## Product Rule

The derivative of the product of two {differentiate) functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function:

It is also possible, of course, to rearrange the terms and express the rule as

Example 5 Find the derivat've of y = (2x + 3)(3jt). Let f{x) = 2x + 3 and g(x) = 3>r. Then it follows -—- that ff{x) = 2 and = 6a, and according to (7.4) the desired derivative is

— [(2x + 3}(3*2)] = (2x + 3){6x) f (3x2)(2) = 18x2 + 18*

This result can be checked by first multiplying out f(x)g(x) and then taking the derivative of the product polynomial. The product polynomial is in this case f(x)g(x) = (2a + 3)(3x2)-6x3 i 9x2f and direct differentiation does yield the same derivative,

The important point to remember is that the derivative of a product of two functions is not the simple product of the two separate derivatives. Instead, it is a weighted sum of /'(*) and g (x)i the weights being g(x) and /U), respectively. Since this difilers from what intuitive generalization leads one lo expect, let us produce a proof for (7.4). According to (6.13), the value of the derivative of /(x)^(.y) when x — N should be d f(x)g(x) ~ f(N)g(N)

But, by adding and subtracting j\x)g(N) in the numerator (thereby leaving the original magnitude unchanged), we can transform the quotient on the right of (7,5) as follows:

Substituting this for the quotient on the right of (7.5) and taking its limit, we then get

— [f(x)g(x)] = lim f(x) lim M } * dx *=v -c^.v x — N

+ lim g(N) lim J(X) ~ {(N) (7.5') a-».v x-* .'V x — A

The four limit expressions in (7.5') are easily evaluated. The first one is /(N), and the third is g(N) (limit of a constant). The remaining two are, according to (6.13), respectively, g'{N) and f(N). Thus (7.5') reduces to

-^[/MgW] = fWgXW +g(N)f,{N) <7,5") ax .r=.\

And, since Nrepresents any value of x, (7.5") remains valid il we replace every N symbol by x. This proves the rule.

As an extension of the rule to the case of three functions, we have

In words, the derivative of the product of three functions is equal to the product of the second and third functions times the derivative of the first, plus the product of the first and third functions times the derivative of the second, plus the product of the first and second functions times the derivative of the third. This result can be derived by the repeated application of (7.4). Hirst treat the product g(_r)/i(x) as a single function, say, 0(x), so that the original product of three functions will become a product of two functions, f(x)<t>(x). To this, (7.4) is applicable. After the derivative of f(x)\$(x) is obtained, we may reapply (7.4) to the product g(x)/>(A-) = (f>(x) to get tp'(x). Then (7.6) will follow. The details arc left to you as an exercise.

The validity of a rule is one thing; its serviceability is something else. Why do we need the product rule when we can resort to the alternative procedure of multiplying out the two functions f{x) and g(x) and then taking the derivative of the product directly? One answer to this question is that the alternative procedure is applicable only to specific (numerical or parametric) functions, whereas the product rule is applicable even when the functions are given in the genera/ form. Let us illustrate with an economic example. ## Procrastination Killer

Procrastination in probably the number one cause of failure in life and business. You Can Change Your Life Forever and Discover Success by Overcoming Procrastination. Learn how to defeat procrastination and transform your life into success.

Get My Free Ebook