Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 225: 80

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

Let $u=x^2-x$. Then, $\frac{du}{dx}=2x-1$. We have $f(x)=\arctan(u)$. Using the chain rule, $f'(x)=\frac{df}{dx}$ $=\frac{df}{du}\cdot \frac{du}{dx}$ $=\frac{1}{1+u^2}\cdot ({2x-1})$ $=\frac{1}{1+(x^2-x)^2}\cdot (2x-1)$ $=\frac{2x-1}{1+(x^2-x)^2}$ Thus, $f'(x)=\frac{2x-1}{1+(x^2-x)^2}$