Differentiate \tan^{-1}\frac{3x-x^{3}}{1-3x^{2}},\left | x \right |< \frac{1}{\sqrt{3}}\; w.r.t\; \, \tan^{-1}\frac{x}{\sqrt{1-x^{2}}}

 

 

 

 
 
 
 
 

Answers (1)

Differentiate    \tan^{-1}\frac{3x-x^{2}}{1-3x^{2}}
Let  y= \tan^{-1}\left (\frac{3x-x^{2}}{1-3x^{2}} \right )
put  x= \tan \theta                     \therefore y= \tan^{-1}\left(\frac{3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }\right)
\therefore \left | x \right |< \frac{1}{\sqrt{3}} \Rightarrow \frac{-1}{\sqrt{3}}< x< \frac{1}{\sqrt{3}}\\y = \tan^{-1}\tan 3\theta = 3\theta = 3\tan^{-1}x---(i)                          \Rightarrow \frac{-1}{\sqrt{3}}< \tan \theta < \frac{1}{\sqrt{3}}               
\Rightarrow \frac{-\pi }6< \theta < \frac{\pi }6\; \; \therefore \frac{-\pi }{2}< 3\theta < \frac{\pi }{2}
Also Let z= \tan^{-1}\frac{x}{\sqrt{1-x^{2}}}\; \; put \; x= \sin \alpha
\Rightarrow z= \tan^{-1}\frac{\sin \alpha }{\sqrt{1-\sin^2 \alpha }}= \tan^{-1}(\frac{\sin \alpha }{\cos \alpha })\Rightarrow \tan^{-1}\tan \alpha
         z=\alpha = \sin^{-1}x---(ii)
Now by (i) and (ii) , \frac{dy}{dx}= 3\times \frac{1}{1+x^{2}}  and
                              \frac{dz}{dx}= \frac{1}{\sqrt{1-x^{2}}}
\therefore \frac{dy}{dz}=\frac{dy}{dx} \times \frac{dx}{dz}= 3 \times \frac{\sqrt{1-x^{2}}}{1+x^{2}}

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