Question

State True or False:

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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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

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

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

Hence the given statement is true.