If f(x)=e^x g(x), where g(0)=7, and g'(0)= 1, find'(0)

Answer 1

Begin by finding f'(x) using the product rule:

f(x) = uv

f'(x) = u'v + uv'

Therefore, because e^x is its own derivative,

f'(x) = e^xg(x) + e^xg'(x)

Plug in 0 for x to get:

f'(0) = e⁰g(0) + e⁰g'(0)

Because e⁰=1 and the other values are given,

f(0) = (1*7) + (1*1)

f'(0) = 7 + 1

f'(0) = 8

Answer 2

If you mean "Find f '(0)" then you would use the product rule because f(x) is the product of two functions.

For f(x) = u•v, then f '(x) = u'v + uv'

u = e^x

u' = e^x

v = g(x)

v' = g'(x)

So f '(x) = e^x•g(x) + e^x•g'(x) and

f '(0) = e⁰•g(0) + e⁰•g'(0) = 1•7 + 1•1 = 8