Product Rule of Differentiation

By the product rule (f·g)′ = f′·g + f·g′. So you differentiate the first factor and multiply it by the unchanged second, then add the unchanged first factor times the derivative of the second. Example: for x²·e^x, f = x² with f′ = 2x and g = e^x with g′ = e^x. This gives (x²·e^x)′ = 2x·e^x + x²·e^x = e^x·(2x + x²). It is important that you do not simply multiply the two derivatives; that is the most common mistake. Both summands belong, because in the limit both factors change simultaneously.

Because when differentiating a product both factors change simultaneously and both change contributions must be taken into account. In the difference quotient of f·g one adds and subtracts an intermediate term, which creates two contributions: one from the change of f with g held fixed, one from the change of g with f held fixed. This gives (f·g)′ = f′·g + f·g′. Taking only f′·g′ would omit the mixed contributions and give a wrong result. A simple test: for f = g = x, (x·x)′ = (x²)′ = 2x, whereas f′·g′ = 1·1 = 1, which is obviously wrong.

The product rule applies when two functions are multiplied together, that is for an expression of the form f(x)·g(x), for example x²·sin x. The chain rule, by contrast, applies to nested functions where one function is inserted into another, that is f(g(x)), for example sin(x²). The key is the structure: a times sign between two independent functions points to the product rule, an inner function in the argument of an outer one to the chain rule. Often both rules appear combined in one problem, for example x²·sin(3x); then you first apply the product rule and use the chain rule for the nested factor.

The product rule extends to three factors. For f·g·h, (f·g·h)′ = f′·g·h + f·g′·h + f·g·h′. So you differentiate one factor at a time and leave the other two unchanged, then add all three terms. The pattern continues for any number of factors: in each summand exactly one factor is differentiated. Alternatively you can combine two factors into one and apply the ordinary product rule twice. Example: (x·sin x·e^x)′ = 1·sin x·e^x + x·cos x·e^x + x·sin x·e^x. Make sure to form only a single derivative in each term.

A good check is to test the result at a known point with an expected property. For (x²·sin x)′ = 2x·sin x + x²·cos x the slope at x = 0 should be zero, because there the function itself is zero and has a local extremum: 2·0·0 + 0²·1 = 0, which fits. In addition you can evaluate the product in special cases, for example when one factor is constant; then the product rule must give the same as simple differentiation with a constant prefactor. Checking at a rounded numerical value with the difference quotient also confirms whether your symbolic derivative is correct.