Question

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

• A. $\frac{x^2}{(x\sin x+\cos x{)}^2}$
• B. $\frac{-x^2}{(x\sin x+\cos x{)}^2}$
• C. $\frac{x^2(\cos x-\sin x)}{(x\sin x+\cos x{)}^2}$
• D. $\frac{x^2(\cos x+\sin x)}{(x\sin x+\cos x{)}^2}$

Answer 1

The correct option is A $\frac{x^2}{(x\sin x+\cos x{)}^2}$

$\frac{d}{dx}\left(\frac{\sin x-x\cos x}{x\sin x+\cos x}\right)$

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

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

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

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

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