# If prove that

Question:

If prove that $\frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}=\sqrt{a^{2}-x^{2}}$.

Solution:

Let $y=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}$

On differentiating $y$ with respect to $x$, we get

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

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

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\frac{d}{d x}\left(x \times \sqrt{a^{2}-x^{2}}\right)+a^{2} \frac{d}{d x}\left(\sin ^{-1} \frac{x}{a}\right)\right]$

We have (uv)' $=v u^{\prime}+u v^{\prime}$ (product rule)

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

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$ and $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}} \times 1+x\left\{\frac{1}{2}\left(a^{2}-x^{2}\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(a^{2}-x^{2}\right)\right\}\right.$

$\left.+a^{2}\left\{\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \frac{d}{d x}\left(\frac{x}{a}\right)\right\}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}+\frac{x}{2}\left(a^{2}-x^{2}\right)^{-\frac{1}{2}}\left\{\frac{d}{d x}\left(a^{2}\right)-\frac{d}{d x}\left(x^{2}\right)\right\}+\frac{a^{2}}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}}\left\{\frac{1}{a} \frac{d}{d x}(x)\right\}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}+\frac{x}{2 \sqrt{a^{2}-x^{2}}}\left\{\frac{d}{d x}\left(a^{2}\right)-\frac{d}{d x}\left(x^{2}\right)\right\}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}} \frac{d}{d x}(x)\right]$

However, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$ and derivative of a constant is 0

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}+\frac{x}{2 \sqrt{a^{2}-x^{2}}}\{0-2 x\}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}} \times 1\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}+\frac{x(-2 x)}{2 \sqrt{a^{2}-x^{2}}}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{\sqrt{a^{2}-x^{2}}}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}+\frac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}+\sqrt{a^{2}-x^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2} \times 2 \sqrt{a^{2}-x^{2}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}$

Thus, $\frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}=\sqrt{a^{2}-x^{2}}$