Differentiate the function $\sec^{-1} \left( \frac{1}{4x^3 - 3x} \right), \quad 0 < x < \frac{1}{\sqrt{2}}$ with respect to $x$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{-3}{\sqrt{1 - x^2}}$ ##
Let $y = \sec^{-1} \left( \frac{1}{4x^3 - 3x} \right) \quad \dots(i)$
On putting $x = \cos \theta$ in Eq. (i), we get
$y = \sec^{-1} \left( \frac{1}{4\cos^3 \theta - 3\cos \theta} \right)$
$= \sec^{-1} \left( \frac{1}{\cos 3\theta} \right) \quad [∵\cos 3\theta = 4\cos^3 \theta - 3\cos \theta]$
$= \sec^{-1} (\sec 3\theta) = 3\theta \quad \left[ ∵\sec \theta = \frac{1}{\cos \theta} \right]$
$= 3 \cos^{-1} x \quad [∵\theta = \cos^{-1} x]$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = \frac{d}{dx} (3 \cos^{-1} x)$
$= 3 \cdot \frac{-1}{\sqrt{1 - x^2}}$