The product rule is one of the differentiation rules you need to know. This rule is used when you are differentiating the products of two functions.
Product rule formula and function notation
There is a formula that you can follow to use the product rule. It is important to try and remember this formula as it is not usually in exam formula booklets. If when u and v are functions of x then the product rule formula is:
\[\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\]
This can also be written in function notation,
If \(f(x) = g(x)h(x)\) then \(f'(x) = g(x)h'(x) + h(x)g'(x)\)
Examples of the product rule
To understand the product rule further, let's look through some examples of using it.
If \(y = 5xe^2\) find \(\frac{dy}{dx}\):
First, you can start by looking at your product rule formula and finding each aspect of the formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
If \(y = 5xe^2\), \(u = 5x\) and \(v = e^2\)
To find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\) you can differentiate your u and v:
\(\frac{du}{dx} = 5\) \(\frac{dv}{dx} = 0\)
Now that you have all aspects of your formula, you can solve it to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = (5x)(0) + (e^2)(5) = 5e^2\]
You may also be asked to differentiate a trigonometric function using the product rule.
If \(y = (4\sin{x})e^2\) find \(\frac{dy}{dx}\)
As before, you can start by finding each part of the formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
If \(y = (4\sin{x})e^2\), \(u = 4\sin x\) and \(v = e^2\)
Now you can differentiate your u and v to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):
\(\frac{du}{dx} = 4\cos x\) \(\frac{dv}{dx} = 0\)
Finally, you can substitute each part into your formula to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = (4\sin x)(0) + (e^2)(4\cos x)\]
\[\frac{dy}{dx} = 4e^2\cos x\]
You may even be asked to use the product rule in function notation, so let's work through an example.
If \(f(x) = 2x^2(x^2 + 4)\) find \(f'(x)\)
Once again, you can start by breaking down the formula for the product rule in function notation and finding each part.
\(f'(x) = g(x)h'(x) + h(x)g'(x)\)
If \(f(x) = 2x^2(x^2 + 4)\), \(g(x) = 2x^2\) and \(h(x) = x^2 + 4\)
Next, you can differentiate g(x) and h(x) to find the derivatives, g'(x) and h'(x):
\(g' = 4x\) \(h'(x) = 2x\)
Now that you have each part of the formula, you can solve it to find f'(x):
\(f'(x) = (2x^2)(2x) + (x^2 + 4)(4x) = 2(4x^3 + 8x)\)
If \(y = \ln x(x^2)\) find \(\frac{dy}{dx}\)
To begin with, let's look at the product rule formula and find each aspect of it:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
If \(y = \ln x(x^2)\), \(u = \ln x\) and \(v = x^2\)
Let's differentiate the terms to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):
\(\frac{du}{dx} = \frac{1}{x}\) \(\frac{dv}{dx} = 2x\)
Now you can put each part into the formula to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
\[\frac{dy}{dx} = (\ln{x})(2x) + (x^2)(\frac{1}{x}) = 2x \ln{x} + x\]
Product Rule - Key takeaways
The product rule is one of the differentiation rules.
The product rule can be used when differentiating the products of two functions.
When using the product rule, you can either use the formula in y form or in the function notation form.
You may also need to differentiate trigonometric functions using the product rule.
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