Question

find the derivative of $f(x) = \frac{1}{(2x^5 + \sin x)^3}$.\
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$\bigcirc$ $f(x) = \frac{-3}{(10x^4 + \cos x)^4}$\
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$\bigcirc$ $f(x) = \frac{-3(10x^4 + \cos x)}{(2x^5 + \sin x)^2}$\
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$\bigcirc$ $f(x) = \frac{-3(10x^4 + \cos x)}{(2x^5 + \sin x)^4}$\
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$\bigcirc$ $f(x) = \frac{-3}{(2x^5 + \sin x)^4}$

Explanation:

Step1: Rewrite function for easier differentiation

$f(x) = (2x^5 + \sin x)^{-3}$

Step2: Apply chain rule: $\frac{d}{dx}[u^n] = nu^{n-1}u'$

Let $u = 2x^5 + \sin x$, $n=-3$.
$f'(x) = -3(2x^5 + \sin x)^{-4} \cdot \frac{d}{dx}(2x^5 + \sin x)$

Step3: Compute derivative of $u$

$\frac{d}{dx}(2x^5 + \sin x) = 10x^4 + \cos x$

Step4: Substitute back and simplify

$f'(x) = \frac{-3(10x^4 + \cos x)}{(2x^5 + \sin x)^4}$

Answer:

$\boldsymbol{f'(x)=\frac{-3(10x^{4}+\cos x)}{(2x^{5}+\sin x)^{4}}}$ (the third option)