Question

If $y=x\log |x+\sqrt{(1+x^2)}|-\sqrt{1+x^2}$ then $\frac{dy}{dx}=$

Answer 1

$y=x\log (x+\sqrt{1+x^2})-\sqrt{1+x^2}$

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

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

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

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