If $f(x)=\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$, then $f^{\prime}(x)=$

If $f(x)=\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$, then $f^{\prime}(x)=$
  1. $\frac{-1}{2 \sqrt{1-x^{2}}}$
  2. $\frac{1}{\sqrt{1-x^{2}}}$
  3. $\frac{-1}{2 \sqrt{1+x^{2}}}$
  4. $\frac{1}{2 \sqrt{1+x^{2}}}$

Solution

$\frac{d}{d x}\left[\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right)\right]=\frac{1}{\sqrt{1-\left(\sqrt{\frac{1-x}{2}}\right)^{2}}} \cdot \frac{d}{d x}\left(\sqrt{\frac{1-x}{2}}\right)$ $=\frac{1}{\sqrt{1-\frac{1-x}{2}}} \times \frac{1}{2 \sqrt{\frac{1-x}{2}}} \times \frac{d}{d x}\left(\frac{1-x}{2}\right)=\frac{1}{\sqrt{\frac{2-1+x}{2}}} \times \frac{1}{\sqrt{2} \cdot \sqrt{1-x}} \times \frac{-1}{2}$ $=\frac{\sqrt{2}}{\sqrt{1+x}} \times \frac{1}{\sqrt{2} \cdot \sqrt{1-x}} \times \frac{-1}{2}=\frac{-1}{2 \sqrt{1-x^{2}}}$ This problem can also be solved as follows $\begin{array}{l} f(x)=\sin ^{-1}\left(\sqrt{\frac{1-x}{2}}\right) \\ \text { Put } x=\cos \theta \Rightarrow \sqrt{\frac{1-x}{2}}=\sqrt{\frac{1-\cos \theta}{2}}=\sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2}}=\sin \frac{\theta}{2} \\ \therefore f(x)=\sin ^{-1}\left(\sin \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{\cos ^{-1} x}{2} \\ \therefore f^{\prime}(x)=\frac{-1}{2 \sqrt{1-x^{2}}} \end{array}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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