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# Generalising differentiation of g(x)e^f(x))? watch

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1. Hi, I'm having trouble with differentiating almost anything involving e (and ln sometimes, but I can mostly manage that).

Say for example I have y=(2x2+1)(e4x), how would I differentiate this? Even better, what generalised formula can I use to just sub in the numbers? I know that when y=ef(x), dy/dx = f'(x)ef(x) [sorry for mixed notation], but how do I factor in another function of x at the start, or more importantly, what process happens to the function at the start.

In other words, if I had (axb+c)(e(dx+f)) [f and d are variables] then what formula, using the same variables, would represent the differentiated version?
2. (Original post by samzeman)
Hi, I'm having trouble with differentiating almost anything involving e (and ln sometimes, but I can mostly manage that).

Say for example I have y=(2x2+1)(e4x), how would I differentiate this? Even better, what generalised formula can I use to just sub in the numbers? I know that when y=ef(x), dy/dx = f'(x)ef(x) [sorry for mixed notation], but how do I factor in another function of x at the start, or more importantly, what process happens to the function at the start.

In other words, if I had (axb+c)(e(dx+f)) [f and d are variables] then what formula, using the same variables, would represent the differentiated version?
Product rule. means

and since then that just means so but that's just trivial.

Why would you have and not just ? Unless you mean which is what I assumed you meant.

Anyway, as I said, if then
3. (Original post by RDKGames)
Product rule. means

and since then that just means so but that's just trivial.

Anyway, as I said, if then
Okay, cool, that's probably another form of some rule that I should definitely know. But I just prefer formulae. Thank you!

(Original post by RDKGames)
Why would you have and not just ? Unless you mean which is what I assumed you meant.
My mistake aha. I wrote this initial post being angry and tired at homework. It's because of edx+f being a part of it and I just mentally did the same again but wronger. lol. thanks again
4. There is a method you will learn usually in your second year of a level maths which i forgot the name of but it goes like this

Differentiating f(x) you get f '(x)

Differentiating g(x) you get g '(x)

Differentiating f(x)+g(x) you get f '(x)+g '(x)

Differentiating f(x)g(x) you get f '(x)g(x)+g '(x)f(x)

Differentiating e^[f(x)] you get f '(x)e^[f(x)]

SO

To differentiate g(x)e^(f(x)) you need to use the fourth one.

which = g '(x)e^(f(x)) + f '(x)e^[f(x)]

An easier way to see it :

e^[f(x)] you already know how to differentiate. so instead of looking at it as

g(x)e^[f(x)] just look at is as f(v)f(u) and to differentiate that you differentiate the first one times the second one PLUS differentiate second one time the first.

Hope that helps

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Updated: October 11, 2017
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