Derivative problem that I thought would be easy

[Loopas]

I've attached a picture of the problem.

The problem asks from f'(0) of the function:

f(x)=e^(2x)*g(x)

g(0)=-5 and g'(0)=3 are given.

So the answer I came to was:

f'(x)=2e^(2x)*g'(x)

However, when I work out the numbers I get:

f'(0)=6

This is not right. So I figured I made a mistake a mistake when taking the derivative of f(x). Now I'm stuck because I'm not sure how to derive the function when g(x) isn't explicitly given.

 

[Dick]

I've attached a picture of the problem.

The problem asks from f'(0) of the function:

f(x)=e^(2x)*g(x)

g(0)=-5 and g'(0)=3 are given.

So the answer I came to was:

f'(x)=2e^(2x)*g'(x)

However, when I work out the numbers I get:

f'(0)=6

This is not right. So I figured I made a mistake a mistake when taking the derivative of f(x). Now I'm stuck because I'm not sure how to derive the function when g(x) isn't explicitly given.

Yes, you made a mistake when taking the derivative of f(x). Use the product rule.

 

[SqueeSpleen]

The product rule say that:
$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$
You did:
$(f(x)g(x))'=f'(x)g'(x)$
Which is incorrect.

 

[Loopas]

Wow, thanks guys. Sometimes sleep deprivation can do some crazy things.