The Principles of Derivatives - Product Rule
In the article “The Principles of Derivatives- Secant Line, Tangent Line, and Power Rule”, we learned that a derivative is a function that represents another function’s rate of change. This was explained through the knowledge of the Secant Line and the Tangent Line. We also learned about the product rule which is a way to find the derivative of simple equations that have a power in them such as x^2. Well, now it is time to learn how to differentiate more complex functions using the product rule.
The product rule is used to differentiate the product of two or more expressions. For example, if we have the function y=(x^3+3x^2+1)(x+1), we can use the product rule here to find the derivative since this function involves a product of two expressions. Though we can multiply the two individual expressions together and differentiate the function using the power rule, it can be easier and much faster to find the derivative of the function using the product rule in complex expressions like this one. The product rule states that the derivative of a function that contains two or more expressions is found using the formula f’(x)*g(x)+f(x)(g’x).
What does this formula mean? First, let’s define f(x) as the first expression and g(x) as the second expression. Well, it means that the derivative of a product rule is found by taking the derivative of f(x) and multiplying that by the expression g(x) itself, and then multiplying the derivative of g(x) and multiplying that by the expression f(x) itself, and adding the two resulting products together. If this is confusing in any way or a bit hard to understand, don’t worry, we will do an example together.
In this example, let’s use the equation y=(x^3+3x^2+1)(x+1) take (x^3+3x^2+1) as f(x) and (x+1) as g(x). So, first, we would differentiate what we just defined as f(x) using the product rule. This would give us that f’(x) as (3x^2+6x). We then do the same thing for g(x) and find that g’(x) is equal to 1. Then, we would multiply f’(x) and g(x) which would be (3x^2+6x)(x+1), and also multiply f(x) and g’(x) which would be (x^3+3x^2+1)(1), and then you would add the two resulting expressions. The derivative of the function y=(x^3+3x^2+1)(x+1) is equal to ((3x^2+6x)(x+1)+(x^3+3x^2+1)(1)) which simplifies to 4x^3+12x^2+6x+1.
One last important thing to note about the product rule is that it doesn’t have to be used in cases where the two expressions can easily be simplified. For example, if an equation is y=(x^3)(x^2), you can easily simplify this equation to be x^5 and use the product rule to differentiate. Well, now that you know how the product rule works, you are one step further in the journey in calculus. I hope these examples have given you a better insight into how the product rule works.
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