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(…

Explanation:

Step1: Apply chain - rule

Let (u = e^{3x}+\sin(2x)), so (f(x)=4u^{4}). By the chain - rule (\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}). First, find (\frac{df}{du}): If (f = 4u^{4}), then (\frac{df}{du}=16u^{3}).

Step2: Find (\frac{du}{dx})

We know that if (u = e^{3x}+\sin(2x)), then (\frac{du}{dx}=\frac{d(e^{3x})}{dx}+\frac{d(\sin(2x))}{dx}). Using the chain - rule, (\frac{d(e^{3x})}{dx}=3e^{3x}) and (\frac{d(\sin(2x))}{dx}=2\cos(2x)). So (\frac{du}{dx}=3e^{3x}+2\cos(2x)).

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

By the chain - rule (\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}), substituting (u = e^{3x}+\sin(2x)), (\frac{df}{du}=16u^{3}) and (\frac{du}{dx}=3e^{3x}+2\cos(2x)) we get: (f^{\prime}(x)=4(e^{3x}+\sin(2x))^{3}(3e^{3x}+2\cos(2x)))

Answer:

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