Exercises - The Chain Rule

  1. Find the derivative of each, assuming $a$ is a constant.

    1. $\displaystyle{\frac{d^3}{dt^3} e^{-at}}$

    2. $\displaystyle{\frac{d}{dx} \sec e^x}$

    3. $\displaystyle{\frac{d^2}{dx^2} \left( x^2 - a^2 \right)^{2/3}}$

    4. $\displaystyle{\frac{d}{dt} \sqrt[3]{e^2-\sin 3}}$

    1. $\displaystyle{-a^3 e^{-at}}$

    2. $\displaystyle{e^x \sec e^x \tan e^x}$

    3. $\displaystyle{\frac{4(x^2-3a^2)}{9(x^2-a^2)^{4/3}}}$

    4. $\displaystyle{0}$

  2. Use the observation that $\displaystyle{\lim_{x \rightarrow c} \frac{f(x) - f(c)}{g(x) - g(c)} = \lim_{x \rightarrow c} \frac{\displaystyle{\frac{f(x) - f(c)}{x-c}}}{\displaystyle{\frac{g(x) - g(c)}{x-c}}}}$ to evaluate the following limits.

    1. $\displaystyle{\lim_{t \rightarrow \ln 4} \frac{e^{2t} - 16}{e^t - 4}}$

    2. $\displaystyle{\lim_{\theta \rightarrow \pi/8} \frac{\cos 4\theta}{2\sin 2\theta - \sqrt{2}}}$

    3. $\displaystyle{\lim_{y \rightarrow 4} \frac{\sqrt{y^2 + 9} - 5}{\sqrt{3y - 3} - 3}}$

    Each limit equals some $f'(x)/g'(x)$:

    1. $\displaystyle{8}$

    2. $\displaystyle{-\sqrt{2}}$

    3. $\displaystyle{\frac{8}{5}}$

  3. Find numbers $a$, $b$, and $c$, so that $f$ is continuous everywhere and differentiable everywhere except $x=2$

    $$f(x) = \left\{ \begin{array}{ll} \sin 4x & x \lt 0\\ ax^2 + bx + c & 0 \le x \le 2\\ e^{2-x} & x \gt 2 \end{array} \right. $$

    $\displaystyle{a = -\frac{7}{4}, \, b = 4, \, c = 0}$

  4. Find $f'(x)$ if $f(x) = (3x-1)^5 (x^2 + 1)^8$.

    $(3x-1)^4 (x^2+1)^7 (63-16x+15)$

  5. Find $\displaystyle{\frac{dy}{dx}}$ if $y = \displaystyle{\frac{x}{\sqrt{x^2+1}}}$.

    $\displaystyle{\frac{1}{(x^2 + 1)^{3/2}}}$

  6. Find the first derivative of $\displaystyle{h(x) = \frac{x^2}{(x+1)^2 (4x+1)^2}}$.

    $\displaystyle{\frac{2x(1-2x)(1+2x)}{(x+1)^3 (4x+1)^3}}$

  7. Let the position of a car decelerating to a stop be $s(t) = 100 - 10t^2$ feet at $t$ seconds. How fast is the velocity changing per foot traveled when the car is halfway to its stopping point?

    $\displaystyle{\frac{\sqrt{2}}{5}}$ ft/s per foot