The derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) with respect to tan^-1x is:

Given:

The derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) with respect to tan-1x is:

Concept:

The derivative of f(x) with respect to g(x) is \(\rm \frac{f'(x)}{g'(x)}\)

Calculation:

Let \(f(x)=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and g(x) = tan-1x

Put x = tanθ in f(x) then

\(\rm f(x)=\tan^{-1}\left(\frac{sec\theta-1}{tan\theta}\right)\)

\(\rm f(x)=\tan^{-1}\left(\frac{1-cos\theta}{sin\theta}\right)\)

\(\rm f(x)=\tan^{-1}\left(\frac{sin^2\frac{\theta}{2}}{2sin\frac{\theta}{2}cos\frac{\theta}{2}}\right)\)

\(\rm f(x)=\tan^{-1}\left(\tan\frac{\theta}{2}\right)\)

\(\rm f(x)=\frac{\theta}{2}\)

\(\rm f(x)=\frac12tan^{-1}x\)

On differentiating .

\(\rm f'(x)=\frac{1}{2}\cdot\frac{1}{1+x^2}\)

we have g(x) = tan-1x

On differentiating 

\(\rm g'(x)=\frac{1}{1+x^2}\)

Now the derivative of f(x) with respect to g(x) is

\(\rm \frac{f'(x)}{g'(x)}=\frac{\frac{1}{2}\cdot\frac{1}{1+x^2}}{\frac{1}{1+x^2}}=\frac{1}{2}\)

Hence the option (2) is correct.