I'm trying to get the derivative of 3xe^(2x)

I know the product rule and I know the derivative of e^2x is 2e^(2x).

So I was going to use the product rule, but I've only ever done that with two terms. Here, there are three.

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- March 25th 2009, 10:00 AMktprietoDerivatives - product rule with three terms?
I'm trying to get the derivative of 3xe^(2x)

I know the product rule and I know the derivative of e^2x is 2e^(2x).

So I was going to use the product rule, but I've only ever done that with two terms. Here, there are three. - March 25th 2009, 10:10 AMPinkk
You have a function where and

- March 25th 2009, 12:18 PMmatheagle
the 3 is just a constant, BUT if you have three functions then

. - March 25th 2009, 12:18 PMSuperjones
thats a product rule and a chain rule

3xe^(2x)

3e^(2x)

for the first. take derivative of 3x which is just 3 and keep second part. than. chain rule e^(2x) so e^ whatever is just e^ the whatever. so e^(2x) than multiply by 2 since the derivative of 2x is 2. keep the first part.

So you get:

3e^(2x) + 3xe^(2x)*2 which is 3e^(2x) + 6xe^(2x)

you can then simplify by pulling 3e^(2x) out of the answer

3e^(2x)*(1+2x) but that wouldnt really help unless your graphing and looking for critical numbers - March 25th 2009, 01:06 PMtom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/d...ghtHandExp.png

You have the chain rule - generally this pattern -

http://www.ballooncalculus.org/chain_rule.png

... wrapped inside the product rule...

http://www.ballooncalculus.org/prod_rule.png

... more visible here if we enclose some of the balloons in some more...

http://www.ballooncalculus.org/asy/d...htHandExp1.png

Hope this helps, or doesn't further confuse.

Don't integrate - balloontegrate!

Balloon Calculus: worked examples from past papers