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David Terr
Ph.D. Math, UC Berkeley

 

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11.3. The Product Rule

There are several useful rules for helping us compute derivatives of functions. One of the most important one is the product rule, also known as Leibniz' rule, named after the 17th century German mathematician Gottfried Lebiniz, who discovered it. The product rule tells us how to compute the derivative of a product of two functions in terms of the original functions and their derivatives. Specifically, it says the following:

  • (11.3.1a) D(fg) = f Dg + g Df

In Leibniz' notation, the product rule reads as follows:

  • (11.3.1b) d(fg) / dx = f dg/dx + g df/dx.

In Newton's primed notation, it reads as follows:

  • (11.3.1c) (fg)' = fg' + gf'.

We will prove the Product Rule in two different ways. The first method uses the definition (11.2.1) of the derivative. The second method, while conceptually much simpler, involves the concept of differentials, which we will need to define. We will prove (11.3.1) first using the first method.

Let f and g be two differentiable functions and let F(x) = f(x)g(x). We need to show that F'(x) = f(x) g'(x) + g(x) f'(x). We use (11.2.1) to compute F'(x). We have

proof of product rule

 

Our second proof of the product rule uses the concept of differentials. A differential is an infinitesimally small change in a quantity. At this point, you are probably wondering what in the world does this mean? The best way to understand differentials is to look back at our definition (11.2.1) of the derivative. We will rewrite this equation using Leibniz' notation. In place of h, we write Δx, which stands for "change in x", and in place of f'(x) we write df / dx. How can we interpret f(x + h) - f(x) = f(x + Δx) - f(x)? This is the amount by which f(x) changes when x is increased by Δx. Thus, we may write it as Δf. With this new notation, equation (11.2.1) becomes

differential derivative definition

Now the differential quantities df and dx should begin to make sense. We think of dx as an infinitesimal change in the variable x and df as the corresponding infinitesimal change in f(x). The quantities Δf and Δx approach the zero-size quantities df and dx respectively, but the ratio df / dx, which is to be thought of as the limit of Δf / Δx as Δx goes to zero, approaches a finite value, which may be nonzero.

So how can we prove the product rule using differentials? When written in Leibniz' differential notation, the product rule takes on the form (11.3.1b). Now we multiply both sides of this equation by the differential quantity dx, obtaining

  • (11.3.2) d(fg) = f dg + g df.

This is a statement about differentials which is equivalent to the product rule. So how can we prove it? Consider the following diagram:

pictorial proof of product rule

Figure 11.3.3: Pictorial Proof of Product Rule

 

As can be seen from the diagram, the area of the large rectangle is (f + df)(g + dg) = fg + f dg + g df + df dg. But since df and dg are infinitesimal quantities, the product df and dg, though also infinitesimal, is infinitesimally smaller than either df or dg, so we may disregard it. Thus we have (f + df)(g + dg) = fg + d(fg) = fg + f dg + g df. Subtracting fg from both sides, we see that d(fg) = f dg + g df, once again, proving the product rule.

 

As the following examples show, the produt rule allows us to compute the derivative of a product of polynomials withouth having to first evaluate the product of the polynomials, which can often save us a lot of work.

 

Example 1: For the given polynomials p(x) and q(x), compute the derivative of f(x) = p(x)q(x) and evaluate it at x=0.

  • (a) p(x) = x + 3; q(x) = x + 4
  • (b) p(x) = 3x + 7; q(x) = 2x - 5
  • (c) p(x) = x2 - 2x + 4; q(x) = x - 2
  • (d) p(x) = 2x2 - 7x + 6; q(x) = 4x2 - 3x + 8

Solution:

  • (a) We have f'(x) = p(x)q'(x) + q(x)p'(x) = (x + 3)(1) + (x + 4)(1) = x + 3 + x + 4 = 2x + 7. Thus, f'(0) = 7.
  • (b) We have f'(x) = p(x)q'(x) + q(x)p'(x) = (3x + 7)(2) + (2x - 5)(3) = 6x + 14 + 6x - 15 = 12x - 1. Thus, f'(0) = -1.
  • (c) We have f'(x) = p(x)q'(x) + q(x)p'(x) = (x2 - 2x + 4)(1) + (x - 2)(2x - 2) = x2 - 2x + 4 + 2x2 - 6x + 4 = 3x2 - 8x + 8. Thus, f'(0) = 8.
  • (d) We have f'(x) = p(x)q'(x) + q(x)p'(x) = (2x2 - 7x + 6)(8x - 3) + (4x2 - 3x + 8)(4x - 7). Thus, f'(0) = (6)(-3) + (8)(-7) = -18 -56 = -74.

 

We can use the product rule to extend the power rule (11.2.6) to negative powers of x, as the following theorem shows.

 

Theorem 11.3.4 (Power Rule for Negative Powers): Let n be a positive integer and let f(x) = x-n. Then we have f'(x) = -nx-(n+1).

Proof: We let g(x) = xn and apply the product rule to 1 = f(x)g(x). We have

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

= (x-n)(nxn-1) + xnf'(x)

= nx-1 + xnf'(x).

Dividing both sides by xn, we obtain 0 = nx-(n+1) + f'(x), implying f'(x) = -nx-(n+1).

QED

 

Note that this case is really just an extension of Equation (11.2.6), which we can see if we substitute n for -n and let n be negative. In fact, it can be shown that (11.2.6) holds for all real n, and in fact, for complex n as well!

 

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