# How do you differentiate f(x)=arctan(1/(2x-1)) using the chain rule?

##### 1 Answer
Nov 23, 2015

$f ' \left(x\right) = - \frac{1}{2 {x}^{2} - 2 x + 1}$

#### Explanation:

According the Chain Rule, $\frac{d}{\mathrm{dx}} \left[\arctan \left(u\right)\right] = \frac{u '}{1 + {u}^{2}}$

If you want explanation of how to find this, please ask.

Therefore, f'(x)=overbrace(d/dx[1/(2x-1)])^(color(blue)("write as " (2x-1)^(-1)))/(1+(1/(2x-1))^2.

Simplify and get: $f ' \left(x\right) = - \frac{1}{2 {x}^{2} - 2 x + 1}$