Question
Find $\frac{\text{dy}}{\text{dx}},\ \text{if y}=\sin^{-1}\text{x}+\sin^{-1} \sqrt{1-\text{x}^2},\ -1\leq\text{x}\leq1$

Answer

It is given that, $\text{y}=\sin^{-1}\text{x}+\sin^{-1}\sqrt{1-\text{x}^2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\text{x}+\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})+\frac{\text{d}}{\text{dx}}(\sin^{-1}\sqrt{1-\text{x}^2})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\sqrt{1-(\sqrt{1-\text{x}^2})^2}}.\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\text{x}}.\frac{1}{2\sqrt{1-\text{x}^2}}.\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{2\text{x}\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{x}^2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=0$

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