Question

lf $y=\log \left(\log \right(x+\sqrt{1+x^2}\left)\right)$ then $\frac{dy}{dx}=$

• A. $\frac{1}{x+\sqrt{1+x^2}}$
• B. $\frac{x}{\log (x+\sqrt{1+x^2})}$
• C. $\frac{-1}{\log (x+\sqrt{1+x^2})}$
• D. $\frac{1}{\sqrt{1+x^2}\log (x+\sqrt{1+x^2})}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{1+x^2}\log (x+\sqrt{1+x^2})}$

We have, $y=\log \left(\log \right(x+\sqrt{1+x^2}\left)\right)$
Using chain rule of differentiation,
$\frac{dy}{dx}=\frac{1}{\log (x+\sqrt{1+x^2})}\times \frac{1}{(x+\sqrt{1+x^2})}\times (1+\frac{2x}{2\sqrt{1+x^2}})$

$=\frac{1}{\log (x+\sqrt{1+x^2})}\times \frac{1}{(x+\sqrt{1+x^2})}\times \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right)$
$=\frac{1}{\sqrt{1+x^2}\log (x+\sqrt{1+x^2})}$

Hence, $\frac{dy}{dx}=\frac{1}{\sqrt{1+x^2}\log (x+\sqrt{1+x^2})}$