Question

Let f(x,y) = 8e^{2xy}. Evaluate the following partial derivatives at (1,-1). As an option, answers may be written using e and an exponent instead of entering large or small numerical values.
Find fxx, fxy and fyy

Answer 1

Okay, first we need to find the first "general" partial with respect to x, as well as respect to y. What would that be?

Answer 2

Is it fx=2y(8e^(2xy))?

Answer 3

Yes, that's correct for the first partial with respect to x. What is first partial with respect to y? Then you can take the second derivatives you need to and evaluate.

Answer 4

alright, so fy=2x(8e^2y)?

Answer 5

Right.. now if you take the partial with respect to x and derive it again, you find:
\[f_{xx}(x,y)\]
As well as deriving the partial with respect to why according to y again:
\[f_{yy}(x,y)\]

Answer 6

Would it be fx=fxx?

Answer 7

And fy=fyy?

Answer 8

No, \[f_x(x,y) = \frac{\delta}{\delta x}\]
\[f_{xx}(x,y) = \frac{\delta}{\delta x} \frac{\delta f}{\delta x} = \frac{\delta^2f}{\delta x^2} = \frac{\delta}{\delta x} \left( f_x(x,y)\right)\]

Answer 9

In other words, fxx(x,y) is analogous to your f''(x), except it has two variables instead of one(and one of them is "constant" when taking the derivative)

Answer 10

I only know of your final statement there

Answer 11

thanks to khan's academy

Answer 12

A first-order partial of a two variable function, f(x,y), is the derivative of one of those variables(say, x), assuming that the other variable remains constant(y). In comparison, the second-order partial(fxx(x,y) is the second derivative of one variable(x) assuming the other variable is constant(y).

Answer 13

uh

Answer 14

Wait wait You lost me now

Answer 15

I'm basically putting into my own words the definition of partial derivatives, both first and second derivatives. If you are confused, how would you describe it so I can see where you might be getting confused.

Answer 16

Alright, sos heres what i did so far

Answer 17

wrong one

Answer 18

this one

Answer 19

Those are all correct answers as far as I can tell. All that's left is the mixed partial:
\[f_{xy}(x,y)\]
Now, if everything is nice, this also happens to equal:
\[f_{yx}(x,y)\]
But we won't worry about it. Instead we can just take the partial we already have with respect to x:
\[f_x(x,y)\]
And take the partial of *that* with respect to y. As some extra practice, you can take the partial with respect to y and take the derivative of that with respect to x and see if you come up with the same answer both ways.

Answer 20

Ok. I understand. Thank you very much, and also to those who have taken a look at this question.

Answer 21

oops. what did i do wrong?

Answer 22

Your first step should look like:
\[ f_{xy}=2(8e^{2xy}) + 2y(8e^{2xy})(2x) \]

Answer 23

why is the end 2x?

Answer 24

\[f_x(x,y) = 2y8e^{2xy},
u = 2y, v = 8e^{2xy}, u_y = 2, v_y = (2x)8e^{2xy}\]\[ f_{xy}(x,y) = uv_y + vu_y = 2(8e^{2xy}) + 2y(8e^{2xy})(2x)\]

Answer 25

i opposite it was opposite variables? that what im interpreting ti as from my notes

Answer 26

What about fyy?

Answer 27

@heril ?

Answer 28

Oh

Answer 29

An arthimetic error. Sorry to disturb you...

Answer 30

I figured it out