Chapter 8 — Quiz

10 questions covering implicit differentiation, inverse-function and logarithmic differentiation, and related rates. Answer each, then open the solution. Each answer cites the section to review.

1. In implicit differentiation, when a term containing $y$ is differentiated with respect to $x$, you must:

  • A) treat $y$ as a constant
  • B) apply the chain rule, multiplying by $y'$
  • C) drop the term
  • D) set $y = 0$
Answer**B.** Because $y$ is a hidden function $y(x)$, every $y$-term sprouts a factor of $\dfrac{dy}{dx}$. (§8.2.)

2. For $x^2 + y^2 = 1$, $\dfrac{dy}{dx} =$

  • A) $-x/y$
  • B) $-y/x$
  • C) $x/y$
  • D) $-2x$
Answer**A.** $2x + 2y\,y' = 0 \Rightarrow y' = -x/y$. (§8.2, the circle.)

3. Using implicit differentiation, $\dfrac{d}{dx}\arctan x =$

  • A) $1/(1-x^2)$
  • B) $1/(1+x^2)$
  • C) $1/\sqrt{1+x^2}$
  • D) $1/\sqrt{1-x^2}$
Answer**B.** From $x = \tan y$, $1 = \sec^2 y\,y'$, and $\cos^2 y = 1/(1+x^2)$. (§8.5.)

4. For $y = x^x$, $\dfrac{dy}{dx} =$

  • A) $x\cdot x^{x-1}$
  • B) $x^x\ln x$
  • C) $x^x(\ln x + 1)$
  • D) $x^{x+1}$
Answer**C.** Logarithmic differentiation: $\ln y = x\ln x$, $y'/y = \ln x + 1$. (§8.6.)

5. The single most destructive error in a related-rates problem is to:

  • A) draw a diagram
  • B) substitute the instant's numerical values before differentiating
  • C) differentiate with respect to $t$
  • D) check units at the end
Answer**B.** Plugging in numbers first freezes the variables (their rates become $0$), destroying the very motion. Differentiate first; substitute at step 6. (§8.7.)

6. A $10$-ft ladder slides; the base moves away from the wall at $1$ ft/s. How fast does the top descend when the base is $6$ ft from the wall?

  • A) $0.5$ ft/s
  • B) $0.75$ ft/s
  • C) $1$ ft/s
  • D) $1.5$ ft/s
Answer**B.** $x^2+y^2=100$, at $x=6$, $y=8$. $\left|\dfrac{dy}{dt}\right| = \dfrac{x}{y}\dfrac{dx}{dt} = \dfrac{6}{8}(1) = 0.75$ ft/s. (§8.8.)

7. In $\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}$ for a sphere, the factor $4\pi r^2$ is the sphere's:

  • A) volume
  • B) diameter
  • C) surface area
  • D) circumference
Answer**C.** $S = 4\pi r^2$; the identity $dV = S\,dr$ adds a thin shell each instant. (§8.14.)

8. Water flows into an inverted cone at a constant rate. As the water gets deeper, the level rises:

  • A) faster, because the cone is taller
  • B) at a constant rate
  • C) slower, because the surface widens with depth
  • D) the level cannot be determined
Answer**C.** Since $\dfrac{dh}{dt} \propto 1/h^2$ for a cone, the widening surface spreads the same inflow more thinly. (§8.9.)

9. The inverse-derivative formula is:

  • A) $(f^{-1})'(x) = 1/f'(x)$
  • B) $(f^{-1})'(x) = 1/f'(f^{-1}(x))$
  • C) $(f^{-1})'(x) = f'(f^{-1}(x))$
  • D) $(f^{-1})'(x) = -f'(x)$
Answer**B.** From $x = f(y)$, $1 = f'(y)\,y'$, and $y = f^{-1}(x)$. (§8.5.)

10. A balloon's radius increases at $1$ cm/s. When the radius is $10$ cm, the surface area is changing at:

  • A) $80\pi$ cm²/s
  • B) $400\pi$ cm²/s
  • C) $8\pi$ cm²/s
  • D) $20\pi$ cm²/s
Answer**A.** $S = 4\pi r^2 \Rightarrow \dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt} = 8\pi(10)(1) = 80\pi$ cm²/s. (§8.14.)

Scoring Guide

  • 9–10 correct — Excellent. You can differentiate implicitly, derive inverse and logarithmic derivatives, and set up related rates fluently. Move on to Chapter 9.
  • 7–8 correct — Solid. Revisit any missed section (noted in each answer) before the chapter test.
  • 5–6 correct — Partial. Re-read §8.2 (the implicit method) and §8.7 (the seven-step related-rates procedure), then redo the exercises.
  • Below 5 — Rework the chapter's worked examples (§8.8–§8.14) line by line, paying special attention to the chain-rule factor $y'$ and the "differentiate-then-substitute" order.