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Find \(\large \frac {dy}{dx} \) in the following: $ y = sec^{-1} \left(\large\frac{1}{2x^2 - 1}\right), 0 < x < \; { \frac{1}{\sqrt2} } \\ $

$\begin{array}{1 1} \large\frac{dy}{dx}=\frac{-2}{\sqrt{1+x^2}} \\\large\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}} \\ \large\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}} \\ \large\frac{dy}{dx}=\frac{2}{\sqrt{1-x^3}}\end{array} $

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Toolbox:
  • $2\cos^2\theta-1=\cos 2\theta$
Step 1:
Put $x=\cos\theta\Rightarrow \theta=\cos^{-1}x$
$y=\sec^{-1}\big(\large\frac{1}{2\cos^2\theta-1}\big)$
We know that $2\cos^2\theta-1=\cos 2\theta$
$y=\sec^{-1}\big(\large\frac{1}{\cos 2\theta}\big)$
Step 2:
$y=\sec^{-1}(\sec 2\theta)$
$\;\;=2\theta$
$\;\;=2\cos^{-1}x$
$\large\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$
answered May 8, 2013 by sreemathi.v
 

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