How do you differentiate a scalar field with exponentials?

[buddingscientist]

Hey again,

well i just studying several vaiable calculus, and encountered the problem of finding the gradient of the scalar field:

f = ye^(xy)

now I could successfully find the i component (y^2.e^(yx))
but I am having some trouble with the j component.

f = ye^{(yx)}

if my understanding is correct to differentiate this wrt to y, we treat everything else (x) as if it were a constant.
now when i encouner stuff like this, i plug in an arbitrary number for x, such as 2, and continute like that

f = ye^{(2y)}

now my intuition says df/dy = 2ye^{(2y)}
and plugging x back in for the 2:
df/dy = xye^{(xy)}

but this is obviously incorect, with both the solutions and a calculator giving the answer of:
(y.x + 1)e^(xy) or yxe^(xy) + e^(xy)

i have no doubts its correct but what is the procedure to get the additional exponential term? and under what conditions is it +2, +3, etc?

i tried searching the internet but just found examples wthout the leading variable (in this case y in front of the e).
and another thing, is my approach of answering partial DE's okay? (e finding something wrt y, replacing the x's and z's with integers, then diff'ing?
i guess i find it difficult just looking at something like f = ye^{(xy)} and instantly finding df/dx and df/dy. are there ay other approaches out there?



thanks for reading

 

[dextercioby]

It really doesn't matter how u denote (this in the case in which u denote) the variable(s) kept constant in the partial differentiation,the important thing is to apply the rules correctly.
Unfortunately u didn't...
Compute the derivative of the product
f(x)g(x)

wrt to "x".Then use this rule to CORRECTLY differentiate your formula...

Daniel.

 

[buddingscientist]

ahh that's right, good old product rule.
deadset it's been at least 3 years since I've needed to use it lol

thanks a lot Daniel