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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Differentiate \( \tan^{-1} \{ \large\frac{\sqrt{1+x^2}-1}{x} \}\) w.r.t. \( x\).

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Toolbox:
  • $1+\tan^2\theta=\sec^2\theta$
  • $\large\frac{d}{dx}$$(\tan^{-1}x)=\large\frac{1}{1+x^2}$
  • $\sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
Step 1:
$y=\tan^{-1}\big[\large\frac{\sqrt{1+x^2}-1}{x}\big]$
Put $x=\tan\theta$
$\Rightarrow 1+x^2=1+\tan^2\theta=\sec^2\theta$
$\therefore y=\tan^{-1}\bigg[\large\frac{\sec\theta-1}{\tan\theta}\bigg]$
$\Rightarrow y=\tan^{-1}\bigg[\large\frac{(1/\cos\theta)-1}{\sin\theta/\cos\theta}\bigg]$
$\Rightarrow y=\tan^{-1}\bigg[\large\frac{1-\cos\theta}{\sin\theta}\bigg]$
Step 2:
But $1-\cos \theta=2\sin^2\large\frac{\theta}{2}$
$\sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
$y=\tan^{-1}\bigg[\large\frac{2\sin^2\large\frac{\theta}{2}}{2\sin\large\frac{\theta}{2}\cos\large\frac{\theta}{2}}\bigg]$
$\therefore y=\tan^{-1}[\tan\large\frac{\theta}{2}]$
$y=\large\frac{\theta}{2}$
Step 3:
But $\theta=\tan^{-1}x$
Hence $y=\large\frac{1}{2}$$\tan^{-1}x$
$\large\frac{dy}{dx}=\frac{1}{2}\frac{1}{1+x^2}$
$\quad\;\;=\large\frac{1}{2(1+x^2)}$
answered Sep 30, 2013 by sreemathi.v
 

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