Exam 4: More About Derivatives

Let $f ( x ) = \ln ( \operatorname { csch } x )$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \sinh \left( x ^ { 2 } \right)$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \ln ( \tanh x )$ Then $f ^ { \prime } ( x )$ is

The linear approximation $L ( x )$ to $f ( x ) = \sqrt [ 3 ] { x }$ near $x _ { 0 } = 8$ is

Let $y = \log _ { 3 } ( \cot x )$ Then $y ^ { \prime }$ is

Using implicit differentiation on $e ^ { y - x } = x , y ^ { \prime }$ is

Let $f ( x ) = \cot \left( e ^ { x } \right)$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \ln ( \operatorname { sech } x )$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \ln \left( \frac { x + 3 } { x - 2 } \right)$ Then $f ^ { \prime } ( x )$ is

Let $y = \ln \left( 2 x + y ^ { 2 } \right)$ Using implicit differentiation, $y ^ { \prime }$ is

Let $f ( x ) = \ln ( \cosh x )$ Then $f ^ { \prime } ( x )$ is

Let $y = \frac { u } { e ^ { u } }$ and $u = \frac { 3 } { x } .$ By the Chain Rule, $\frac { d y } { d x }$ is

Let $y = \frac { u - 2 } { u + 3 }$ and $u = 2 x + 1$ By the Chain Rule, $\frac { d y } { d x }$ is

Let $y = 3 u e ^ { u }$ and $u = 2 x + 3$ By the Chain Rule, $\frac { d y } { d x }$ is

Let $f ( x ) = \ln ( \tan x )$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \log _ { 2 } \sqrt { x ^ { 2 } + 5 }$ Then $f ^ { \prime } ( x )$ is

Let $y = \ln \left( \frac { x } { y } \right)$ Using implicit differentiation, $y ^ { \prime }$ is

Let $y = u + \frac { 1 } { u }$ and $u = x ^ { 2 } + x - 1$ By the Chain Rule, $\frac { d y } { d x }$ is

Let $y = \cos \left( x ^ { 2 } - x + 1 \right)$ . Then $y ^ { \prime }$ is

Let $f ( x ) = 3 ^ { - x } \sinh x$ Then $f ^ { \prime } ( x )$ is