# What is the critical points of $f(x,y) = e^{\sin x\cos y}$?

I try to find local extreme values and saddle point(s) of the $f(x,y) = e^{\sin x\cos y}$.

But, when I take the partial derivative, I can't figure out to find critical points.

$$f(x,y) = e^{\sin x\cos y}$$ $$f_x(x,y) = \cos x\cos y\, e^{\sin x\cos y} = 0$$ $$f_y(x,y) = -\sin x\sin y\,e^{\sin x\cos y} = 0$$

How does that work?

• Latex hint: \sin and \cos produce more readable tex.
– JMJ
May 13, 2017 at 20:11
• Notice $e^y >0$ May 13, 2017 at 20:11

$$\cos x \cos y =0$$ $$\sin x \sin y =0$$

So $\cos x = \sin y =0$ or $\sin x = \cos y = 0$

So $x=\frac{\pi}{2} + n\pi$ and $y=m\pi$

Or $y=\frac{\pi}{2} + l \pi$ and $x=k\pi$

These are points $(\frac{\pi}{2} + n\pi, m\pi)$ and $(k\pi, \frac{\pi}{2} + l \pi)$ with $k,l,m,n \in \mathbb Z$

Hint: You need $\cos x \cos y=0$, and also $\sin x \sin y=0$. However, $\cos x$ and $\sin x$ are never zero at the same time.

• But if we give x to $\frac{\pi}{2}$ and y to $\pi$, the solutions of these derivatives are 0. This is critical point but I want to formulate it. May 13, 2017 at 20:17
• You need to consider some cases. First let cos(x) = 0 implies sin(y) = 0, this gives a critical points.
– user392395
May 13, 2017 at 20:26
• Next take cos(y) = 0 implies sin(x)=0 which gives same critical points but reversed in x and y.
– user392395
May 13, 2017 at 20:27