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OpenStudy (mtalhahassan2):

Determine the derivative of each of the following: y=xe^3x

OpenStudy (astrophysics):

Product rule

OpenStudy (mtalhahassan2):

I know but how

OpenStudy (astrophysics):

What's the derivative of e^(3x)?

OpenStudy (astrophysics):

Note product rule is \[f'g+g'f\]

OpenStudy (mtalhahassan2):

the derivative of e^(3x) be 3

OpenStudy (astrophysics):

Remember \[\frac{ d }{ dx } e^x = e^x\]

OpenStudy (astrophysics):

So you're really applying the chain rule

OpenStudy (astrophysics):

\[\frac{ d }{ dx } e^{3x} = e^{3x} \times (3x)'\]

OpenStudy (astrophysics):

\[y'=x'(e^{3x})+(e^{3x})'x\] I don't want to do it all for you, but this is the product rule

HanAkoSolo (jamierox4ev3r):

I do believe that the chain rule is also involved here, since there is a function within a function.

OpenStudy (mtalhahassan2):

so how are we goona do this question

OpenStudy (astrophysics):

Go over what I said, I even mentioned it requires chain rule

HanAkoSolo (jamierox4ev3r):

aha i see :o

OpenStudy (mtalhahassan2):

oh you are confusing me know

HanAkoSolo (jamierox4ev3r):

just checking. basically, you're finding the derivative of ^3x. Carry on @Astrophysics :)

OpenStudy (mtalhahassan2):

one is saying we need chain rule and other is saying no

OpenStudy (astrophysics):

You never replied lol, where are you stuck it's basically done already, and no, you need both product rule and chain rule. No one said anything about not using it :P

OpenStudy (mtalhahassan2):

ok i goted

HanAkoSolo (jamierox4ev3r):

I'd be willing to type this out in a super clear manner, if you'd like. You sure you got it?

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