$

     Why not B: Forgot to include the $\frac{1}{3}$ from the inner derivative.
   • C
     $\frac{1}{1+x^2/9}$

     Why not C: Correct unsimplified form but the inner derivative $\frac{1}{3}$ was not applied.
   • D
     $\frac{x}{3(1+x^2)}$

     Why not D: Divided by $x$ instead of multiplying by the inner derivative.
   Explanation

   Let $u=x/3$, so $u^{\prime }=1/3$. $\frac{dy}{dx}=\frac{u^{\prime }}{1+u^2}=\frac{1/3}{1+(x/3{)}^2}=\frac{1/3}{1+x^2/9}=\frac{1/3}{(9+x^2)/9}=\frac{9}{3(9+x^2)}=\frac{3}{9+x^2}$.

   Key takeaway

   Apply the $\arctan$ derivative formula then simplify the compound fraction from the inner derivative.

8. Question 8 · Medium

   If $y={\cos }^3(x^2)$, find $\frac{dy}{dx}$.

   • A
     $-3{\cos }^2(x^2)\sin (x^2)$

     Why not A: Forgot to apply the chain rule to the innermost $x^2$.
   • B
     $-6x{\cos }^2(x^2)\sin (x^2)$Correct
   • C
     $-6x{\sin }^2(x^2)\cos (x^2)$

     Why not C: Swapped $\cos$ and $\sin$ in the result.
   • D
     $3{\cos }^2(x^2)\sin (x^2)\cdot 2x$

     Why not D: Sign error: derivative of $\cos$ is $-\sin$, not $+\sin$.
   Explanation

   Apply chain rule twice. Outer: $3{\cos }^2(x^2)$. Middle: $\cdot (-\sin (x^2))$. Inner: $\cdot 2x$. Combining: $\frac{dy}{dx}=3{\cos }^2(x^2)\cdot (-\sin (x^2))\cdot 2x=-6x{\cos }^2(x^2)\sin (x^2)$.

   Key takeaway

   Nested functions need the chain rule applied at each layer outward to inward.

9. Question 9 · Medium

   Find $\frac{d}{dx}[\ln (\sin x)]$.

   • A
     $\frac{1}{\sin x}$

     Why not A: Applied $\ln$ derivative without the inner derivative $\cos x$.
   • B
     $\cot x$Correct
   • C
     $\tan x$

     Why not C: Inverted the fraction: got $\sin x/\cos x$ instead of $\cos x/\sin x$.
   • D
     $\frac{\sin x}{\cos x}$

     Why not D: Same inversion error, left unsimplified.
   Explanation

   $\frac{d}{dx}[\ln (u)]=\frac{u^{\prime }}{u}$ with $u=\sin x$, $u^{\prime }=\cos x$. So $\frac{d}{dx}[\ln (\sin x)]=\frac{\cos x}{\sin x}=\cot x$.

   Key takeaway

   $\dfrac{d}{dx}[\ln(\sin x)] = \cot x$ — a standard form arising from chain rule applied to $\ln$.

10. Question 10 · Hard

    Find $\frac{dy}{dx}$ by implicit differentiation: $x^3+y^3=6xy$.

    • A
      $\frac{x^2-2y}{2x-y^2}$

      Why not A: Dropped the factor of 3 from both terms.
    • B
      $\frac{2y-x^2}{y^2-2x}$Correct
    • C
      $\frac{3x^2-6y}{6x-3y^2}$

      Why not C: Correct unsimplified answer — did not reduce by 3.
    • D
      $\frac{x^2}{y^2}$

      Why not D: Ignored the $6xy$ term on the right entirely.
    Explanation

    Differentiate both sides: $3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}$. Collect $\frac{dy}{dx}$ terms: $3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^2$. Factor: $\frac{dy}{dx}(3y^2-6x)=6y-3x^2$. Divide and simplify by 3: $\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}$.

    Key takeaway

    For implicit curves with products on the right, apply the product rule and then collect $\frac{dy}{dx}$ terms.

11. Question 11 · Hard

    If $f$ is a differentiable function and $g(x)=[f(x){]}^3$, find $g^{\prime }(2)$ given that $f(2)=-2$ and $f^{\prime }(2)=5$.

    • A
      $60$Correct
    • B
      $-60$

      Why not B: Sign error: $[f(2){]}^2=(-2{)}^2=4$, which is positive.
    • C
      $-12$

      Why not C: Forgot to multiply by $f^{\prime }(2)$, using $3[f(2){]}^2=3(4)=12$ with wrong sign.
    • D
      $125$

      Why not D: Used $[f^{\prime }(2){]}^3$ instead of applying the chain rule.
    Explanation

    Chain rule: $g^{\prime }(x)=3[f(x){]}^2\cdot f^{\prime }(x)$. At $x=2$: $g^{\prime }(2)=3[f(2){]}^2\cdot f^{\prime }(2)=3(-2{)}^2\cdot 5=3\cdot 4\cdot 5=60$. Note $(-2{)}^2=4>0$, so the result is positive.

    Key takeaway

    Chain rule applied to $[f(x)]^n$ gives $n[f(x)]^{n-1} f'(x)$ — substitute the known values carefully, noting that squaring removes the sign.

12. Question 12 · Hard

    Let $f(x)=\arcsin (x)$. Using the derivative of an inverse function, find $f^{\prime }(1/2)$.

    • A
      $\sqrt{3}$

      Why not A: Computed $\sqrt{1-(1/2{)}^2}$ but did not take the reciprocal.
    • B
      $\frac{2}{\sqrt{3}}$Correct
    • C
      $2$

      Why not C: Used $1/x$ instead of $1/\sqrt{1-x^2}$ for the derivative.
    • D
      $\frac{1}{2}$

      Why not D: Substituted $x=1/2$ into $f$ rather than $f^{\prime }$.
    Explanation

    $f^{\prime }(x)=\frac{1}{\sqrt{1-x^2}}$. At $x=\frac{1}{2}$: $f^{\prime }\left(\frac{1}{2}\right)=\frac{1}{\sqrt{1-\frac{1}{4}}}=\frac{1}{\sqrt{3/4}}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$.

    Key takeaway

    $\dfrac{d}{dx}[\arcsin x] = \dfrac{1}{\sqrt{1-x^2}}$ — rationalize if needed when substituting specific values.