The Student Room Group
Reply 1
let u=x+1 and differentiate e^u (remember you'll have to multiply by u')
Reply 2
e^f(x) differentiates to f'(x)e^f(x) by the chain rule.
Reply 3
it doesnt change i think ? or maybe im getting confused
Reply 4
heya use y=e^f(x)
dy' = f'(x)e^f(x)

hope this helps; so therefore it wld be i think.... e^x+1
Reply 5
e^(x+1) differentiated is simply e^(x+1)

substitution method of letting u = x+1 is only for integrating, but you would only use that in the case where you have a non-linear function being integrated.

e^(x+1) integrated is e^(x+1)
Reply 6
in case this is helpful:

f(x) = e^ax
f'(x) = ae^ax

the power of e always remains constant, instead, you multipy e to its power by the derivative of the power.

correction: e^(x+1) integrated is e^(x+1) + c
Reply 7
JJbwa
substitution method of letting u = x+1 is only for integrating...

Why?

Letting u=x+1 is useful to see why the derivative of e^(x+1) is e^(x+1). It's especially useful for newcomers to calculus.
Reply 8
oh maybe, yeah. ive used substitution in integration, but never diff, but if its useful, it makes sense.
Reply 9
notnek_01
Why?

Letting u=x+1 is useful to see why the derivative of e^(x+1) is e^(x+1). It's especially useful for newcomers to calculus.


Yeh i agree. Its just a way of doing it and setting it out so you can trace your steps really, not needed once you're used to it but it helps to start with.
Also, you can write e^x+1 as (e^x) multiplied by (e^1), so it's equal to e(e^x). Since e is just a constant, it doesn't affect the differentiation, since differentiating kf(x) gives kf'(x) if k is a constant. Thus, differentiating it, it remains as ee^x=e^x+1 :smile:
BenSpurgen
Also, you can write e^x+1 as (e^x) multiplied by (e^1), so it's equal to e(e^x). Since e is just a constant, it doesn't affect the differentiation, since differentiating kf(x) gives kf'(x) if k is a constant. Thus, differentiating it, it remains as ee^x=e^x+1 :smile:


Out of all the above methods, this is the most useful one. By the time you've wrote down all these different substitutions it could of been solved by spotting this.
IMO Its always best to spot what the answer should be or should look like, if this isn't possible, then proceed down the route of mathematical tricks.
Reply 12
y=e^x+1
ln y = x+1 ln e (ln e = 1)
ln y = 1x+1
(1/y)* dy/dx = 1
dy/dx = y = e^x+1
dkdeath
y=e^x+1
ln y = x+1 ln e (ln e = 1)
ln y = 1x+1
(1/y)* dy/dx = 1
dy/dx = y = e^x+1


Isn't that sortof overkill? You don't usually need to roll out the chain-rule in that fashion until you're differentiating x^x, that sortof thing...
Reply 14
Even easier than that, it involves no chain rule:

e^(x+1)=e(e^x)
Reply 15
Original post by Swayum
e^f(x) differentiates to f'(x)e^f(x) by the chain rule.

I salute you, sir!
Reply 16
Original post by hukdealz
I salute you, sir!


You're about 7 years too late!
Reply 17
I've now closed this thread due to current responses.