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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}},$ prove that $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}$
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Answer

Givne, $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\bigg[\frac{\sqrt{1-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x}\sin^{-1}\text{x})-(\text{x}\sin^{-1}\text{x})\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})}{(\sqrt{1-\text{x}^2})^2}\bigg]$
[Using quotient rule, product rule, chain rule]
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}-\big(\text{x}\sin^{-1}\text{x}\big)\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(1-\text{x}^2\big)}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big\}-\frac{\text{x}\sin{-1}\text{x}(-2\text{x})}{2\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\text{x}+\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\sqrt{1-\text{x}^2}\sin^{-1}}{1}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{(1-\text{x}^2)\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}-\text{x}^2\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}\ \Big\{\text{Since, given y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big\}$

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