Question
Prove that $\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
✓
Answer
$\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
$\text{L.H.S}=\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\text{x}\frac{\text{d}}{\text{dx}}\sqrt{\text{a}^2-\text{x}^2}+\sqrt{\text{a}^2-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\frac{\text{a}^2}{2}\times\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\text{x}}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$
[Using product rule, chain rule]
$=\frac{1}{2}\bigg[\text{x}\times\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)+\sqrt{\text{a}^2-\text{x}^2}\Big] \\ +\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2}}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\frac{\text{x}(-2\text{x})}{2\sqrt{\text{a}^2-\text{x}^2}}+\sqrt{\text{a}^2-\text{x}^2}\Big]+\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\bigg[\frac{-2\text{x}^2+2\big(\text{a}^2-\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{1}{2}\bigg[\frac{2\big(\text{a}^2-2\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2+\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\big(\text{a}^2-\text{a}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{(\text{a}^2-\text{x}^2)}{\sqrt{\text{a}^2-\text{x}^2}}$
$=\sqrt{\text{a}^2-\text{x}^2}$
$=\text{R.H.S}$
$\text{L.H.S}=\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\text{x}\frac{\text{d}}{\text{dx}}\sqrt{\text{a}^2-\text{x}^2}+\sqrt{\text{a}^2-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\frac{\text{a}^2}{2}\times\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\text{x}}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$
[Using product rule, chain rule]
$=\frac{1}{2}\bigg[\text{x}\times\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)+\sqrt{\text{a}^2-\text{x}^2}\Big] \\ +\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2}}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\frac{\text{x}(-2\text{x})}{2\sqrt{\text{a}^2-\text{x}^2}}+\sqrt{\text{a}^2-\text{x}^2}\Big]+\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\bigg[\frac{-2\text{x}^2+2\big(\text{a}^2-\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{1}{2}\bigg[\frac{2\big(\text{a}^2-2\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2+\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\big(\text{a}^2-\text{a}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{(\text{a}^2-\text{x}^2)}{\sqrt{\text{a}^2-\text{x}^2}}$
$=\sqrt{\text{a}^2-\text{x}^2}$
$=\text{R.H.S}$
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