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

**poirot** · May 14th 2011, 05:03 AM

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) $((f(x)-f(a))(g(x)-g(a))/(x-a)^2$ . Is this correct, if so how do I proceed?

**HallsofIvy** · May 14th 2011, 05:32 AM

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
$\lim_{x\to a}\frac{f(x)g(x)- f(a)g(a)}{x- a}$

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)).

**Plato** · May 14th 2011, 05:35 AM

Originally Posted by poirot

I'm not sure how to start this. I said that fg'(a) =lim(x->a) $((f(x)-f(a))(g(x)-g(a))/(x-a)^2$ . Is this correct, if so how do I proceed?

No that is not correct. You want
$\lim _{h \to 0} \frac{{f(a + h)g(a + h) - f(a)g(a)}}{h}.$
Add and subtract $f(a)g(a+h)$ in the numerator.
Because $g$ is continuous at a we know that $\lim _{h \to 0} g(a + h) = g(a)$ .

**poirot** · May 14th 2011, 05:46 AM

Sorry I don't understand your post. How did you begin and why would add something, then subtract it?

**mr fantastic** · May 14th 2011, 05:52 AM

Originally Posted by poirot

Sorry I don't understand your post. How did you begin and why would add something, then subtract it?

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).

**Plato** · May 14th 2011, 05:54 AM

Originally Posted by poirot

Sorry I don't understand your post. How did you begin and why would add something, then subtract it?

Look at reply #2.
$\frac{{f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)}}{{x - a}}$

$=\frac{{f(x) - f(a)}}{{x - a}}g(x) + f(a)\frac{{g(x) - g(a)}}{{x - a}}$ .

**poirot** · May 14th 2011, 09:19 AM

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.

**Plato** · May 14th 2011, 09:26 AM

Originally Posted by poirot

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.