Question

$Find \ \frac {dy}{dx}:$
$y=sec^{-1}(\frac {1}{2x^2-1}),\ 0<x< \frac {1}{\sqrt2}$

Answer 1

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

y = sec-1$(\frac {1}{2x^2-1})$
$⇒$sec y = $\frac {1}{2x^2-1}$
$⇒$cos y = 2x2-1
$⇒$2x2 = 1+cos y
$⇒$2x2 = 2cos2$\frac y2$
$⇒$x = cos $\frac y2$
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(x) = $\frac {d}{dx}$(cos$\frac y2$)
$⇒$1 = -sin$\frac y2$ . $\frac {d}{dx}$($\frac y2$)
$⇒$-$\frac {1}{sin\frac y2}$ = \frac 12$$\frac {dy}{dx}
⇒$$\frac {dy}{dx}x = -$\frac {2}{sin\frac y2}$
⇒$$\frac {dy}{dx}x= -$\frac {2}{\sqrt {1-cos^2 \frac y2}}$
⇒$$\frac {dy}{dx}= $\frac {-2}{\sqrt {1-x^2}}$