Question

Let f(x,y)=(2x−y)^5 Find=

∂2f∂x∂y =
∂3f∂x∂y∂x =
∂3f∂x2∂y = Let f(x,y)=(2x−y)^5 Find=

∂2f∂x∂y =
∂3f∂x∂y∂x =
∂3f∂x2∂y = @Mathematics

Answer 1

wow, they really want you to hunt these dont they

Answer 2

d/dxdy i believe is forst dx then dy; then to 2 it you go thru it again

Answer 3

f(x,y)=(2x−y)^5
fx(x,y)=5(2x−y)^4 * 2 = 10(2x-y)^4
fxy(x,y)=40(2x−y)^3 *-1 = -40(2x-y)^3

that should be one round

Answer 4

Thats correct, thanks!

Answer 5

I do not understand
how to do the next one though, could you list steps for that as well?

Answer 6

the steps are to go thru it each time from the last results, if that makes sense

Answer 7

in other words; if the first step is to Fx, then we find Fx(y) from it

Answer 8

yeah but odnt I use different numbers when taking fx(Y)

Answer 9

d3f/(dxdydx) this one might mean to find dx, then dy, then dx; 3 times

Answer 10

in Fsy you "y" it from the Fx that you determined

Answer 11

in Fxy .. that is

Answer 12

so I just derive the answer for f(x)
?

Answer 13

yes, Fxy would mean:

derive Fx, then from Fx, derive a "y" to get Fxy

Answer 14

partials mean we consider any variable we are not respecting, as a constant

Answer 15

Fx means we consider all ys to be constants

Answer 16

-240(2x-y)^2 is the answer, you have helped a bunch...thanks!

Answer 17

F(x,y)=(2x−y)^5

\[F_x = 5(2x-y)^4*(2x-y)^{'x}\]
\[F_x = 5(2x-y)^4*2\]

whew!! good :)

Answer 18

\[F_x = 10(2x-y)^4\]
\[F_{xy} = 10*4(2x-y)^3\ (2x-y)^{'y}\]
\[F_{xy} = 10*4(2x-y)^3\ (-1)\]
\[F_{xy} = -40(2x-y)^3\]
\[F_{xyx} = -40*3(2x-y)^2\ (2x-y)^{'x}\]
\[F_{xyx} = -40*3(2x-y)^2\ (2)\]
\[F_{xyx} = -240(2x-y)^2 \]