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My brain has completely frozen and I can't remember how to differentiate $e^{x+1}$ Help!

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.

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

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

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.

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

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...

Original post by Swayum

e^f(x) differentiates to f'(x)e^f(x) by the chain rule.

I salute you, sir!

Original post by hukdealz

I salute you, sir!

You're about 7 years too late!

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