# What are the critical points of  f(x,y)=sin(x)cos(y) +e^xtan(y)?

Mar 20, 2018

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

#### Explanation:

We are given $f \left(x , y\right) = \sin \left(x\right) \cos \left(y\right) + {e}^{x} \tan \left(y\right)$

Critical points occur when $\frac{\partial f \left(x , y\right)}{\partial x} = 0$ and $\frac{\partial f \left(x , y\right)}{\partial y} = 0$

$\frac{\partial f \left(x , y\right)}{\partial x} = \cos \left(x\right) \cos \left(y\right) + {e}^{x} \tan \left(y\right)$

$\frac{\partial f \left(x , y\right)}{\partial y} = - \sin \left(x\right) \sin \left(y\right) + {e}^{x} {\sec}^{2} \left(y\right)$

$\sin \left(y\right) \sin \left(x\right) + \cos \left(y\right) \cos \left(x\right) + {e}^{x} \tan \left(y\right) - {e}^{x} {\sec}^{2} \left(y\right) = \cos \left(x - y\right) + {e}^{x} \left(\tan \left(y\right) - {\sec}^{2} \left(y\right)\right) = \cos \left(x - y\right) + {e}^{x} \left(\tan \left(y\right) - \left(1 + {\tan}^{2} \left(y\right)\right)\right) = \cos \left(x - y\right) + {e}^{x} \left(- {\tan}^{2} \left(y\right) + \tan \left(y\right) - 1\right)$

There is no real way to find solutions, but critical points occur when $\cos \left(x - y\right) + {e}^{x} \left(- {\tan}^{2} \left(y\right) + \tan \left(y\right) - 1\right) = 0$

A graph of solutions is here