Derivative using product rule

[PaulKraemer]

Hi,

My book has a problem where they give you the following:

g(2) = -5
g' (2) = 2

They ask you to find:

(4g)' (2)

I used the product rule, letting f(x) = 4 and thus f '(x) = 0

(fg) ' (2) = f(2) * g' (2) + g(2) * f ' (2) = (4 * 2) + ( -5 * 0 ) = 8

The back of the book says the answer is -4

I'm hoping the back of the book is wrong, but if anyone can either confirm this or tell me where I went wrong, I'd really appreciate it.

Thanks in advance,
Paul

 

[Deleted member 4993]

Hi,

My book has a problem where they give you the following:

g(2) = -5
g' (2) = 2

They ask you to find:

(4g)' (2)

I used the product rule, letting f(x) = 4 and thus f '(x) = 0

(fg) ' (2) = f(2) * g' (2) + g(2) * f ' (2) = (4 * 2) + ( -5 * 0 ) = 8

The back of the book says the answer is -4

I'm hoping the back of the book is wrong, but if anyone can either confirm this or tell me where I went wrong, I'd really appreciate it.

Thanks in advance,
Paul

If what you posted as the problem is correct - then what you did is correct.

However, you took long route to the answer.

remember

\(\displaystyle \frac {d[c\cdot f(x)]}{dx} \ = \ c \cdot f'(x)\)

 

[PaulKraemer]

Thanks Subhotosh,

I really appreciate your help!

Paul