Question

Let f(x,y)=(2x−y)^7. Then
∂^2f/∂x∂y =
∂^3f/∂x∂y∂x=
∂^3f/∂x^2∂y =

Answer 1

Im not positive of the notation, new stuff to me, but im guessing its saying partial with respect to x first then to y for the first one. Just to make sure Im getting what its asking.

Answer 2

\[ \frac{ ∂^{2}f }{ ∂x∂y} = ? \]
\[ \frac{ ∂^{3}f }{ ∂x∂y∂x} = ? \]
\[ \frac{ ∂^{3}f }{ ∂x^{2}∂y} = ? \]

Answer 3

@Psymon I think its asking the 2nd partial for the first one/

Answer 4

Yeah, 2nd parital, but with the first one being with respect to x and then the second being with respect to y. Of course I could be wrong, Either way, partial with respect to x first so we can get the idea.

So, if we differentiate with respect to x, we treat y as a constant. In fact, if it makes it less confusing to do, you could replace y with a constant. If I differentiated the function and pretended y was the number 1, I could do something liek this:

\[(2x-1)^{7} = 7(2x-1)^{6}(2) = 14(2x-y)^{6}\]

Not that making it a 1 is necessary, but maybe its a step that makes it less confusing. Now if I do the derivative of this result with respect to y, I pretend x is a constant. SO in the same manner, possibly to avoid confusion, I can replace 2x with 1 to prevnt it from confusing me

\[14(1-y)^{6} = 84(1-y)^{5}(-1) = -84(2x-y)^{5}\]

Answer 5

so for the next one we take −84(2x−y)^5 and do another partial in respect to x?

Answer 6

Right

Answer 7

so -840 (y-2 x)^4.
No what happens on the 3rd one?

Answer 8

now*

Answer 9

or is that the same question?

Answer 10

So the 3rd one we have to backtrack and instead of doing partial to y for the 2nd derivative we do it to x a second time. So the first partial was
\[14(2x-y)^{6}\]Now we just do partial again with respect to x
\[84(2x-y)^{5}(2) = 168(2x-y)^{5}\]Now finally we do the 3rd derivative but with respect to y this time\[840(2x-y)^{4}(-1) = -840(2x-y)^{4}\]

Answer 11

came to same answer haha.
Thanks!

Answer 12

Yep, np : )