Question

If $y={\sin }^{-1}\sqrt{\left(1-x\right)}$, then $\frac{dy}{dx}=$

• A. $\frac{1}{\sqrt{\left(1-x\right)}}$
• B. $\frac{-1}{2\sqrt{\left(1-x\right)}}$
• C. $\frac{1}{\sqrt{x}}$
• D. $\frac{-1}{2\sqrt{x}\sqrt{\left(1-x\right)}}$

Answer 1

The correct option is D

$\frac{-1}{2\sqrt{x}\sqrt{\left(1-x\right)}}$

To find the value of $\frac{dy}{dx}$:

Given,

$y={\sin }^{-1}\sqrt{\left(1-x\right)}$

We know, $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$,

So,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{\sqrt{1-{\left(\sqrt{\left(1-x\right)}\right)}^2}}\left(\frac{1}{2\sqrt{\left(1-x\right)}}\right)\left(-1\right)\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\sqrt{\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{x}}}\mathbf{]}\\ & =& \frac{-1}{2\sqrt{x}\sqrt{\left(1-x\right)}}\end{array}$

Hence, the correct option is D.