Question

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:

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