Find the derivative of $f ( x ) = \left( x ^ { 2 } + 1 \right) ^ { 4 } , \text { find } \mathrm { f } ^ { \prime } ( 0 )$
A)0
B)28
C)4
D)32
E)16
F)24
G)12
H)8

Question

Find the derivative of $f ( x ) = \left( x ^ { 2 } + 1 \right) ^ { 4 } , \text { find } \mathrm { f } ^ { \prime } ( 0 )$ 
A)0
B)28
C)4
D)32
E)16
F)24
G)12
H)8

Quizplus · Accepted Answer

The derivative of $f(x) = (x^2 + 1)^4$ can be found using the chain rule. Let $u = x^2 + 1$, then $f(x) = u^4$. The derivative of $u^4$ with respect to $u$ is $4u^3$, and the derivative of $u = x^2 + 1$ with respect to $x$ is $2x$. Applying the chain rule, $f'(x) = 4u^3 \cdot 2x = 8x(x^2 + 1)^3$. Evaluating $f'(0)$ gives $8 \cdot 0 \cdot (0^2 + 1)^3 = 0$.

Find the Derivative Of f(x)=(x2+1)4, find f′(0)f ( x ) = \left( x ^ { 2 } + 1 \right) ^ { 4 } , \text { find } \mathrm { f } ^ { \prime } ( 0 )f(x)=(x2+1)4, find f′(0)

Find the Derivative Of $f ( x ) = \left( x ^ { 2 } + 1 \right) ^ { 4 } , \text { find } \mathrm { f } ^ { \prime } ( 0 )$