Question

If $y=\ell n\left\{\frac{x+\sqrt{(a^2+x^2)}}{a}\right\}$, then the value of $\frac{dy}{dx}$ is

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

Answer 1

The correct option is C $\frac{1}{\sqrt{a^2+x^2}}$

$y=ln\left\{\frac{x+\sqrt{a^2+x^2}}{a}\right\}$

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

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

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

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

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