if $f(x)=(e^{3x}+sin(2x))^{4}$, then $f(x)=$

Explanation:

Step1: Apply chain - rule

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

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

We know that (u = e^{3x}+\sin(2x)). The derivative of (y = e^{3x}) using the chain - rule: let (t = 3x), then (\frac{d(e^{3x})}{dx}=\frac{d(e^{t})}{dt}\cdot\frac{dt}{dx}=e^{t}\cdot3 = 3e^{3x}). The derivative of (y=\sin(2x)) using the chain - rule: let (s = 2x), then (\frac{d(\sin(2x))}{dx}=\frac{d(\sin(s))}{ds}\cdot\frac{ds}{dx}=\cos(s)\cdot2 = 2\cos(2x)). So (\frac{du}{dx}=3e^{3x}+2\cos(2x)).

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

Since (\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}) and (\frac{df}{du}=4u^{3}), (u = e^{3x}+\sin(2x)), (\frac{du}{dx}=3e^{3x}+2\cos(2x)), we have (f^{\prime}(x)=4(e^{3x}+\sin(2x))^{3}(3e^{3x}+2\cos(2x))).

Answer:

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