If $f(x) = \sqrt{\cos^2 x - 25}$, the $f'(x) = \frac{1}{2\sqrt{\cos^2 x - 25}} \cdot g(x)$. Find $g(x)$.

$-2 \sin x \cos x$

The correct answer is Option (2) → $-2 \sin x \cos x$ ##

Given:

$f(x) = \sqrt{\cos^2 x - 25}$

We are provided with the expression for $f'(x)$:

$f'(x) = \frac{1}{2\sqrt{\cos^2 x - 25}} \cdot g(x)$

Here, $f(x) = (\cos^2 x - 25)^{\frac{1}{2}}$

$f'(x) = \frac{1}{2}(\cos^2 x - 25)^{-\frac{1}{2}} \cdot \frac{d}{dx}(\cos^2 x - 25)$

$f'(x) = \frac{1}{2\sqrt{\cos^2 x - 25}} \cdot \frac{d}{dx}(\cos^2 x - 25)$

Differentiate $\cos^2 x - 25$:

$\frac{d}{dx}(\cos^2 x - 25) = 2\cos x \cdot (-\sin x) = -2\cos x \sin x$

Express $g(x)$:

Since, $f'(x) = \frac{1}{2\sqrt{\cos^2 x - 25}} \cdot g(x)$,

We have:

$g(x) = -2\cos x \sin x$