differentiation

Question: $\frac{d}{dx}\left( {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)$ is equal to



1) $\frac{1}{1+{{x}^{2}}}$
2) $\frac{1}{2(1+{{x}^{2}})}$
3) $\frac{{{x}^{2}}}{2\sqrt{1+{{x}^{2}}}(\sqrt{1+{{x}^{2}}}-1)}$
4) $\frac{2}{1+{{x}^{2}}}$
Solution: Explanation: No Explanation
Differentiation by substitution Higher order derivatives

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