If y=\arctan(\frac{2x−1}{1+x−x^2}), then \frac{dy}{dx} at x=1 is equal to? I have

A computationally simpler method of directly calculating the derivative at x=1 is to write
${\displaystyle \tan y=\frac{2x-1}{1+x-x^2}}$
so that
${\displaystyle \log \tan y=\log (2x-1)-\log (1+x-x^2)}$
Then implicit differentiation yields
${\displaystyle \frac{{\sec }^{2}y}{\tan y}\frac{dy}{dx}=\frac{2}{2x-1}-\frac{1-2x}{1+x-x^2}}$
Since, when x=1, we have ${\displaystyle \tan y=\frac{2-1}{1+1-1}=1}$, it follows from the trigonometric identity ${\displaystyle {\sec }^{2}y=1+{\tan }^{2}y\text{ that }{\sec }^{2}y=1+1^2=2}$; therefore
${\displaystyle 2{\left[\frac{dy}{dx}\right]}_{x=1}=\frac{2}{2-1}-\frac{1-2}{1+1-1}=3}$
hence the answer is 3/2. Note this solution uses two tactics; logarithmic differentiation and implicit differentiation, to make the calculation extremely simple.