critcal points

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August 19th 2009, 03:01 AM
manalive04

critcal points

find all the critical points of the function

f(x,y) = (x-2y)e^(x^2-y^2)

and classify them as local maxima, local minima or saddle points.

How is this differentiated?
August 19th 2009, 03:17 AM
pickslides

Quote:

Originally Posted by manalive04

find all the critical points of the function

f(x,y) = (x-2y)e^(x^2-y^2)

and classify them as local maxima, local minima or saddle points.

How is this differentiated?

You need to find where $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial x}\right) = 0$

To find $\frac{\partial f}{\partial x}$ differentiate $f(x,y)$ with respect to x holding y constant

To find $\frac{\partial f}{\partial y}$ differentiate $f(x,y)$ with respect to y holding x constant
August 19th 2009, 03:32 AM
manalive04

Quote:

Originally Posted by pickslides

You need to find where $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial x}\right) = 0$

To find $\frac{\partial f}{\partial x}$ differentiate $f(x,y)$ with respect to x holding y constant

To find $\frac{\partial f}{\partial y}$ differentiate $f(x,y)$ with respect to y holding x constant

Can you please explain this further - how is it differentiated i cant work it out
August 19th 2009, 08:40 PM
pickslides

You need to use the product rule it is... $f = uv \Rightarrow f'=uv'+vu'$

Differentiating with respect to x (finding $\frac{\partial f}{\partial x}$ ) make

$u = x-2y \Rightarrow u' = 1$

$v = e^{x^2-y^2} \Rightarrow v' = 2x\times e^{x^2-y^2}$

Now apply $f = uv \Rightarrow f'=uv'+vu'$

After this you need to differentiate for y. (finding $\frac{\partial f}{\partial y}$ )

Then solve for $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial x}\right) = 0$