Question

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

Answer 1

$$\begin{gathered} \frac{d}{dx}\left\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right\}=\sqrt{a^2-x^2} \\ \mathrm{LHS}=\frac{d}{dx}\left\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right\} \\ =\frac{d}{dx}\left(\frac{x}{2}\sqrt{a^2-x^2}\right)+\frac{d}{dx}\left(\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right) \\ =\frac{1}{2}\left[x\frac{d}{dx}\sqrt{a^2-x^2}+\sqrt{a^2-x^2}\frac{d}{dx}\left(x\right)\right]+\frac{a^2}{2}\times \frac{1}{\sqrt{1-{\left({\displaystyle \frac{x}{a}}\right)}^2}}\times \frac{d}{dx}\left(\frac{x}{a}\right) \\ =\frac{1}{2}\left[x\times \frac{1}{2\sqrt{a^2-x^2}}\frac{d}{dx}\left(a^2-x^2\right)+\sqrt{a^2-x^2}\right]+\left[\frac{a^2}{2}\right]\times \frac{1}{\sqrt{{\displaystyle \frac{a^2-x^2}{a^2}}}}\times \left(\frac{1}{a}\right) \\ =\frac{1}{2}\left[\frac{x\left(-2x\right)}{2\sqrt{a^2-x^2}}+\sqrt{a^2-x^2}\right]+\left(\frac{a^2}{2}\right)\frac{a}{\sqrt{a^2-x^2}}\times \left(\frac{1}{a}\right) \\ =\frac{1}{2}\left[\frac{-2x^2+2\left(a^2-x^2\right)}{2\sqrt{a^2-x^2}}\right]+\frac{a^2}{2\sqrt{a^2-x^2}} \\ =\frac{1}{2}\left[\frac{2\left(a^2-2x^2\right)}{2\sqrt{a^2-x^2}}\right]+\frac{a^2}{2\sqrt{a^2-x^2}} \\ =\frac{a^2-2x^2}{2\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}} \\ =\frac{a^2-2x^2+a^2}{2\sqrt{a^2-x^2}} \\ =\frac{2a^2-2x^2}{2\sqrt{a^2-x^2}} \\ =\frac{2\left(a^2-x^2\right)}{2\sqrt{a^2-x^2}} \\ =\frac{\left(a^2-x^2\right)}{\sqrt{a^2-x^2}} \\ =\sqrt{a^2-x^2}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved} \end{gathered}$$