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Differentiate each of the following w.r.t $x$: $tan^{-1}\large\frac{3a^2x-x^3}{a^3-3ax^2}$, $\frac{1}{\sqrt 3}< \frac{x}{a} <\frac{-1}{\sqrt 3}$

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Toolbox:
  • $\tan 3\theta=\large\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$
  • $\large\frac{d}{dx}$$(\tan x)=\large\frac{1}{1+x^2}$
Step 1:
Given :$y=\tan^{-1}\big(\large\frac{3a^2x-x^3}{a^3-3ax^2}\bigg)$
Put $x=a\tan\theta\Rightarrow \theta=\tan^{-1}\big(\large\frac{x}{a}\big)$
Then $\large\frac{3a^2a.\tan\theta-a^3\tan^3\theta}{a^3-3a^3\tan^2\theta}=\large\frac{3a^3\tan\theta-a^3\tan^3\theta}{a^3-3a^3\tan^2\theta}$
$\qquad\qquad\qquad\qquad\quad\;\;\;=\large\frac{a^3(3\tan\theta-\tan 3\theta)}{a^3(1-3\tan^2\theta)}$
But $\large\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=$$\tan 3\theta$
$\Rightarrow y=\tan^{-1}\big(\tan 3\theta\big)$
$y=3\theta$
Step 2:
Substituting for $\theta$,we get
$y=3\bigg(\tan^{-1}\big(\large\frac{x}{a}\big)\bigg)$
$\large\frac{dy}{dx}$$=3\bigg[\large\frac{1}{1+(\large\frac{x}{a})^2}\bigg]$
$\;\;\;\;=\large\frac{3a^2}{a^2+x^2}$
answered Jul 1, 2013 by sreemathi.v
 

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