Differentiate \tan^{-1}\left [ \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \right ]     with respect to \cos^{-1}x^{2}.

 

 

 

 
 
 
 
 

Answers (1)

Let y= \tan^{-1}\left [ \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \right ]
and z= \cos^{-1}x^{2}\Rightarrow x^{2}= \cos z
y= \tan^{-1}\left [ \frac{\sqrt{1+\cos z}-\sqrt{1-\cos z}}{\sqrt{1+\cos z}+\sqrt{1-\cos z}} \right ]
\Rightarrow y= \tan^{-1}\left [ \frac{\sqrt{2}\cos \frac{z}{2}-\sqrt{2}\sin \frac{z}{2}}{\sqrt{2}\cos \frac{z}{2}+\sqrt{2}\sin \frac{z}{2}} \right ]
\Rightarrow y= \tan^{-1}\left [ \frac{\cos \frac{z}{2}-\sin \frac{z}{2}}{\cos \frac{z}{2}+\sin \frac{z}{2}} \right ]= \tan^{-1}\left [ \frac{1-\tan \frac{z}{2}}{1+\tan \frac{z}{2}} \right ]
\Rightarrow y= \tan^{-1}\left ( \tan \left ( \frac{\pi }{4}-\frac{z}{2} \right ) \right )= \frac{\pi }{4}-\frac{z}{2}
\therefore \frac{dy}{dz}= \frac{-1}{2}

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