Question

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

• A. $\frac{dy}{dx}=2\sqrt{a^2-x^2}$
• B. $\frac{dy}{dx}=2\sqrt{a^2+x^2}$
• C. $\frac{dy}{dx}=2\sqrt{a^2+x^3}$
• D. $\frac{dy}{dx}=2\sqrt{a^2-x^3}$

Answer 1

The correct option is B $\frac{dy}{dx}=2\sqrt{a^2+x^2}$

$y=\frac{x\sqrt{a^2+x^2}}{x}+\frac{a^{2}log(x+\sqrt{a^2+x^2})}{v}$

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

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

$\therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$=\frac{a^2+2x^2}{\sqrt{a^2+x^2}}+\frac{a^2}{sqrt{a}^2+x^2}$

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

$=2\sqrt{a^2+x^2}$