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# Let $y= \tan^{-1}\bigg[\Large\frac{4x}{1+5x^2}\bigg] \normalsize +\tan^{-1}\bigg[\Large\frac{2+3x}{3-2x}\bigg],$ show that $\Large\frac{dy}{dy}=\frac{5}{1+25x^2}$

Can you answer this question?

Toolbox:
• Write $\large \frac{4x}{1+5x^2}=\large\frac{5x-x}{1+5x^2}, \large\frac{2+3x}{3-2x}=\large\frac{\large\frac{2}{3}+x}{1-\large\frac{2}{3}x}$
• $tan^{-1}\large\frac{x-y}{1+xy}=tan^{-1}x-tan^{-1}y$
$y = tan^{-1}5x-tan^{-1}x+tan^{-1}\large\frac{2}{3}+tan^{-1}x$
differentiate to get the answer.

answered Mar 9, 2013
edited Mar 25, 2013