The Product Rule is a way of differentiating complex expressions that are the product of 2 simpler expressions. This can be expressed algebraically as such:
\(h'(x) = [\textcolor{red}{f(x)}\textcolor{blue}{g(x)}]' = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{blue}{g'(x)}\textcolor{red}{f(x)}\)
- \(h'(x)\) represents the derivative of the product function
- \(\textcolor{red}{f(x)}\) represents the first term/expression
- \(\textcolor{red}{f'(x)}\) represents the derivative of \(f(x)\)
- \(\textcolor{blue}{g(x)}\) represents the second term/expression
- \(\textcolor{blue}{g'(x)}\) represents the derivative of \(g(x)\)
In short, Product Rule, takes the derivative of \(f(x)\) and multiplies it by \(g(x)\). It then adds this value to \(f(x)\) multiplied by the derivative of \(g(x)\).
It's important to note that using the Product Rule isn't the same as multiplying the derivatives of the respective expressions.
Example
Differentiate \(h(x) = (3x + 2)(x-4)\).
First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives:
\(\textcolor{red}{f(x) = 3x + 2}\)
\(\textcolor{red}{f'(x) = 3}\)
\(\textcolor{blue}{g(x) = x - 4}\)
\(\textcolor{blue}{g'(x) = 1}\)
Next, we can differentiate the entire expression:
\(h'(x) = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{blue}{g'(x)}\textcolor{red}{f(x)}\)
\(h'(x) = \textcolor{red}{3}(\textcolor{blue}{x - 4}) + \textcolor{blue}{1}(\textcolor{red}{3x + 2})\)
\(h'(x) = 3x-12 + 3x + 2\)
\(h'(x) = 6x - 10\)
Therefore, we can determine that \(h'(x) = 6x - 10\).
Differentiate the expression \(p(x) = (x^2+1)(1-x)\)
Show Answer
First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives:
\(\textcolor{red}{f(x) = x^2+1}\)
\(\textcolor{red}{f'(x) = 2x}\)
\(\textcolor{blue}{g(x) = 1-x}\)
\(\textcolor{blue}{g'(x) = -1}\)
Next, we can differentiate the entire function:
\(p'(x) = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{blue}{g'(x)}\textcolor{red}{f(x)}\)
\(p'(x) = \textcolor{red}{2x}(\textcolor{blue}{1-x}) \textcolor{blue}{-1}(\textcolor{red}{x^2+1})\)
\(p'(x) = 2x-2x^2-x^2-1\)
\(p'(x) = x^2+2x-1\)
Therefore, we can determine that \(p'(x) = x^2+2x-1\).