product rule

[gymnast24]

Find F'(0) (derivative of F(x) at x=0), given that
F(x) =(2x+e^x)*g(x) and g(0)=9, g'(0)=0.

I figure I have to use the product rule but I don't know how to set it up this is all I've got so far.

fg'+gf'

(2x+e^x)(0)+(f')(9)

 

[Gene]

Are you mixing up F and (2x+e^x) or are you using f=2x+e^x? in either case
F'(0) = (2*0+e^0)(0)+(2+e^0)(9)

 

[soroban]

Hello, gymnast24!

Find \(\displaystyle F'(0)\), given that: \(\displaystyle F(x)\:=\2x\.+\.e^x)\cdot g(x)\) and \(\displaystyle g(0)\,=\,9,\;g'(0)\,=\,0.\)

Never mind your version of the Product Rule (too confusing) . . . just do it!

We have:\(\displaystyle \,F(x)\;=\;(2x\,+\,e^x)\cdot g(x)\)

Product Rule: \(\displaystyle \,F'(x)\:=\2x\,+\,e^x)\cdot g'(x)\,+\,(2\,+\,e^x)\cdot g(x)\)


When \(\displaystyle x\,=\,0\), we have:

\(\displaystyle \;\;\;F'(0)\;=\;(2\cdot0\,+\,e^0)\cdot g'(0)\,+\,(2\,+\,e^0)\cdot g(0)\)

\(\displaystyle \;\;\;F'(x)\;=\;(0\,+\,1)\cdot 0 \,+\,(2\,+\,1)\cdot9 \:=\:0\,+\,3\cdot9\:=\:27\)

 

[Gene]

And I thought I gave too much info. You must be trying catch up with Eliz.
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Gene