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Mathematics 22 Online
OpenStudy (anonymous):

Just starting to learn derivatives... learned that if f(x)=e^x, f'(x)=e^x also. Blew my mind, really. but I decided to test it - because it should also be true that f'(x)=x(e^(x-1)). I came up with the conclusion that either one of them is wrong, or my work is wrong. Care to check my work? x(e^(x-1))=e^x if x=1, e^0=e^1 I'm somewhat confused.

OpenStudy (anonymous):

Thats wrong. The rule \[ \frac{d}{dx} x^n = n x^{n-1} \] does only apply if n DOES NOT DEPEND ON x!

OpenStudy (anonymous):

You can verify that \[ \frac{d}{dx} e^x = e^x \] by using the definition of e^x which is \[ e^x = \sum^\infty_{n=0} \frac{x^n}{n!} \]

OpenStudy (anonymous):

Differentiate that sum term by term,

OpenStudy (anonymous):

So... that rule only works when x is the base?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

And the exponent needs to be a constant!

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