d/dx(tan^(-1)2x)=2/(4x^2+1). Why is there a 2 in the numerator?

d/dx(tan^(-1)2x) = 2/(4x^2+1). Why is there a $2$ in the numerator? I know how to get to the denominator but not the numerator. Can anyone just briefly explain? Nothing lengthy needed. Thanks!

Dec 18, 2017

The factor of $2$ occurs as a results of applying the chain rule.

We have a known result:

$\frac{d}{\mathrm{dx}} {\tan}^{- 1} x = \frac{1}{1 + {x}^{2}}$

Then applying the chain rule we have

$\frac{d}{\mathrm{dx}} {\tan}^{- 1} 2 x = \frac{1}{1 + {\left(2 x\right)}^{2}} \frac{d}{\mathrm{dx}} \left(2 x\right)$
$\text{ } = \frac{2}{1 + 4 {x}^{2}}$

If that was unclear we can perform an explicit substitution, viz:

Let $\left\{\begin{matrix}y = {\tan}^{- 1} 2 x \\ u = 2 x\end{matrix}\right. \implies \left\{\begin{matrix}y = {\tan}^{- 1} u & \null \\ \frac{\mathrm{du}}{\mathrm{dx}} = 2 & \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{1 + {u}^{2}}\end{matrix}\right.$

Then by the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \setminus \frac{\mathrm{du}}{\mathrm{dx}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{1 + {u}^{2}} \setminus 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{2}{1 + {u}^{2}}$

And restoring the substitution we get the result:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{1 + {\left(2 u\right)}^{2}} = \frac{2}{1 + 4 {u}^{2}}$