Question

find the derivative of $f(x) = \frac{2\cos x}{9x^3 + 16}$.\
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$\boldsymbol{f(x) = \frac{-18x^3\sin x - 32\sin x - 54x^2\cos x}{9x^3 + 16}}$\
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$\boldsymbol{f(x) = \frac{-2\sin x}{27x^2}}$\
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$\boldsymbol{f(x) = \frac{18x^3\sin x + 32\sin x + 54x^2\cos x}{(9x^3 + 16)^2}}$\
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$\boldsymbol{f(x) = \frac{-18x^3\sin x - 32\sin x - 54x^2\cos x}{(9x^3 + 16)^2}}$

Explanation:

Step1: Recall quotient rule

For $f(x)=\frac{u(x)}{v(x)}$, $f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}$

Step2: Define u, v, find derivatives

Let $u(x)=2\cos x$, so $u'(x)=-2\sin x$.
Let $v(x)=9x^3+16$, so $v'(x)=27x^2$.

Step3: Substitute into quotient rule

$$\begin{align*} f'(x)&=\frac{(-2\sin x)(9x^3+16)-(2\cos x)(27x^2)}{(9x^3+16)^2}\\ &=\frac{-18x^3\sin x -32\sin x -54x^2\cos x}{(9x^3+16)^2} \end{align*}$$

Answer:

$\boldsymbol{f'(x)=\frac{-18x^3\sin x -32\sin x -54x^2\cos x}{(9x^3+16)^2}}$ (the fourth option)