Question

2nd Partial Derivative: Can someone please break down the finding of the 2nd partial for this first function u1 for me? see attached.

Answer 1

http://i42.tinypic.com/2psr33r.jpg

Answer 2

PARTIALS HAVE CERTAIN ORDERS.... sorry, pinky hit the caps

Answer 3

what are we trying to do with u1?

Answer 4

find fxx and find fyy of the function u1

Answer 5

\[ln(\sqrt{x^2+y^2})=\frac{1}{2}ln(x^2+y^2)\]if that helps us start it

Answer 6

nice, my computer froze :/

Answer 7

to derive this with respect to x; consider the y parts as constants

Answer 8

spose we had:

ln(x^2+6) how would you take the derivative wrt x?

Answer 9

2x/(x^2 + 6) the derivative of the inside divided by the original inside

Answer 10

I've done partials before and I understand that if we are taking the derivative wrt x, then we treat y as a constant as we go. This one though, is confusing to me.

Answer 11

how is it confusing perse.

Answer 12

fxx would be the derivative of that one

Answer 13

i always try to do the product rule instead of the quotient rule; just simpler to perform in my view

Answer 14

2x/(x^2 + 6) = 2x (x^2 + 6)^(-1)

fxx = 2x' (x^2 + 6)^(-1) + 2x (x^2 + 6)'^(-1)


fxx = 2 (x^2 + 6)^(-1) - 2x (x^2 + 6)^(-2)

Answer 15

change the 6s back to ys :)

Answer 16

since there is no real difference between your xs and ys in this case; the fy and fyy is going to look exactly the same

Answer 17

except for a y up top :)

Answer 18

okay, I see. It really is better to use the product rule instead I see. You substituted the constant, 6, in place of the y's to make it simpler, then back substituted them in afer you were done.

Answer 19

right; i just used that 6 as a placeholder for a constant so that I wasnt tempted to see it as something it wasnt

Answer 20

that definitely helps...seeing the y there does lend itself to me making errors.

Answer 21

some suggest renaming the dummy variable as another letter; but even then I tend to want to use it :) so using a dummy constant is best for me ;)

Answer 22

my answer key says that both of these functions are solutions to the diff eqn. I'm able to show it for the 2nd function, u2, but I'm still struggling with showing u1 is a solution.....it got really messy and my sum of both 2nd partials for u1 did not equal zero. I will keep trying. Thanks

Answer 23

u1 = (1/2) ln (x² + y²)
u1 (x) = 2x / (x² + y²)
u1 (xx) = [2 (x² + y²) - 4x² ] / (x² + y²)²
= ( -2x² + 2y² ) / (x² + y²)²

u1 ( y) = 2y/ (x² + y²)
u1( yy) = [2 (x² + y²) - 4y² ] / (x² + y²)²
= ( 2x² - 2y² ) / (x² + y²)²
=> u1 (xx) + u1( yy) = 0

Answer 24

You have no problem with U2, so I just help with U1 :)

Answer 25

Thank you! I was able to compute u2 just fine. It looks like I need to brush up on my differentiation of complex log functions. I will follow the steps that you've listed here ....and practice it several times.