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

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?
(edited 6 years ago)
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. y=f(x)g(x)y=f(x)g(x) means y′=f′(x)g(x)+f(x)g′(x)y'=f'(x)g(x)+f(x)g'(x)

and since g(x)=eh(x)g(x)=e^{h(x)} then that just means g′(x)=h′(x)eh(x)g'(x)=h'(x)e^{h(x)} so y′=f′(x)g(x)+f(x)h′(x)g(x)y'=f'(x)g(x)+f(x)h'(x)g(x) but that's just trivial.

Why would you have xb+cx^{b+c} and not just xbx^b? Unless you mean xb+cx^b+c which is what I assumed you meant.

Anyway, as I said, if y=g(x)ef(x)y=g(x)e^{f(x)} then y′=g(x)f′(x)ef(x)+g′(x)ef(x)y'=g(x)f'(x)e^{f(x)}+g'(x)e^{f(x)}
(edited 6 years ago)
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samzeman
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Original post by RDKGames
Product rule. y=f(x)g(x)y=f(x)g(x) means y′=f′(x)g(x)+f(x)g′(x)y'=f'(x)g(x)+f(x)g'(x)

and since g(x)=eh(x)g(x)=e^{h(x)} then that just means g′(x)=h′(x)eh(x)g'(x)=h'(x)e^{h(x)} so y′=f′(x)g(x)+f(x)h′(x)g(x)y'=f'(x)g(x)+f(x)h'(x)g(x) but that's just trivial.

Anyway, as I said, if y=g(x)ef(x)y=g(x)e^{f(x)} then y′=g(x)f′(x)ef(x)+g′(x)ef(x)y'=g(x)f'(x)e^{f(x)}+g'(x)e^{f(x)}


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 xb+cx^{b+c} and not just xbx^b? Unless you mean xb+cx^b+c 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
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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