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Differentiate $ tan^{-1}\bigg (\Large \frac{\sqrt {1+x^2}-1}{x}\bigg)\;\normalsize w.r.t\;\tan^{-1}x\;when\;x\neq 0$

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Toolbox:
  • Let $u=f(x)$ and $v=g(x)$ be two functions of $x$,then to find the derivative of $f(x)$ w.r.t $g(x)$.
  • (i.e) $\large\frac{du}{dv}=\large\frac{du/dx}{dv/dx}$
Step 1:
Given : $y=\tan^{-1}\bigg(\large\frac{\sqrt{1+x^2}-1}{x}\bigg)$
Let $u=\tan^{-1}\bigg(\large\frac{\sqrt{1+x^2}-1}{x}\bigg)$ and $v=\tan^{-1}x$
Put $x=\tan\theta \Rightarrow \tan^{-1} x=\theta$
$u=\tan^{-1}\bigg(\large\frac{\sqrt{1+\tan^2\theta}-1}{\tan \theta}\bigg)$
But $1+\tan^2\theta=\sec^2\theta$
Therefore $u=\tan^{-1}\bigg(\large\frac{\sec\theta-1}{\tan\theta}\bigg)$
$\Rightarrow \tan^{-1}\bigg(\large\frac{\Large\frac{1}{\cos\theta}-1}{\tan \theta}\bigg)$
$\Rightarrow \tan^{-1}\bigg(\large\frac{1-\cos\theta}{\sin\theta}\bigg)$
But $1-\cos\theta=2\sin^2\large\frac{\theta}{2}$
$\sin\theta=2\sin \large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
$\Rightarrow \tan^{-1}\bigg(\large\frac{2\sin^2\Large\frac{\theta}{2}}{2\sin\Large\frac{\theta}{2}\cos\Large\frac{\theta}{2}}\bigg)$
$\Rightarrow \tan^{-1}\bigg(\tan\large\frac{\theta}{2}\bigg)=\large\frac{\theta}{2}$
Step 2:
Substituting for $\theta$ we get,
$u=\large\frac{1}{2}$$\tan^{-1}x$
Differentiating w.r.t $x$,
Therefore $\large\frac{du}{dx}=\frac{1}{2}\frac{1}{1+x^2}$
$\qquad\qquad\;\;\;\;=\large\frac{1}{2(1+x^2)}$
Consider $v=\tan^{-1}x$
$\large\frac{dv}{dx}=\frac{1}{1+x^2}$
Therefore $\large\frac{du}{dv}=\frac{du/dx}{dv/dx}$
$\Rightarrow \large\frac{1}{2(1+x^2)}$$(1+x^2)$
$\Rightarrow \large\frac{1}{2}$
Therefore $\large\frac{du}{dv}$$=\large\frac{1}{2}$
answered Jul 2, 2013 by sreemathi.v
 

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