Question

Prove that $\frac{d}{dx}(\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a})=\sqrt{a^2-x^2}$.

Answer 1

To prove $\frac{d}{dx}(\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a})=\sqrt{a^2-x^2}$

L.H.S $\frac{d}{x}(\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a})$

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

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

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

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

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

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

$=RHS$