Question

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

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

Answer 1

The correct option is C

$\frac{1}{2\sqrt{x}\sqrt{1-x}}$

Finding the value of $\frac{dy}{dx}$:

Derivative of ${\sin }^{-1}x$is $\frac{1}{\sqrt{1-x^2}}$.

Here, $y={\sin }^{-1}\sqrt{x}$

Differentiating with respect to $x$ both sides:

$$\begin{gathered} \frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(\sqrt{x}\right)}^2}}\times \frac{1}{2\sqrt{x}} \\ =\frac{1}{2\sqrt{x}\sqrt{1-x}} \end{gathered}$$

Hence, option (C) is the correct answer.