# What is the derivative of x = tan (x+y)?

Apr 9, 2015

Consider $y$ as a function of $x$ and get (deriving both sides with respect to $x$):

$1 = \left[\frac{1}{{\cos}^{2} \left(x + y\right)}\right] \cdot \left[1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right]$ (Chain Rule for $\tan$)

Rearranging:

$1 + \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(x + y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(x + y\right) - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(1 - {\cos}^{2} \left(x + y\right)\right) = - {\sin}^{2} \left(x + y\right)$
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