# Differentiating exponentials

1. Jan 29, 2005

### 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 im 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 approachs out there?

2. Jan 29, 2005

### 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.

3. Jan 30, 2005

### buddingscientist

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

thanks alot Daniel
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