Question

Find $\frac{dy}{dx}$ for $y={\log }_e(x+\sqrt{x^2-a^2})$.

Answer 1

Given,

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

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{1+\frac{2x}{2\sqrt{x^2-a^2}}}{x+\sqrt{x^2-a^2}}$

or, $\frac{dy}{dx}=\frac{1}{\sqrt{x^2-a^2}}$.