Question

Find the derivative of $y=\frac{\sin x-x\cos x}{\cos x+x\sin x}$

Answer 1

$y=\frac{\sin x-x\cos x}{\cos x+x\sin x}$

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

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

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

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

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

$=\frac{x^2}{(\cos x+x\sin x{)}^2}[\because {\sin }^2+{\cos }^{2}x=1]$.