# How do you take the derivative of x = tan (x+y)?

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{2} / \left(1 + {x}^{2}\right)$
I refer to http://socratic.org/questions/how-do-you-find-the-derivative-of-tan-x-y-x-1?answerSuccess=1, where we have found that given $x = \tan \left(x - u\right)$; $\frac{\mathrm{du}}{\mathrm{dx}} = {x}^{2} / \left(1 + {x}^{2}\right)$ (I have replaced $y$ by $u$ for convenience). This means that if we substitute $u$ by $- y$, we find that for $x = \tan \left(x + y\right)$; $- \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} / \left(1 + {x}^{2}\right)$, so $\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{2} / \left(1 + {x}^{2}\right)$.