Exam 4: More About Derivatives

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

Let $f ( x ) = \cos ^ { - 1 } ( \sin x )$ Then $f ^ { \prime } ( x )$ is

The linear approximation $L ( x )$ to $f ( x ) = e ^ { - x }$ near $x _ { 0 } = 1$ is

Let $y = \tan ^ { - 1 } ( \ln x )$ Then dy is

Using implicit differentiation on $x ^ { 3 } - 3 x y + y ^ { 3 } = 6 , y\prime$ is

Let $f ( x ) = \tan ^ { - 1 } \left( \log _ { 3 } x \right)$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \tanh ^ { - 1 } ( \ln x )$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = x \sinh ^ { - 1 } x$ Then $f ^ { \prime } ( x )$ is

Let $y = \ln \left( x ^ { y } \right)$ assuming $y \neq 0$ Using implicit differentiation, $y ^ { \prime }$ is

Using implicit differentiation on $x ^ { 2 } + y ^ { 2 } = 2 y , y ^ { \prime }$ is

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

Let $y = \sin ( \cos x )$ Then dy is

Let $y = \csc ( 5 x ) .$ Then $y ^ { \prime }$ is

Let $f ( x ) = \cos ^ { - 1 } \left( x ^ { 4 } \right)$ Then $f ^ { \prime } ( x )$ is

Let $f ( x ) = \sec ^ { - 1 } \left( e ^ { x } \right)$ Then $f ^ { \prime } ( x )$ is

The linear approximation $L ( x )$ to $f ( x ) = \tan x$ near $x _ { 0 } = \frac { \pi } { 4 }$ is

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

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

Let $f ( x ) = \cot ^ { - 1 } \left( e ^ { 2 x } \right)$ Then $f ^ { \prime } ( x )$ is

Let $y = \frac { x } { e ^ { x } } .$ Then dy is