Hey again,(adsbygoogle = window.adsbygoogle || []).push({});

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 im having some trouble with the j component.

[tex]f = ye^{(yx)}[/tex]

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

[tex]f = ye^{(2y)}[/tex]

now my intuition says [tex]df/dy = 2ye^{(2y)}[/tex]

and plugging x back in for the 2:

[tex]df/dy = xye^{(xy)}[/tex]

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 [tex]f = ye^{(xy)}[/tex] and instantly finding df/dx and df/dy. are there ay other approachs out there?

thanks for reading

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# Differentiating exponentials

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