Differentiate the following w.r.t. $\tan^{-1} \left(\frac{2x}{1-x^2}\right)$ Differentiate : $$ \tan^{-1}  \left(\frac {\sqrt {1+x^2}-1}x\right) \quad w.r.t.\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$       
 A: Let $y=\tan^{-1}  \left(\dfrac {\sqrt {1+x^2}-1}x\right)$
and let $u= \tan^{-1} \left(\dfrac{2x}{1-x^2}\right)$.
We want to find $dy/du$. Note that:
$$
\dfrac{dy}{dx} = \dfrac{1}{2(1+x^2)}
$$
similarly for $u$, we obtain:
$$
\dfrac{du}{dx} = \dfrac{2}{1+x^2} \iff \dfrac{dx}{du} = \dfrac{1+x^2}{2}
$$
Hence, by Chain Rule, we obtain:
$$
\dfrac{dy}{du} = \dfrac{dy}{dx} \cdot \dfrac{dx}{du} = \dfrac{1}{2(1+x^2)} \cdot \dfrac{1+x^2}{2} = \dfrac{1}{4}
$$
A: Putting $x=\tan\theta,$  
$$\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$
$$=\quad \tan^{-1} \left(\frac{2\tan\theta}{1-\tan^2\theta}\right) $$
$$=\quad \tan^{-1}(\tan2\theta)=n\pi+2\theta=n\pi+2\tan^{-1}x $$ where $n$ is any integer
$$\frac {\sqrt {1+x^2}-1}x=\frac {\sqrt {1+\tan^2\theta}-1}{\tan\theta}=\frac {\sqrt {1+\tan^2\theta}-1}{\tan\theta}=\frac{1-\cos\theta}{\sin\theta}=\frac{1-\frac{1-\tan^2\frac{\theta}2}{1+\tan^2\frac{\theta}2}}{\frac{2\tan\frac{\theta}2}{1+\tan^2\frac{\theta}2}}=\tan\frac{\theta}2$$
$$\implies\tan^{-1}  \left(\frac {\sqrt {1+x^2}-1}x\right) \quad=\tan^{-1} \left(\tan\frac{\theta}2 \right)=m\pi+\frac{\theta}2=m\pi+\frac12\tan^{-1} x $$  where $m$ is any integer
So, $$\frac{d \tan^{-1}  \left(\frac {\sqrt {1+x^2}-1}x\right) \quad}{d  \tan^{-1} \left(\frac{2x}{1-x^2}\right)}=\frac{d(m\pi+\frac12\tan^{-1} x )}{d (n\pi+2\tan^{-1}x )}=\frac{\frac{d(m\pi+\frac12\tan^{-1} x )}{d(\tan^{-1} x)}}{\frac{d (n\pi+2\tan^{-1}x)}{d(\tan^{-1} x})}$$
A: the question have been solved by assuming tan inverse is “arc tan“.
    put x=tanθ and simplify 
 →arc tan(√1+tan²θ -1/tanθ)
→arc tan(√sec²θ -1/tanθ)
→arc tan(2sin²(θ/2)/sinθ)    (on simplifying)
→arc tan(2sin²(θ/2)/2sinθcosθ)
→arc tan(tan(θ/2))        =θ/2
→but θ=arc tan(x)
hence develops to →(1/2)arc tan(x) & term it as “u“.
also, arc tan(2x/1-x²)= 2arc tan(x)→“v“
finally, du/dx =1/4   (on simplifying with the above “u“ &“v“ )
