Question

1. if ( f(x)=(e^{3 x}+sin (2 x))^{4} ), then ( f^{prime}(x)= ) (a) ( 4left(3 e^{3 x}+2 cos (2 x)

ight)^{3} ) (b) ( 4left(e^{3 x}+sin (2 x)
ight)^{3}left(e^{3 x}+cos (2 x)
ight) ) (c) ( 4left(e^{3 x}+sin (2 x)
ight)^{3}left(3 e^{3 x}+2 sin (2 x)
ight) ) (d) ( 4left(e^{3 x}+sin (2 x)
ight)^{3}left(3 e^{3 x}+2 cos (2 x)
ight) )

Explanation:

Step1: Use the chain rule

If \(y = u^{n}\), then \(y^\prime=nu^{n - 1}u^\prime\). Here \(u = e^{3x}+\sin(2x)\), \(n = 4\).
First, find the derivative of \(u = e^{3x}+\sin(2x)\).
The derivative of \(e^{3x}\) using the chain rule (\((e^{ax})^\prime=ae^{ax}\)) is \(3e^{3x}\), and the derivative of \(\sin(2x)\) using the chain rule (\((\sin(ax))^\prime=a\cos(ax)\)) is \(2\cos(2x)\). So \(u^\prime=3e^{3x}+2\cos(2x)\).

Step2: Apply the chain - rule formula

Since \(y=(e^{3x}+\sin(2x))^{4}\), by the chain rule \(y^\prime = 4(e^{3x}+\sin(2x))^{3}(e^{3x}+\sin(2x))^\prime\).
Substitute \(u^\prime = 3e^{3x}+2\cos(2x)\) into the formula.

Answer:

\(4(e^{3x}+\sin(2x))^{3}(3e^{3x}+2\cos(2x))\), so the answer is A.