differentiate $f(x)=\frac{4sin x}{-2cos x - 9}$. $f(x)=$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4\sin x$, $u'=4\cos x$, $v=-2\cos x - 9$, and $v' = 2\sin x$.

Step2: Apply quotient - rule

$f'(x)=\frac{(4\cos x)(-2\cos x - 9)-(4\sin x)(2\sin x)}{(-2\cos x - 9)^{2}}$.

Step3: Expand the numerator

Expand the numerator: [ \begin{align*} &(4\cos x)(-2\cos x - 9)-(4\sin x)(2\sin x)\ =&-8\cos^{2}x-36\cos x - 8\sin^{2}x\ =&-8(\cos^{2}x+\sin^{2}x)-36\cos x \end{align*} ] Since $\sin^{2}x+\cos^{2}x = 1$, the numerator becomes $-8 - 36\cos x$.

Answer:

$\frac{-8 - 36\cos x}{(-2\cos x - 9)^{2}}$