Find, if


Find $\frac{d y}{d x}$, if $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}},-1 \leq x \leq 1$


It is given that, $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$

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

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

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

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 x \sqrt{1-x^{2}}}(-2 x)$

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

$\therefore \frac{d y}{d x}=0$


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