Derivative of arctan...

**fleabass123** · February 28th 2010, 03:13 PM

I'm currently working this problem

pretty unsuccessfully. I tried to use the derivative of arctan and the general power rule and as a result I got:

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

**ANDS!** · February 28th 2010, 03:30 PM

You should have three segments:

$\frac{1}{1+(4x^2-1)} \cdot \frac{1}{2(\sqrt{4^2-1})} \cdot 8x$

**Keithfert488** · February 28th 2010, 03:30 PM

Using the chain rule, and the derivative of arctan, we'll get

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

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

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

So you were right =)

**skeeter** · February 28th 2010, 03:33 PM

Originally Posted by fleabass123

I'm currently working this problem

pretty unsuccessfully. I tried to use the derivative of arctan and the general power rule and as a result I got:

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

actually, what you arrived at is correct.

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

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

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

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

**fleabass123** · February 28th 2010, 04:14 PM

I see. Thanks for the help, everybody!

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