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Differentiate the following w.r.t $x$ : $\sec^{-1}\bigg(\Large \frac{1}{4x^3-3x}\bigg),\normalsize 0 < x <\Large \frac{1}{\sqrt 2}$

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

1 Answer

Toolbox:
  • $4\cos^3\theta-3\cos\theta=\cos 3\theta$
  • $\large\frac{dy}{dx}$$\big(\cos ^{-1}x)=\large\frac{-1}{\sqrt{1-x^2}}$
Step 1:
Given : $y=\sec^{-1}\bigg[\large\frac{1}{4x^3-3x}\bigg]$$,0 < x <\large\frac{1}{\sqrt 2}$
Put $x=\cos \theta\Rightarrow \theta=\cos^{-1}x$
Then $4x^3-3x=4\cos^3\theta-3\cos\theta$
$\qquad\qquad\quad\;\;\;=\cos 3\theta$
Therefore $y=\sec^{-1}\big[\large\frac{1}{\cos 3\theta}\big]$
But $\large\frac{1}{\cos 3\theta}=$$\sec 3\theta$
Therefore $y=\sec^{-1}(\sec 3\theta)$
$\Rightarrow 3\theta$
Step 2:
Substituting for $\theta$ we get,
$y=3\cos^{-1}x$
$\large\frac{dy}{dx}$$=3\big(\large\frac{-1}{\sqrt{1-x^2}}\big)$
$\quad\;=\large\frac{-3}{\sqrt{1-x^2}}$
answered Jul 1, 2013 by sreemathi.v
 

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