Question

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

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

Answer 1

The correct option is B $\frac{1}{\sqrt{x^2-a^2}}$

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

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

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

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