If $g ( x ) = e ^ { ( 4 x + \sin x ) } \text {, then } g ^ { \prime } ( x ) = ( 4 + \cos x ) \cdot e ^ { ( 4 x + \sin x ) }$

Name: If $g ( x ) = e ^ { ( 4 x + \sin x ) } \text {, then } g ^ { \prime } ( x ) = ( 4 + \cos x ) \cdot e ^ { ( 4 x + \sin x ) }$
Brand: Quizplus
Rating: 4.8 (38 reviews)

Question

Quizplus · Accepted Answer

The derivative of $g(x) = e^{(4x + \sin x)}$ with respect to $x$ is found using the chain rule, which gives $g'(x) = (4 + \cos x) \cdot e^{(4x + \sin x)}$, because the derivative of the exponent $4x + \sin x$ with respect to $x$ is $4 + \cos x$.

If g(x)=e(4x+sin⁡x), then g′(x)=(4+cos⁡x)⋅e(4x+sin⁡x)g ( x ) = e ^ { ( 4 x + \sin x ) } \text {, then } g ^ { \prime } ( x ) = ( 4 + \cos x ) \cdot e ^ { ( 4 x + \sin x ) }g(x)=e(4x+sinx), then g′(x)=(4+cosx)⋅e(4x+sinx)

If $g ( x ) = e ^ { ( 4 x + \sin x ) } \text {, then } g ^ { \prime } ( x ) = ( 4 + \cos x ) \cdot e ^ { ( 4 x + \sin x ) }$