differentiate the following function. y = (9 + sin x)/(9x + cos x) y =

Explanation:

Step1: Recall quotient - rule

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

Step2: Find $u'$ and $v'$

Differentiate $u$ with respect to $x$: $u'=\frac{d}{dx}(9+\sin x)=0 + \cos x=\cos x$. Differentiate $v$ with respect to $x$: $v'=\frac{d}{dx}(9x+\cos x)=9-\sin x$.

Step3: Apply quotient - rule

$y'=\frac{\cos x(9x + \cos x)-(9+\sin x)(9 - \sin x)}{(9x+\cos x)^{2}}$. Expand the numerator: [ \begin{align*} &\cos x(9x+\cos x)-(9 + \sin x)(9-\sin x)\ =&9x\cos x+\cos^{2}x-(81-\sin^{2}x)\ =&9x\cos x+\cos^{2}x - 81+\sin^{2}x\ =&9x\cos x+(\cos^{2}x+\sin^{2}x)-81\ =&9x\cos x + 1-81\ =&9x\cos x-80 \end{align*} ]

Answer:

$\frac{9x\cos x - 80}{(9x+\cos x)^{2}}$