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October 21st, 2014, 03:12 PM   #1
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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!
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October 21st, 2014, 03:29 PM   #2
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$\displaystyle \frac{d}{dx} \arctan(u) = \frac{u'}{1+u^2}$

$\displaystyle u = ax$

$\displaystyle \frac{d}{dx} \arctan(ax) = \frac{a}{1+(ax)^2}$

fyi ...

$\displaystyle \frac{a}{x^2 + \frac{1}{a^2}} = \frac{a^3}{a^2x^2 + 1}$
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October 21st, 2014, 03:51 PM   #3
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Okay, don't know where I got the incorrect method from. Thanks again!
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