Find dy dx if y= ^ -1 (1 2 x^ 2 -1 ), 0<x<1 2 - Vidyadip

Question

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

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Answer

It is given that $y=\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$
$\Rightarrow \sec y=\frac{1}{2 x^{2}+1}$
$\Rightarrow$ cos y = 2x² + 1
$\Rightarrow$ 2x² = 1 + cos y
$\Rightarrow$ 2x² = 2cos² $\frac{y}{2}$
$\Rightarrow$ x = cos $\frac{y}{2}$
Differentiating w.r.t x, we get
$\frac{d}{d x}(x)=\frac{d}{d x}\left(\cos \frac{y}{2}\right)$
$\Rightarrow 1=-\sin \frac{\mathrm{y}}{2} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{2}\right)$
$\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos ^{2} \frac{y}{2}}}$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}$

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