# How do I find the derivative of y= x arctan (2x) - (ln (1+4x^2))/4?

Jan 28, 2015

You can use the Product Rule and Chain Rule to solve your problem.
1) Consider first $x \arctan \left(2 x\right)$; you have the product of $x$ and $\arctan \left(2 x\right)$ and the function of a function, $\arctan \left(2 x\right)$ which requires the chain rule (derive the $\arctan$ and then multiply by the derivative of the argument $2 x$).
You get:
$y ' = 1 \cdot \arctan \left(2 x\right) + x \cdot \frac{1}{1 + {\left(2 x\right)}^{2}} \cdot 2$
2) Secondly you have the $- \ln \frac{1 + 4 {x}^{2}}{4}$ bit. Here, again, you have the chain rule to deal with the $\ln$ of a function (the argument $1 + 4 {x}^{2}$):
You get:
$y ' = - \frac{1}{4} \cdot \frac{1}{1 + 4 {x}^{2}} \cdot 8 x = - \frac{2 x}{1 + {\left(2 x\right)}^{2}}$

Collecting the two parts you get:
$y ' = \arctan \left(2 x\right) + \frac{2 x}{1 + {\left(2 x\right)}^{2}} - \frac{2 x}{1 + {\left(2 x\right)}^{2}} = \arctan \left(2 x\right)$

Hope it helps