Question

Calculate all four second-order partial derivatives of f(x,y)=sin(2xy).

** No work needed ** Just results
fxx(x,y)=

fxy(x,y)=

fyx(x,y)=

fyy(x,y)= Calculate all four second-order partial derivatives of f(x,y)=sin(2xy).

** No work needed ** Just results
fxx(x,y)=

fxy(x,y)=

fyx(x,y)=

fyy(x,y)= @Mathematics

Answer 1

how do you expect to get results without working them thru tho?

Answer 2

fx = 2y cos(2xy)

fxx = -2y 2y sin(2xy) = -4y^2 sin(2xy)

Answer 3

fx = 2y cos(2xy)

fxy = 2y' cos(2xy)+2y cos(2xy)'
= 2 cos(2xy)- 2y 2x sin(2xy)
= 2 cos(2xy)- 4xy sin(2xy)

Answer 4

fx means ys act like constants

fy means xs act like constants

Answer 5

the rest of the rules that you are used to still apply; it might help to rewrite the "constant" one as a B or C or some such to stimulate you out of habit

Answer 6

Yeah keeping the constant is where i struggle

Answer 7

fy = sin(2Cy).
fy = 2C cos(2Cy)

fy = 2x cos(2xy)

Answer 8

fyx = 2x cos(2xB)

fyx = -2x 2B sin(2xB)

fyx = -2x 2y sin(2xy)

Answer 9

fy = 2x cos(2xy)

fyy = 2C cos(2Cy)

fyy = -2C 2C sin(2Cy)

fyy = -2x 2x sin(2xy)

hopefullt i kept track of them

Answer 10

fyx is wrong, i forgot to product rule it out

Answer 11

fyx = 2x cos(2xB)

fyx = 2x' cos(2xB) +2x cos(2xB)'

fyx = 2 cos(2xB) -2x 2B sin(2xB)

fyx = 2 cos(2xB) -2x 2y sin(2xy)

maybe

Answer 12

i tend to struggle with it more when I try to cover the variables; just aint used to it :) but many find it useful

Answer 13

Alright awesome, This helped alot!