Prove that if f and g are differentiable at a then fg is differentiable at a.

I'm not sure how to start this. I said that fg'(a) =lim(x->a) . Is this correct, if so how do I proceed?

- May 14th 2011, 06:03 AMpoirotprove that if f and g are differentiable at a then fg is differentiable at a
Prove that if f and g are differentiable at a then fg is differentiable at a.

I'm not sure how to start this. I said that fg'(a) =lim(x->a) . Is this correct, if so how do I proceed? - May 14th 2011, 06:32 AMHallsofIvy
I assume you mean using the basic definition of the derivative rather than the product rule.

No, what you have written is NOT correct. The derivative of a product is NOT the product of the derivatives. The derivative would be given by

The crucial point you need is that f(x)g(x)- f(a)g(a)= f(x)g(x)- f(a)g(x)+ f(a)g(x)- f(a)g(a)= (f(x)- f(a))g(x)+ f(a)(g(x)- g(a)). - May 14th 2011, 06:35 AMPlato
- May 14th 2011, 06:46 AMpoirot
Sorry I don't understand your post. How did you begin and why would add something, then subtract it?

- May 14th 2011, 06:52 AMmr fantastic
You are questioning what is actually a standard approach in many calculus textbooks for deriving the product rule.

I suggest you Google the derivation or go to your institute's library and find a Calculus textbook that contains the derivation (eg. Sallas and Hille or Thomas and Finney). - May 14th 2011, 06:54 AMPlato
- May 14th 2011, 10:19 AMpoirot
Ok, so as x tend to a g(x) tends to g(a) and so we get g(a)f'(a)+f(a)g'(a). I.e the familiar product rule. One question, at the begining it is as though fg(x) =f(x)g(x) in the numerator. Is this using the algebra of limits, Limit of fg =limit of f *limit of g? Thanks for your help.

- May 14th 2011, 10:26 AMPlato
- May 15th 2011, 03:55 AMpoirot
Yes sorry I misunderstood the notation.