User:Natkuhn/sandbox

Proof of the product rule

A rigorous proof of the product rule can be given using the definition of the derivative as a limit, and the basic properties of limits.

Let h(x) = f(x) g(x), and suppose that f and g are each differentiable at x₀. (Note that x₀ will remain fixed throughout the proof). We want to prove that h is differentiable at x₀ and that its derivative h'(x₀) is given by f'(x₀) g(x₀) + f(x₀) g'(x₀).

Let Δh = h(x₀+Δx) - h(x₀); note that although x₀ is fixed, Δh depends on the value of Δx, which is thought of as being "small."

The function h is differentiable at x₀ if the limit

$\lim _{\Delta x\to 0}{\Delta h \over \Delta x}$

exists; when it does, h'(x₀) is defined to be the value of the limit.

As with Δh, let Δf = f(x₀+Δx) - f(x₀) and Δg = g(x₀+Δx) - g(x₀) which, like Δh, also depend on Δx. Then f(x₀+Δx) = f(x₀) + Δf and g(x₀+Δx) = g(x₀) + Δg.

It follows that h(x₀+Δx) = f(x₀+Δx) g(x₀+Δx) = (f(x₀) + Δf) (g(x₀)+Δg); applying the distributive law, we see that

$h(x_{0}+\Delta x)=f(x_{0}+\Delta x)g(x_{0}+\Delta x)=f(x_{0})g(x_{0})+\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g$ (

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While it is not necessary for the proof, it can be helpful to understand this product geometrically as the area of the rectangle in this diagram:

To get the value, of Δh, subtract h(x₀)=f(x₀)+g(x₀) from equation (*). This removes the area of the white rectangle, leaving three rectangles:

$\Delta h=\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g$

To find h'(x₀), we need to find the limit as Δx goes to 0 of

${\frac {\Delta h}{\Delta x}}={\frac {\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g}{\Delta x}}={\frac {\Delta f}{\Delta x}}g(x_{0})+f(x_{0}){\frac {\Delta g}{\Delta x}}+{\frac {\Delta f\Delta g}{\Delta x}}$ (

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The first two terms of the right-hand side of this equation correspond to the areas of the blue rectangles; the third corresponds to the area of the gray rectangle. Using the basic properties of limits and the definition of the derivative, we can tackle this term-by term. First,

$\lim _{\Delta x\to 0}\left({\frac {\Delta f}{\Delta x}}g(x_{0})\right)=f'(x_{0})g(x_{0})$.

Similarly,

$\lim _{\Delta x\to 0}\left(f(x_{0}){\frac {\Delta g}{\Delta x}}\right)=f(x_{0})g'(x_{0})$.

The third term, corresponding to the small gray rectangle, winds up being negligible (i.e. going to 0 in the limit) because Δf Δg "vanishes to second order." Rigorously,

$\lim _{\Delta x\to 0}{\frac {\Delta f\Delta g}{\Delta x}}=\lim _{\Delta x\to 0}\left({\frac {\Delta f}{\Delta x}}\Delta g\right)=\lim _{\Delta x\to 0}{\frac {\Delta f}{\Delta x}}\cdot \lim _{\Delta x\to 0}{\Delta g}=f'(x_{0})\lim _{\Delta x\to 0}{\Delta g}$

The limit of Δg is 0 because g is continuous; without appealing to that fact, we can say that

$\lim _{\Delta x\to 0}{\Delta g}=\lim _{\Delta x\to 0}\left({\frac {\Delta g}{\Delta x}}\Delta x\right)=\lim _{\Delta x\to 0}{\frac {\Delta g}{\Delta x}}\cdot \lim _{\Delta x\to 0}\Delta x=g'(x_{0})\cdot 0=0$

We have shown that the limit of each of the three terms on the right-hand side of equation (**) exists, hence

$\lim _{\Delta x\to 0}{\frac {\Delta h}{\Delta x}}$

exists and is equal to the sum of the three limits. Thus, the product h(x) is differentiable at x₀ and its derivative is given by

${\begin{aligned}h'(x_{0})&=\lim _{\Delta x\to 0}{\frac {\Delta h}{\Delta x}}\\&=\lim _{\Delta x\to 0}\left({\frac {\Delta f}{\Delta x}}g(x_{0})\right)+\lim _{\Delta x\to 0}\left(f(x_{0}){\frac {\Delta g}{\Delta x}}\right)+\lim _{\Delta x\to 0}\left({\frac {\Delta f\Delta g}{\Delta x}}\right)\\&=f'(x_{0})g(x_{0})+f(x_{0})g'(x_{0})+0\\&=f'(x_{0})g(x_{0})+f(x_{0})g'(x_{0})\\\end{aligned}}$

as was to be shown.

Bad proof of the product rule

A rigorous proof of the product rule can be given using the properties of limits and the definition of the derivative as a limit of Newton's difference quotient.

If

$h(x)=f(x)g(x),\,$

and ƒ and g are each differentiable at the fixed number x, then

$h'(x)=\lim _{w\to x}{h(w)-h(x) \over w-x}=\lim _{w\to x}{f(w)g(w)-f(x)g(x) \over w-x}.\qquad \qquad (1)$

Now the difference

$f(w)g(w)-f(x)g(x)\qquad \qquad (2)$

is the area of the big rectangle minus the area of the small rectangle in the illustration.

The region between the smaller and larger rectangle can be split into two rectangles, the sum of whose areas is

$f(x){\Bigg (}g(w)-g(x){\Bigg )}+g(w){\Bigg (}f(w)-f(x){\Bigg )}.\qquad \qquad (3)$

Therefore the expression in (1) is equal to

$\lim _{w\to x}\left(f(x)\left({g(w)-g(x) \over w-x}\right)+g(w)\left({f(w)-f(x) \over w-x}\right)\right).\qquad \qquad (4)$

Assuming that all limits used exist, (4) is equal to

$\left(\lim _{w\to x}f(x)\right)\left(\lim _{w\to x}{g(w)-g(x) \over w-x}\right)+\left(\lim _{w\to x}g(w)\right)\left(\lim _{w\to x}{f(w)-f(x) \over w-x}\right).\qquad \qquad (5)$

Now

$\lim _{w\to x}f(x)=f(x)$

This holds because f(x) remains constant as w → x.

$\lim _{w\to x}g(w)=g(x)\,$

This holds because differentiable functions are continuous (g is assumed differentiable in the statement of the product rule).

Also:

$\lim _{w\to x}{f(w)-f(x) \over w-x}=f'(x)$    and    $\lim _{w\to x}{g(w)-g(x) \over w-x}=g'(x)$

because f and g are differentiable at x;

We conclude that the expression in (5) is equal to

$f(x)g'(x)+g(x)f'(x).\,$