differentiation of expontentials

[Becky4paws]

I have a feeling I am way off on this one

f(x) = (x+1)e^-2x where x=0
f'(x) = (x+1)e^-2x d/dx(-2x)
f'(x) = (x+1)e^-2x(-2)
f'(x) = -2(x+1)e^-2x
f'(x) = (-2x-2)e^-2x

where x = 0
f'(0) = -2(0) -2 e^-2(0)
f'(0) = -2e^0
f'(0) = -2

 

[nezenic]

You need the product rule as well as the chain rule.

\(\displaystyle f(x) = u(x)*v(x)\)

\(\displaystyle f'(x) = u(x)*v'(x) + v(x)*u'(x)\)

Your u(x) would be (x+1) and v(x) would be e^(-2x). And don't forget the chain rule as well.

 

[skeeter]

\(\displaystyle \L f(x) = (x+1)e^{-2x}\)

\(\displaystyle \L f'(x) = (x+1)(-2e^{-2x}) + (e^{-2x})(1)\)

\(\displaystyle \L f'(x) = e^{-2x}[-2(x+1) + 1]\)

\(\displaystyle \L f'(x) = e^{-2x}(-2x - 1) = -e^{-2x}(2x + 1)\)

 

[Becky4paws]

Where x = 0

f'(x)=-e^-2x(2x+1)
f'(0)=-e^0(1)
f'(0) = -1(1)
f'(0) = -1