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## Product Rule for Differentiation

January 16th, 2011

The product rule allows us to differentiate the product of two or more functions provided we know how to differentiate each of the functions seperatley.

If we can differentiate two functions f and g the derivative is

$\frac{d}{dx}(f(x)g(x) = \frac{df(x)}{dx}g(x) + f(x)\frac{dg(x)}{x}$

This rule can be repeated so that we can differentiate the product of more than two functions for exampe

$\frac{d}{dx}(f(x)g(x)h(x) = \frac{df(x)}{dx}g(x)h(x) + f(x)\frac{dg(x)}{dx}h(x) + f(x)g(x)\frac{dh(x)}{dx}$

### Example

Suppose we want to differentiate $f(x) = x^2sin(x)$

First we notice that f is the product of two functions we know how to differentiate, $x^2$ and sin(x).

The derivative of x2 is 2x and the derivative of sin(x) is cos(x) so putting these into the rule we get

$\frac{df(x)}{dx} = \frac{d(x^2)}{dx}sin(x) +x^2\frac{d}{dx}(sin(x))$

So putting in the derivative for sin and x2; we get

$\frac{df(x)}{dx} = 2xsin(x) + x^2cos(x)$

### Proof of the Product Rule

Completely rigourus proofs of most rules with differentiation can be quite complicated and I don’t this site covers some of the concepts in enough detail, but we can give a basic outline of a proof which you can expand as you learn higher levels of maths (mainly analysis).

So as we did when deriving from first principles we want to the limit of the gradient between two point on the curve as they get closer together.

$\frac{d}{dx}(f(x)g(x)) = \lim_{t\to 0}\frac{f(x+t)g(x+t) - f(x)g(x)}{t}$

We now add and subtract f(x+t)g(x) because we can cunningly use it to get the derivatives work and because it is the same as adding 0 it is perfectly valid.

$\frac{d}{dx}(f(x)g(x)) = \lim_{t\to0}\frac{f(x+t)g(x+t) -f(x+t)g(x) + f(x+t)g(x) - f(x)g(x)}{t}$

Now we bracket in such a way that we can see the derivatives we want

$\frac{d}{dx}(f(x)g(x)) = \lim_{t\to0}\frac{f(x+t)[g(x+t) -g(x)]}{t} +\frac{g(x)[f(x+t) - f(x)]}{t}$

Spotting the limits that make dg/dx and df/dx(and making use of the fact that we can multiply limits) we get

$\frac{d}{dx}(f(x)g(x)) = \lim_{t\to0}f(x+t)\frac{dg(x)}{dx} +g(x)\frac{df(x)}{dx}$

and finally taking the limit of f(x+t) to be f(x) (since f is differentiable)

$\frac{d}{dx}(f(x)g(x)) = f(x)\frac{dg(x)}{dx} + g(x)\frac{df(x)}{dx}$

which is the required result!

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