# Derivatives - product rule with three terms?

• Mar 25th 2009, 09:00 AM
ktprieto
Derivatives - 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.
• Mar 25th 2009, 09:10 AM
Pinkk
You have a function $h(x)=f(x)g(x)$ where $f(x)=3x$ and $g(x)=e^{2x}$

$h'(x)=f'(x)g(x)+f(x)g'(x)$

$h'(x)=(3)e^{2x}+3x(2e^{2x})$

$h'(x)=3e^{2x}+6xe^{2x}$
• Mar 25th 2009, 11:18 AM
matheagle
the 3 is just a constant, BUT if you have three functions then

$(fgh)'=f'gh+fg'h+fgh'$.
• Mar 25th 2009, 11:18 AM
Superjones
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
• Mar 25th 2009, 12:06 PM
tom@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