Local extremes of a trig/ln function

I am having trouble finding the local extremes for a particular function.

f(xy)=e^(2x)cosy

I have so far:

f'(x)= 2e^(2x)cosy

f'(y)=-e^(2x)cosy

f''(x)=4e^(2x)cosy

f''(y)=-e^(2x)cosy

f''(xy)=-2e^(2x)siny

Now I'm setting both f'(x) and f'(y) = 0, but this is where I'm having trouble. Does that mean that 2e^(2x)=0, cosy=0, -e^(2X)=0, and siny=0?

Thanks for any help that can be provided.

2 Attachment(s)

Re: Local extremes of a trig/ln function

f(x,y) as a surface:

as seen f(x,y)=0 can be considered as extreme value.

http://mathhelpforum.com/attachment....1&d=1350153665

Local extremes perhaps can have meaning for x=constant:

http://mathhelpforum.com/attachment....1&d=1350156578(Nerd)

Re: Local extremes of a trig/ln function

Thanks for the reply. I'm trying to understand what you mean. So the extremes exist @ x=n(pi)/2?

1 Attachment(s)

Re: Local extremes of a trig/ln function

http://mathhelpforum.com/attachment....1&d=1350155212

As seen on the contours of f(x,y), there is no locality where f'x, f'y can be 0 at the same time ...except at x=-inf where the plane of f(x,y) is just a flat plane at z=0(Shake)

Re: Local extremes of a trig/ln function

Hello, MajMinor!

We have:

. .

. .

. .

. .

. .

Since is always positive, is always negative.

. . Hence, all extremes are *saddle points.*

Re: Local extremes of a trig/ln function

Got it. I understand now. Thanks, guys!