In studying the fields and concepts of Calculus and Physics one finds that derivatives are used commonly in many different fields and types of applications. A derivative is simply the resulting change that occurs in a function with respect to its independent variable up to a certain limit.

There are several rules we can use to find the derivative of a function including the Power, Quotient, Chain and the Product rules. The Product rule is applicable when you are finding the derivative of a function that can be identified as being the product of two other functions, each of which must be differentiable.

The Product Rule states:

Let F(x) = (j(x))(k(x))

F’(x) = d/dx[(j(x))(k(x))] = j(x)[k’(x)] + k(x)[j’(x)]

With this being expressed in words, the Product Rule states that the derivative of function F (a product consisting of function j and function k each with respect to x) is the summation of function j multiplied by the derivative of function k plus function k multiplied by the derivative of function j.

Product Rule differentiation can also be used with other rules of differentiation such as the Power, Quotient and Chain rules for more complicated functions as appropriate.

Here is an example to further explain how the Product Rule works.

Let’s find the derivative of F(x) = (2x^{1/2 }– 4x) x^{2}.

First, we will rewrite this function in the form of the Product rule.

F’(x) = (2x^{1/2 }– 4x) d/dx(x^{2}) + (x^{2}) d/dx(2x^{1/2 }– 4x)

^{ }

The Power rule can now be used to find the individual derivatives of functions j(x) and k(x) and these values can then be substituted back in to the formula.

F’(x) = (2x^{1/2 }– 4x)(2x) + (x^{2})(x ^{-1/2 }– 4 )

Now each of the products can be solved and added to obtain the final derivative.

F’(x) = 4x^{3/2 }– 8x^{2} + x^{3/2 }– 4x^{2}

F’(x) = 5x^{3/2 }– 12x^{2}

To illustrate how multiple rules of differentiation can be used in conjunction with one another, let’s look at a more complex example that also incorporates the use of the Quotient rule.

Find the derivative of F(x) = 1 / (x^{-1/2})(1– x^{2}).

We will begin by rewriting this function into the form of its derivative using the Quotient rule.

F’(x) = [ (x^{-1/2})(1– x^{2}) d/dx[1] – (1) d/dx[(x^{-1/2})(1– x^{2})] ] / [(x^{-1/2})(1– x^{2})]^{2}

Since the derivative of a constant is zero, the first term of the numerator is also zero. For the second term in the numerator, since it is the product of two differentiable functions, we can use the Product Rule to find its derivative.

F’(x) = [ –(1) d/dx[(x^{-1/2})(1– x^{2})] ] / [(x^{-1/2})(1– x^{2})]^{2}

d/dx[(x^{-1/2})(1– x^{2})] = (x^{-1/2}) d/dx(1– x^{2}) + (1– x^{2}) d/dx(x^{-1/2})

d/dx[(x^{-1/2})(1– x^{2})] = (x^{-1/2})( –2x) + (1– x^{2})( –1/2x^{-3/2})

d/dx[(x^{-1/2})(1– x^{2})] = –2x^{1/2} –1/2x^{-3/2} + 1/2x^{1/2}

^{ }

d/dx[(x^{-1/2})(1– x^{2})] = –3/2x^{1/2} –1/2x^{-3/2}

We will now substitute this derivative into F’(x) and use algebra to solve.

F’(x) = [ –(1) (–3/2x^{1/2} –1/2x^{-3/2})] / [(x^{-1/2})(1– x^{2})]^{2}

F’(x) = [ (3/2x^{1/2} +1/2x^{-3/2})] / (x^{-1/2 }– x^{3/2})^{2}

F’(x) = (3/2x^{1/2} +1/2x^{-3/2}) / [ (x^{-1/2 }– x^{3/2}) (x^{-1/2 }– x^{3/2}) ]

F’(x) = (3/2x^{1/2} +1/2x^{-3/2}) / [ x^{-1 }– x^{1} – x^{1} + x^{3}]

F’(x) = (3/2x^{1/2} +1/2x^{-3/2}) / ( x^{-1 }– 2x^{}+ x^{3})

Resource:

Stewart, James. (2001). Calculus Concepts And Contexts (5^{th} Ed.). Pacific Grove, CA: Wadsworth Group.

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