My Math Forum Algebra issues with derivative

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 October 21st, 2014, 03:12 PM #1 Newbie   Joined: Mar 2013 Posts: 24 Thanks: 0 Algebra issues with derivative I keep finding this in my solutions book for this problem, and even wolfram alpha spits this out as the answer to this derivative using substitution to derive it for reasons I can't comprehend, when it doesn't appear to be necessary. $\frac{d}{dx}(tan^{-1}(ax)) = \frac{a}{a^2x^2 + 1}$ However, shouldn't I also be able to write this derivative as $\frac{d}{dx}(tan^{-1}(ax)) = \frac{a}{x^2 + \frac{1}{a^2}}$? and if not, why is the second way incorrect? What would be a step-by-step approach without using this seemingly random substitution? Thanks in advance for the help!
 October 21st, 2014, 03:29 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 $\displaystyle \frac{d}{dx} \arctan(u) = \frac{u'}{1+u^2}$ $\displaystyle u = ax$ $\displaystyle \frac{d}{dx} \arctan(ax) = \frac{a}{1+(ax)^2}$ fyi ... $\displaystyle \frac{a}{x^2 + \frac{1}{a^2}} = \frac{a^3}{a^2x^2 + 1}$
 October 21st, 2014, 03:51 PM #3 Newbie   Joined: Mar 2013 Posts: 24 Thanks: 0 Okay, don't know where I got the incorrect method from. Thanks again!

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