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 Calculus Calculus Math Forum

 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! Tags algebra, derivative, issues Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post CherryPi Calculus 10 April 21st, 2012 10:23 PM Brightstar Geometry 3 January 19th, 2012 12:39 PM tinyone Calculus 5 September 11th, 2010 08:00 AM raiseit Math Events 1 June 1st, 2010 01:38 AM

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