Question

How do we derive the product rule?

Answer 1

Please see below.

Explanation:

Let `f(x)=g(x)h(x)`.

Now from definition `(df)/(dx)=Lt_(deltax->o)(f(x+deltax)-f(x))/(deltax)`

also `(dg)/(dx)=Lt_(deltax->o)(g(x+deltax)-g(x))/(deltax)`

and `(dh)/(dx)=Lt_(deltax->o)(h(x+deltax)-h(x))/(deltax)`

As `f(x)=g(x)h(x)` and `f(x+deltax)=g(x+deltax)h(x+deltax)`

`(df)/(dx)=Lt_(deltax->o)(f(x+deltax)-f(x))/(deltax)`

= `Lt_(deltax->o)(g(x+deltax)h(x+deltax)-g(x)h(x))/(deltax)`

= `Lt_(deltax->o)(g(x+deltax)h(x+deltax)color(red)(-g(x)h(x+deltax)+g(x)h(x+deltax))-g(x)h(x))/(deltax)`

= `Lt_(deltax->o)(g(x+deltax)h(x+deltax)-g(x)h(x+deltax))/(deltax)+Lt_(deltax->o)(g(x)h(x+deltax)-g(x)h(x))/(deltax)`

= `Lt_(deltax->o)[h(x+deltax)(g(x+deltax)-g(x))/(deltax)]+g(x)Lt_(deltax->o)[(h(x+deltax)-h(x))/(deltax)]`

= `h(x)(dg)/(dx)+g(x)(dh)/(dx)`

Answer 2

Please see below.

Explanation:

I use the product rule in the form:

`d/dx(f(x)g(x)) = f'(x)g(x)+f(x)g'(x)`

(Because multiplication and addition are commutative, it can also be written in other orders.)

Proof:

Suppose that `f` and `g` are differentiable at `x`.

Let `P(x) = f(x)g(x)`.

(We will show that `P'(x)= f'(x)g(x)+f(x)g'(x)`.)

Then

`P'(x) = lim_(hrarr0)(P(x+h)-P(x))/h`

`= lim_(hrarr0)(f(x+h)g(x+h) - f(x)g(x))/h`

Now we will add `0` in the numerator. We'll use `-f(x)g(x+h)+f(x)g(x+h)`. (See note (1) below.)

`= lim_(hrarr0)(f(x+h)g(x+h) -f(x)g(x+h)+f(x)g(x+h) - f(x)g(x))/h`

Now regroup:

`= lim_(hrarr0)((f(x+h) -f(x))g(x+h)+f(x)(g(x+h) - g(x)))/h`

Rewrite as the sum of two ratios and factor.

`= lim_(hrarr0)((f(x+h) -f(x))g(x+h))/h + (f(x)(g(x+h) - g(x)))/h`

`= lim_(hrarr0)((f(x+h) -f(x)))/hg(x+h) + lim_(hrarr0)f(x)((g(x+h) - g(x)))/h`

We now have limits of 4 things to evaluate.

`lim_(hrarr0)((f(x+h) -f(x)))/h = f'(x)`
`lim_(hrarr0)g(x+h)=g(x)` (see note (2) below).
`lim_(hrarr0)f(x) = f(x)`
`lim_(hrarr0)((g(x+h) -g(x)))/h = g'(x)`

Finally we write the total limit:

`= f'(x)g(x)+f(x)g'(x)`

That is:

`P(x) = f(x)g(x)` has derivative

`P'(x) = f'(x)g(x)+f(x)g'(x)`

Notes

(1) To prove the product rule in a different order we would add `0` in the form `-f(x+h)g(x)+f(x+h)g(x)` and we'd end with `f(x)g'(x)+f'(x)g(x)`

(2) By hypothesis `g` is differentiable at `x`, so `g` is continuous at `x` and therefore `lim_(hrarr0)g(x+h)=g(x)`.