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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?

Thanks in advance.

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

2 Answers 2

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

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 $k,l,m,n \in \mathbb Z$

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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.

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  • $\begingroup$ 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. $\endgroup$
    – K. Talha
    May 13, 2017 at 20:17
  • $\begingroup$ You need to consider some cases. First let cos(x) = 0 implies sin(y) = 0, this gives a critical points. $\endgroup$
    – user392395
    May 13, 2017 at 20:26
  • $\begingroup$ Next take cos(y) = 0 implies sin(x)=0 which gives same critical points but reversed in x and y. $\endgroup$
    – user392395
    May 13, 2017 at 20:27

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