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Question:

Find $\frac{d y}{d x}$

$y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0<x<\frac{1}{\sqrt{2}}$

Solution:

The given relationship is $y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)$

$y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)$

$\Rightarrow \sec y=\frac{1}{2 x^{2}-1}$

$\Rightarrow \cos y=2 x^{2}-1$

$\Rightarrow 2 x^{2}=1+\cos y$

$\Rightarrow 2 x^{2}=2 \cos ^{2} \frac{y}{2}$

$\Rightarrow x=\cos \frac{y}{2}$

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}(x)=\frac{d}{d x}\left(\cos \frac{y}{2}\right)$

$\Rightarrow 1=-\sin \frac{y}{2} \cdot \frac{d}{d x}\left(\frac{y}{2}\right)$

$\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos ^{2} \frac{y}{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}$

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