Question

Differentiate
$\frac{\sin x-x\cos x}{x\sin x+\cos x}$

Answer 1

Let $f\left(x\right)=\frac{\sin x-x\cos x}{x\sin x+\cos x}$ which is of the form $\frac{u}{v}$

$f^{\prime }\left(x\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

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

$=\frac{(x\sin x+\cos x)(\cos x-\cos x+x\sin x)-(\sin x-x\cos x)(x\cos x+\sin x-\sin x)}{{(x\sin x+\cos x)}^2}$

$=\frac{(x\sin x+\cos x)\left(x\sin x\right)-(\sin x-x\cos x)\left(x\cos x\right)}{{(x\sin x+\cos x)}^2}$

$=\frac{x^2{\sin }^{2}x+x\sin x\cos x-x\sin x\cos x+x^2{\cos }^{2}x}{{(x\sin x+\cos x)}^2}$

$=\frac{x^2{\sin }^{2}x+x^2{\cos }^{2}x}{{(x\sin x+\cos x)}^2}$

$=\frac{x^2({\sin }^{2}x+{\cos }^{2}x)}{{(x\sin x+\cos x)}^2}$

$=\frac{x^2}{{(x\sin x+\cos x)}^2}$ since ${\sin }^{2}x+{\cos }^{2}x=1$