Question

Find dy/dx : y = sec^-1 (1/2x^2-1), 0< x <1/√2

Answer 1

Let $x=\cos \theta$. Use the identity,

$2\cos^2 \theta-1=\cos 2\theta$.

Now,

$y=\sec^{-1}(\frac{1}{2\cos^2 \theta-1}) =\sec^{-1}(\frac{1}{\cos 2\theta}) \\ y=\sec^{-1}(\sec 2\theta) \\ y= 2\theta$

Now,

$\frac{dy}{dx}= 2\frac{d \theta }{dx}\\ \frac{dy}{dx}= \frac{2}{\frac{d x }{d\theta }} \\ \frac{dy}{dx}= \frac{2}{\frac{d \cos \theta }{d\theta }} \\ \frac{dy}{dx}= \frac{2}{-\sin \theta} \\ \frac{dy}{dx}= -\frac{2}{\sqrt{1-x^2}}$