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Mathematics 15 Online
OpenStudy (anonymous):

the deriviative of e^2x

OpenStudy (turingtest):

do you know the derivative of e^x ?

OpenStudy (anonymous):

e^x

OpenStudy (anonymous):

use u substitution

OpenStudy (turingtest):

great, now just use the chain rule do you know that?

OpenStudy (anonymous):

how is that chain rule?

OpenStudy (turingtest):

we haven't done it yet... but this problem requires it

OpenStudy (anonymous):

the chain rule is f'(gx)(gx)'

OpenStudy (turingtest):

call \(2x=f(x)\) now we can look at this like\[\large e^{f(x)}\]the derivative is then\[\large e^{f(x)}f'(x)\]

OpenStudy (anonymous):

but i dont get how e^2x is a chain rule its not a function inside a fucntion

OpenStudy (anonymous):

is it 2e^2x

OpenStudy (anonymous):

yes it is.

OpenStudy (turingtest):

the 2x is the function inside the function e^(whatever)

OpenStudy (anonymous):

so e^-x would be -e^x

OpenStudy (anonymous):

-e^-x*

OpenStudy (anonymous):

\[y=a^{f(x)}\] \[y'=a^{f(x)}*f'(x)*\ln a\]

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