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.
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 =)
actually, what you arrived at is correct.
$\displaystyle u = \sqrt{4x^2-1}$
$\displaystyle u' = \frac{4x}{\sqrt{4x^2-1}}$
$\displaystyle \frac{d}{dx} \tan^{-1}(u) = \frac{u'}{1+u^2}$
$\displaystyle \frac{4x}{\sqrt{4x^2-1}} \cdot \frac{1}{1 + (4x^2-1)} = \frac{1}{x\sqrt{4x^2-1}}$