If $f(x) = \cos^{-1} \sqrt{x}, 0 < x < 1$, which of the following is equal to $f'(x)$?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{-1}{2\sqrt{x(1-x)}}$ ##
$f(x) = \cos^{-1} \sqrt{x}$.
On differentiating, we get:
$f'(x) = \frac{-1}{\sqrt{1-(\sqrt{x})^2}} \frac{d}{dx}\sqrt{x}$
$= \frac{-1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}$
$= \frac{-1}{2\sqrt{x(1-x)}}$