If tan−1x+tan−1y=π2, then dydx at (2,3) is

$\Rightarrow \frac{1}{1 + x^{2}} + \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0$

$\Rightarrow \frac{1}{1 + y^{2}} \frac{d y}{d x} = \frac{- 1}{1 + x^{2}}$

$\Rightarrow \frac{d y}{d x} = \frac{- (1 + y^{2})}{1 + x^{2}}$

$\Rightarrow {(\frac{d y}{d x})}_{(2, 3)} = \frac{- (1 + 9)}{1 + 4}$

$= \frac{- 10}{5} = - 2$

If $\tan^{- 1} x + \tan^{- 1} y = \frac{π}{2}$ , then $\frac{d y}{d x}$ at $(2, 3)$ is