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

1 Answer

  • 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}$$=\large\frac{du/dx}{dv/dx}$
  • Quotient rule : $\large\frac{d}{dx}\big(\large\frac{u}{v}\big)=\large\frac{v.(\Large\frac{du}{dx})-\normalsize u.(\Large\frac{dv}{dx})}{v^2}$
Step 1:
Given : $\large\frac{x}{\sin x}$
Let $u=\large\frac{x}{\sin x}$ and Let $v=\sin x$
Differentiating w.r.t $x$ by applying the quotient rule.
$\large\frac{du}{dx}=\large\frac{\sin x(1)-x.\cos x}{\sin^2 x}$
$\quad\quad=\large\frac{\sin x-x.\cos x}{\sin^2 x}$
Step 2:
Consider $v=\sin x$
Differentiating w.r.t $x$ we get,
$\large\frac{dv}{dx}=$$\cos x$
Therefore $\large\frac{du}{dv}$$=\large\frac{du/dx}{dv/dx}$
$\qquad\qquad\;\;\;=\large\frac{\Large\frac{\sin x-x.\cos x}{\sin^2 x}}{\cos x}$
$\qquad\qquad\;\;\;=\large\frac{\big(\Large\frac{\sin x}{\cos x}-\frac{x\cos x}{\cos x}\big)}{\sin^2 x}$
$\large\frac{du}{dx}=\large\frac{\tan x-x}{\sin^2 x}$
answered Jul 2, 2013 by sreemathi.v

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