Derivative Exercises

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Here we’ll practice derivative rules.

Let $Y(z) = -2 z^2 + \sqrt [5]{z} -3 \sin (z)$ . Compute:

Rewrite $\sqrt [5]{z}$ as $z^{1/5}$ .

$Y'(z) =\answer {-4z+\frac {1}{5 (z)^{4/5}}-3 \cos (z)}$

Let $f(x) = 3x^5+\frac {1}{\sqrt [3]{x}-3 \cos (x)}$ . Compute:

Rewrite $\frac {1}{\sqrt [3]{x}}$ as $x^{-1/3}$ .

$f'(x) =\answer [given]{15x^4-\frac {(1/3)x^{-2/3} + 3\sin (x)}{(x^{1/3} - 3\cos (x))^2}}$

Let $R(v) = \left (\sqrt {v}-\frac {2}{\sqrt [4]{v}}\right ) \left (3 \sin (v)-\frac {4}{v^5}\right )$ . Compute: $\dd {v}R(v) =\answer [given]{\left (\sqrt {v}-\frac {2}{{v}^{1/4}}\right ) \left (\frac {20}{v^6}+3 \cos (v)\right ) + \left (\frac {1}{2 {v}^{5/4}}+\frac {1}{2 \sqrt {v}}\right ) \left (3 \sin (v)-\frac {4}{v^5}\right )}$

Let $p(n) = \left (-2 \sqrt [5]{n}-\frac {4}{\sqrt [4]{n}}\right ) \left (3 n^4-2 \sin (n)\right )$ . Compute: $p'(n) =\answer [given]{\left (\frac {1}{{n}^{5/4}}-\frac {2}{5 {n}^{4/5}}\right ) \left (3 n^4 - 2 \sin (n)\right )+\left (-2 {n}^{1/5}-\frac {4}{{n}^{1/4}}\right ) \left (12 n^3 - 2 \cos (n)\right )}$

Let $R(u) = \frac {5 u^4+3 {u}^{1/3}}{-5 {u}^{1/3}}$ . Compute: $R'(u) =\answer {\frac {5 u^4 + 3 {u}^{1/3}}{15 u {u}^{1/3}}-\frac {20 u^3+\frac {1}{{u}^{2/3}}}{5 {u}^{1/3}}}$

Let $C(t) = \frac { \sqrt {t}-\frac {3}{\sqrt [4]{t}}}{-2 \sqrt {t}}$ . Compute: $C'(t) =\answer [given]{\frac {-9}{8 {t}^{7/4}}}$

Let $r(z) = -\cot \left (2 \sqrt {2} {z}^{1/4}\right )$ . Compute:

$\dd {z}r(z) =\answer {\frac {\csc ^2\left (2 \sqrt {2} {z}^{1/4}\right )}{\sqrt {2} {z}^{3/4}}}$

Let $y(v) = 2 \sin (2 v)-4 \sin ^9(2 v)$ . Compute:

$y'(v) =\answer [given]{4 \cos (2 v)-72 (\sin (2 v))^8 \cos (2 v)}$

Let $Q(z) = \frac {A(z)}{B(z)}$

We have the following table of values for $A$ and $B$ .

$\begin {array}{c|c|c|c|c} z & A(z) & A'(z) & B(z) & B'(z)\\ \hline 1 & 5 & -9 & 4 & -7\\ 2 & -3 & 3 & -10 & 5\\ 3 & 2 & 6& 6 &-2\\ 4 & -7 & 10& 9 &-4\\ 5 & -1& -7 &-1& -8\\ \end {array}$

Use this data to compute $\eval {\frac {\d }{\d z} Q(z)}_{z = 2}$ .

$Q'(2) = \answer {-0.15}$

Let $a(x)=g(x)h(x)$ and $b(x) = x^2g(x)+(h(x))^2$ and $c(x)=g(h(x))$

We have the following table of values for $g$ and $h$ .

$\begin {array}{c|c|c|c|c} x & g(x) & g'(x) & h(x) & h'(x)\\ \hline 1 & 2 & 2 & 6 & -3\\ 2 & 5 & 2 & 5 & 2\\ 3 & 4 & 3 & 1 & -1\\ 4 & -1 & 0 & 2 & 5\\ 5 & 3 & 1 & 4 & 9\\ \end {array}$

Use this data to compute $a'(3)$ .

$a'(3) = \answer {-1}$

Use this data to compute $b'(2)$ .

$b'(2) = \answer {48}$

Use this data to compute $c'(5)$ .

$c'(5) = \answer {0}$

Let $A(x) = \frac {\sqrt [3]{x}}{B(x)}$ . Assume that $B(8) = 3$ and $B'(8) = \frac {1}{8}$ . Calculate $A'(8)$ .

$A'(8) = \answer {0}$

Use implicit differentiation to find $\frac {dy}{dx}$ where $x = y^2.$ $\frac {dy}{dx} = \answer {\frac {1}{2y}}$

Use implicit differentiation to find $\frac {dy}{dx}$ where $y^2x = \sin (xy).$

$\frac {dy}{dx} = \answer {\frac {y(y-\cos (xy))}{x(-2y+\cos (xy))}}$

Use implicit differentiation to find $\frac {dy}{dx}$ where $\sin (x+y)+y = 0.$ $\frac {dy}{dx} = \answer {\frac {-\cos (x+y)}{1+\cos (x+y)}}$

Use implicit differentiation to find $\frac {dy}{dx}$ where $y = \frac {x-2}{y+3}.$

$\frac {dy}{dx} = \answer {\frac {y+3}{7+x+6y+y^2}}$

The image below shows the graph of $f$ , $f'$ , and $f''$ . Determine which graph corresponds to which function.

$\begin{align*} f(x) &= \answer{B}\\ f'(x) &= \answer{A}\\ f''(x) &= \answer{C}\\ \end{align*}$

The image below shows the graph of $f$ , $f'$ , and $f''$ . Determine which graph corresponds to which function.

$\begin{align*} f(x) &= \answer{A}\\ f'(x) &= \answer{C}\\ f''(x) &= \answer{B}\\ \end{align*}$

The image below shows the graph of $f$ , $f'$ , and $f''$ . Determine which graph corresponds to which function.

$\begin{align*} f(x) &= \answer{A}\\ f'(x) &= \answer{B}\\ f''(x) &= \answer{C}\\ \end{align*}$

The image below shows the graph of $f$ , $f'$ , and $f''$ . Determine which graph corresponds to which function.

$\begin{align*} f(x) &= \answer{B}\\ f'(x) &= \answer{C}\\ f''(x) &= \answer{A}\\ \end{align*}$