My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

LinkBack Thread Tools Display Modes
October 21st, 2014, 03:12 PM   #1
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!
Polaris84 is offline  
October 21st, 2014, 03:29 PM   #2
Math Team
skeeter's Avatar
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 ...

$\displaystyle \frac{a}{x^2 + \frac{1}{a^2}} = \frac{a^3}{a^2x^2 + 1}$
skeeter is online now  
October 21st, 2014, 03:51 PM   #3
Joined: Mar 2013

Posts: 24
Thanks: 0

Okay, don't know where I got the incorrect method from. Thanks again!
Polaris84 is offline  

  My Math Forum > College Math Forum > Calculus

algebra, derivative, issues

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Are there issues with this? CherryPi Calculus 10 April 21st, 2012 10:23 PM
Geometry Issues! Brightstar Geometry 3 January 19th, 2012 12:39 PM
ISSUES with circles... tinyone Calculus 5 September 11th, 2010 08:00 AM
Solvability issues raiseit Math Events 1 June 1st, 2010 01:38 AM

Copyright © 2019 My Math Forum. All rights reserved.