This is the user sandbox of Natkuhn. A user sandbox is a subpage of the user's user page. It serves as a testing spot and page development space for the user and is not an encyclopedia article. Create or edit your own sandbox here.
Finished writing a draft article? Are you ready to request review of it by an experienced editor for possible inclusion in Wikipedia? Submit your draft for review!
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
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
(
)
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:
To find h'(x0), we need to find the limit as Δx goes to 0 of
(
)
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,
.
Similarly,
.
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,
The limit of Δg is 0 because g is continuous; without appealing to that fact, we can say that
We have shown that the limit of each of the three terms on the right-hand side of equation (**) exists, hence
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
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
and ƒ and g are each differentiable at the fixed number x, then
Now the difference
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[1]
Therefore the expression in (1) is equal to
Assuming that all limits used exist, (4) is equal to
Now
This holds because f(x) remains constant as w → x.
This holds because differentiable functions are continuous (g is assumed differentiable in the statement of the product rule).
Also:
and
because f and g are differentiable at x;
We conclude that the expression in (5) is equal to
^The illustration disagrees with some special cases, since – in actuality – ƒ(w) need not be greater than ƒ(x) and g(w) need not be greater than g(x). Nonetheless, the equality of (2) and (3) is easily checked by algebra.