Question

if (f(x)=(e^{3x}+sin(2x))^{4}), then (f(x)=)
(a) (4(3e^{3x}+2cos(2x))^{3})
(b) (4(e^{3x}+sin(2x))^{3}(e^{3x}+cos(2x)))
(c) (4(e^{3x}+sin(2x))^{3}(3e^{3x}+2sin(2x)))
(d) (4(e^{3x}+sin(2x))^{3}(3e^{3x}+2cos(2x)))

Explanation:

Step1: Apply chain - rule

Let $u = e^{3x}+\sin(2x)$, then $f(x)=u^{4}$. By the chain - rule $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$. First, find $\frac{df}{du}$. Since $f(u) = u^{4}$, then $\frac{df}{du}=4u^{3}=4(e^{3x}+\sin(2x))^{3}$.

Step2: Find $\frac{du}{dx}$

We know that if $y = e^{3x}$, by the chain - rule $\frac{dy}{dx}=e^{3x}\cdot3 = 3e^{3x}$, and if $z=\sin(2x)$, by the chain - rule $\frac{dz}{dx}=\cos(2x)\cdot2 = 2\cos(2x)$. So, $\frac{du}{dx}=\frac{d}{dx}(e^{3x}+\sin(2x))=3e^{3x}+2\cos(2x)$.

Step3: Calculate $f^{\prime}(x)$

By the chain - rule $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}=4(e^{3x}+\sin(2x))^{3}(3e^{3x}+2\cos(2x))$.

Answer:

D. $4(e^{3x}+\sin(2x))^{3}(3e^{3x}+2\cos(2x))$