# Derivative of arctan...

• Feb 28th 2010, 03:13 PM
fleabass123
Derivative of arctan...
I'm currently working this problem
https://webwork.math.lsu.edu/webwork...d407a3c4c1.png

pretty unsuccessfully. I tried to use the derivative of arctan and the general power rule and as a result I got: https://webwork.math.lsu.edu/webwork...cb2ccf7b71.png

But this is wrong. Any thoughts? Thanks to anybody that helps. :D

• Feb 28th 2010, 03:30 PM
ANDS!
You should have three segments:

$\displaystyle \frac{1}{1+(4x^2-1)} \cdot \frac{1}{2(\sqrt{4^2-1})} \cdot 8x$
• Feb 28th 2010, 03:30 PM
Keithfert488
Using the chain rule, and the derivative of arctan, we'll get

$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} (\arctan u)=\frac{{u}'}{1+u^2}$

With $\displaystyle u=\sqrt{4x^2-1}$

$\displaystyle \frac{\frac{4x}{\sqrt{4x^2-1}}}{1+4x^2-1}=\frac{4x}{4x^2\cdot \sqrt{4x^2-1}}=\frac{1}{x\cdot \sqrt{4x^2-1}}$

So you were right =)
• Feb 28th 2010, 03:33 PM
skeeter
Quote:

Originally Posted by fleabass123
I'm currently working this problem
https://webwork.math.lsu.edu/webwork...d407a3c4c1.png

pretty unsuccessfully. I tried to use the derivative of arctan and the general power rule and as a result I got: https://webwork.math.lsu.edu/webwork...cb2ccf7b71.png

But this is wrong. Any thoughts? Thanks to anybody that helps. :D

actually, what you arrived at is correct.

$\displaystyle u = \sqrt{4x^2-1}$

$\displaystyle u' = \frac{4x}{\sqrt{4x^2-1}}$

$\displaystyle \frac{d}{dx} \tan^{-1}(u) = \frac{u'}{1+u^2}$

$\displaystyle \frac{4x}{\sqrt{4x^2-1}} \cdot \frac{1}{1 + (4x^2-1)} = \frac{1}{x\sqrt{4x^2-1}}$
• Feb 28th 2010, 04:14 PM
fleabass123
I see. Thanks for the help, everybody!