In Exercise 48 in Section 4.2, we saw an example for which the erroneous version of the product rule, [ƒ(x)g(x)]′= ƒ′(x)g′(x), does hold. As a more general version of this example, calculate [ƒ(x)g(x)]′and ƒ′(x)g′(x) for ƒ(x) = a/(n - x)ⁿ and g(x) = bxⁿ for n ≠ 0, and verify that they are equal.

Data from Exercise 48 in Section 4.2

There are special cases in which the derivative of ƒ(x)g(x) does equal ƒ′(x)g′(x). Calculate both [ƒ(x)g(x)]′and ƒ′(x)g′(x) for ƒ(x) = a/(1 - x) and g(x) = bx, and verify that the two expressions are equal.