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Mathematics 6 Online
OpenStudy (anonymous):

find the derivative of e^(2x) using the product rule

OpenStudy (anonymous):

\[(f \prime) e ^{2x}\]

OpenStudy (anonymous):

I know its 2e^(2x), but i'm not sure why. I know e^x is equal to its derivative but thats about it

OpenStudy (anonymous):

the derivative of \[e^{f(x)}\] is \[e^{f(x)}*f'(x)\]

OpenStudy (anonymous):

Do you know the derivative for \[f(x)=2x\]

OpenStudy (anonymous):

its 2

OpenStudy (anonymous):

Correct

OpenStudy (anonymous):

And that makes sense I just didn't know the rule that you just used. Should be handy. Is it only good for e?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Awesome thanks. Good to remember. I haven't explicitly been taught that, but i guess i was just suposed to figure it out since it said to use the product rule,which i have learned

OpenStudy (anonymous):

oh duh i just figured out another way to look it as e^2x = e^x * e^x

OpenStudy (anonymous):

The problem is that mj632 is using the chainrule but the question strickly says the use of the product rule. Here e\[e ^{2x}=e ^{x}*e ^{x}\] and aplying the product rule you'll get \[e ^{x}e ^{x}+e ^{x}e ^{x}=e ^{2x}+e ^{2x}=2e ^{2x}\]

OpenStudy (anonymous):

awesome. thanks!

OpenStudy (anonymous):

nice you figured it out yourself already:)

OpenStudy (anonymous):

ha, that tends to happen. thanks though. Still gave ya a medal

OpenStudy (anonymous):

haha no probs and thanks^^

OpenStudy (anonymous):

Ah, sorry about that.

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