# How do you implicitly differentiate -1=xy-tan(x+2y) ?

Jun 17, 2017

Differentiate both sides of the equation with respect to $x$:

$0 = y + x y ' - \frac{1 + 2 y '}{\cos} ^ 2 \left(x + 2 y\right)$

solve for $y '$

$0 = \left(y + x y '\right) {\cos}^{2} \left(x + 2 y\right) - \left(1 + 2 y '\right)$

$y ' \left(2 - x {\cos}^{2} \left(x + 2 y\right)\right) = y {\cos}^{2} \left(x + 2 y\right) - 1$

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