True or false? Justify your answer: (fg)'=f'g' ?

Question

I'm fairly certian that they are not equal becasue you can easily take the derivitive of the first, but not the second, and you cant distribute the prime sign like that anyway... But I'm having trouble proving it for my calc two class. I'm a little rusty, can anyone prove it to me or direct me to a good website that can help me out?

Answer 1

I have NO idea... :-)

Answer 2

Your right. There is special rule for the derivative of a product

(fg)' = f ' *g + g' *  f .

The third responder provided a counterexample.

To prove that (fg)' = f'g' is false all you need is one counterexample that shows that the statement doesn't work for all functions.

Answer 3

this doesn't work.

say f = 2x

and g = x

fg = 2x^2

f' = 2

g' = 1

f'g' = 2

(fg)' = 4x

4x does not = 2

so they are not the same.

Answer 4

Well, set f(x) = 1 and g(x) = x.

Then f*g(x) = x.

(fg)'(x) = 1

f'(x) = 0 and g'(x) = 1 so f'g'(x) = 0

This shows that the equality in question is false.

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As Jon F said, the correct form is given by the product rule:

(fg)' = f' g + f g'