MATHS: Exponential Derivative

cheechm · 26 Apr 2010 at 18:28

Just a quick one: How would I go about doing this?

f(x) = e^e^3x

I need to find f'(x), which I got to be:

f'(x) = e^3e^3x.3e^3x

That's using the substitution method, however it might not be right.

Any help?

Cheers

Amleto · 26 Apr 2010 at 18:35

d/dx{ e^u(x) } = du/dx . d/du { e^u(x) } = 3e^3x . d/du{ e^u } = 3e^3x . e^e^3x

aside:
d/dx {exp( f(x) )} = f'(x)exp( f(x) )

cheechm · 26 Apr 2010 at 18:48

Thank you!

Munkeh · 26 Apr 2010 at 18:49

Wolfram Alpha is your friend - http://www.wolframalpha.com/input/?i=d/dx(+e^e^(3x))

Yes it doesn't help you solve it, but at least you can check you have got it right.

p4radox · 26 Apr 2010 at 19:02

Amleto said:
d/dx{ e^u(x) } = du/dx . d/du { e^u(x) } = 3e^3x . d/du{ e^u } = 3e^3x . e^e^3x

aside:
d/dx {exp( f(x) )} = f'(x)exp( f(x) )

This.

OP: in case you were wondering, it's an application of the chain rule.

Darg · 26 Apr 2010 at 19:08

All I know is the exponential doesn't change.. and I just did my masters maths exam today.. I really should be able to answer that.. I think it's all quickly leaking out the back of my head already.

TheMagicPirate · 26 Apr 2010 at 19:38

Amleto said:
d/dx{ e^u(x) } = du/dx . d/du { e^u(x) } = 3e^3x . d/du{ e^u } = 3e^3x . e^e^3x

aside:
d/dx {exp( f(x) )} = f'(x)exp( f(x) )

I hope I'm not the only person that goes :confused:

when I look at this lol

ennogs · 26 Apr 2010 at 22:10

TheMagicPirate said:
I hope I'm not the only person that goes when I look at this lol

No you are not alone. I only got up to O level maths lol.

cheechm · 26 Apr 2010 at 22:54

ennogs said:
No you are not alone. I only got up to O level maths lol.

I'm only Lower Sixth...

DavieJG · 26 Apr 2010 at 23:02

goody, you can look forward to integration by parts