The value of $\frac{d}{dx}\left\lbrack \left( \frac{ an^{2}{2x - an^{2}x}}{1 - an^{2}{2x an^{2}x}} ight)\cot{3x} ight brack$is

Question

The value of $\frac{d}{dx}\left\lbrack \left( \frac{	an^{2}{2x - 	an^{2}x}}{1 - 	an^{2}{2x	an^{2}x}} ight)\cot{3x} ightbrack$is

Dr. Abubacker · Accepted Answer

(c)
Let $y = \frac{	an^{2}{2x - 	an^{2}x}}{1 - 	an^{2}{2x\ tan^{2}x}}$
$= \frac{(	an{2x - 	an{x)}}}{(1 + 	an{2x	an{x)}}}.\frac{(	an{2x + 	an x})}{(1 - 	an{2x	an x)}}$
$= 	an{(2x - x)	an{(2x + x)}}$
$= 	an{x	an\ 3x}$
On differentiating w.r.t. $x,\$we get
$\frac{d}{dx}(y.\cot{3x) = \frac{d}{dx}(	an{x	an{3x\cot{3x)}}}}$
$= \frac{d}{dx}(	an{x)}$
$= \sec^{2}x$

Q. The value of $\frac{d}{dx}\left\lbrack \left( \frac{\tan^{2}{2x - \tan^{2}x}}{1 - \tan^{2}{2x\tan^{2}x}} \right)\cot{3x} \right\rbrack$is

Explanation

Reviewed by Dr. Abubacker

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