Question

question 14
find the derivative of $f(x) = 3x^5\sin x + 8\cos x$.
\\(\boldsymbol{f(x) = 15x^4\sin x + 3x^5\cos x - 8\sin x}\\)
\\(\boldsymbol{f(x) = 15x^4\sin x - 3x^5\cos x + 8\sin x}\\)
\\(\boldsymbol{f(x) = -15x^4\sin x\cos x + 8\sin x}\\)
\\(\boldsymbol{f(x) = 15x^4\sin x\cos x - 8\sin x}\\)

Explanation:

Step1: Differentiate $3x^5\sin x$ (product rule)

Recall product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x)+u(x)v'(x)$. Let $u(x)=3x^5$, $v(x)=\sin x$.
$u'(x)=15x^4$, $v'(x)=\cos x$.
So $\frac{d}{dx}[3x^5\sin x] = 15x^4\sin x + 3x^5\cos x$

Step2: Differentiate $8\cos x$

Use derivative rule: $\frac{d}{dx}[\cos x]=-\sin x$.
$\frac{d}{dx}[8\cos x] = 8(-\sin x) = -8\sin x$

Step3: Combine the two derivatives

$f'(x) = 15x^4\sin x + 3x^5\cos x -8\sin x$

Answer:

$f'(x)=15x^{4}\sin x + 3x^{5}\cos x - 8\sin x$