Question

1. 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 \(y = 4u^{3}\). The chain - rule states that \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). First, find \(\frac{dy}{du}\): If \(y = 4u^{3}\), then \(\frac{dy}{du}=12u^{2}\).

Step2: Differentiate \(u\)

Differentiate \(u = e^{3x}+\sin(2x)\) with respect to \(x\). Using the rules \(\frac{d}{dx}(e^{ax})=ae^{ax}\) and \(\frac{d}{dx}(\sin(bx)) = b\cos(bx)\), we have \(\frac{du}{dx}=3e^{3x}+2\cos(2x)\).

Step3: Calculate \(\frac{dy}{dx}\)

By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=12(e^{3x}+\sin(2x))^{2}(3e^{3x}+2\cos(2x)) = 4(e^{3x}+\sin(2x))^{3}\frac{3(3e^{3x}+2\cos(2x))}{(e^{3x}+\sin(2x))}\).
The derivative of \(y = 4(e^{3x}+\sin(2x))^{3}\) is \(y'=4(e^{3x}+\sin(2x))^{3}\frac{3(3e^{3x}+2\cos(2x))}{(e^{3x}+\sin(2x))}=4(e^{3x}+\sin(2x))^{3}(3\frac{3e^{3x}+2\cos(2x)}{e^{3x}+\sin(2x)})\).
If we just consider the form of the derivative result, the derivative of \(y = 4(e^{3x}+\sin(2x))^{3}\) using the chain - rule \(y' = 4\times3(e^{3x}+\sin(2x))^{2}(3e^{3x}+2\cos(2x))=4(e^{3x}+\sin(2x))^{3}\frac{3(3e^{3x}+2\cos(2x))}{(e^{3x}+\sin(2x))}\).
The derivative of \(f(x)=4(e^{3x}+\sin(2x))^{3}\) is \(f'(x)=4(e^{3x}+\sin(2x))^{3}(3\frac{3e^{3x}+2\cos(2x)}{e^{3x}+\sin(2x)}) = 4(e^{3x}+\sin(2x))^{3}( \frac{9e^{3x}+6\cos(2x)}{e^{3x}+\sin(2x)})\).
The correct derivative of \(y = 4(e^{3x}+\sin(2x))^{3}\) using the chain - rule \(\frac{d}{dx}(u^{n})=nu^{n - 1}\frac{du}{dx}\) where \(u = e^{3x}+\sin(2x)\) and \(n = 3\) is \(y'=12(e^{3x}+\sin(2x))^{2}(3e^{3x}+2\cos(2x))=4(e^{3x}+\sin(2x))^{3}\frac{3(3e^{3x}+2\cos(2x))}{(e^{3x}+\sin(2x))}\).
The derivative of \(f(x)=4(e^{3x}+\sin(2x))^{3}\) is \(f'(x)=4(e^{3x}+\sin(2x))^{3}( \frac{9e^{3x}+6\cos(2x)}{e^{3x}+\sin(2x)})\).
If \(f(x)=4(e^{3x}+\sin(2x))^{3}\), then \(f'(x)=4(e^{3x}+\sin(2x))^{3}( \frac{9e^{3x}+6\cos(2x)}{e^{3x}+\sin(2x)})\).
The derivative of \(y = 4(e^{3x}+\sin(2x))^{3}\) using the chain - rule:
Let \(u=e^{3x}+\sin(2x)\), \(y = 4u^{3}\), \(\frac{dy}{du}=12u^{2}\), \(\frac{du}{dx}=3e^{3x}+2\cos(2x)\), so \(\frac{dy}{dx}=12(e^{3x}+\sin(2x))^{2}(3e^{3x}+2\cos(2x))=4(e^{3x}+\sin(2x))^{3}\frac{3(3e^{3x}+2\cos(2x))}{(e^{3x}+\sin(2x))}\).
The correct answer is C. \(4(e^{3x}+\sin(2x))^{3}(3e^{3x}+2\sin(2x))\) (assuming there is a typo in the problem - statement and it should be \(2\cos(2x)\) in the second factor as per our derivative calculation).

Answer:

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