# Differentiate $\large \frac{x}{\sin x}\;\normalsize w.r.t\;\sin x.$

Toolbox:
• Let $u=f(x)$ and $v=g(x)$ be two function of $x$,then to find the derivative of $f(x)$ w.r.t $g(x)$ (i.e) $\large\frac{du}{dv}$$=\large\frac{du/dx}{dv/dx} • Quotient rule : \large\frac{d}{dx}\big(\large\frac{u}{v}\big)=\large\frac{v.(\Large\frac{du}{dx})-\normalsize u.(\Large\frac{dv}{dx})}{v^2} Step 1: Given : \large\frac{x}{\sin x} Let u=\large\frac{x}{\sin x} and Let v=\sin x Differentiating w.r.t x by applying the quotient rule. \large\frac{d}{dx}\big(\large\frac{u}{v}\big)=\large\frac{vu'-uv'}{uv} \large\frac{du}{dx}=\large\frac{\sin x(1)-x.\cos x}{\sin^2 x} \quad\quad=\large\frac{\sin x-x.\cos x}{\sin^2 x} Step 2: Consider v=\sin x Differentiating w.r.t x we get, \large\frac{dv}{dx}=$$\cos x$
Therefore $\large\frac{du}{dv}$$=\large\frac{du/dx}{dv/dx}$
$\qquad\qquad\;\;\;=\large\frac{\Large\frac{\sin x-x.\cos x}{\sin^2 x}}{\cos x}$
$\qquad\qquad\;\;\;=\large\frac{\big(\Large\frac{\sin x}{\cos x}-\frac{x\cos x}{\cos x}\big)}{\sin^2 x}$
$\large\frac{du}{dx}=\large\frac{\tan x-x}{\sin^2 x}$