## Calculus Tutorial

*Product Rule Differentiation (Single Variable)*

#### Intro

The product rule is used when you are finding the derivative of a function that has one dependent variable expression multiplied by another dependent variable expression. Compare this function:

f(x) = x^{2} + 2x^{5}

…with this function:

g(x) = x^{4}cos(x)

Function g(x) can be thought of as two separate functions multiplied together:

h(x) = x^{4}

j(x) = cos(x)

g(x) = h(x)j(x) = x^{4}cos(x)

That is why taking the derivative of g(x) means taking the derivative of the **product** of more than one function, which is why we will use the product rule.

#### Sample Problem

#### Solution

Remember that a function like f(x) is thought of as more than one function multiplied together. The product rule is as follows given a function f(x):

f(x) = g(x)h(x)

f'(x) = g'(x)h(x) + g(x)+h'(x)

In this example, our g(x) and h(x) are defined when there are two distinct expressions of the dependent variable; in other words, we cannot reduce or combine x^{3} and ln(x) together, so these can be considered two different functions of x. Following the formula, we can break f(x) like so:

f(x) = x^{3}(ln(x)) = g(x)h(x)

g(x) = x^{3}

h(x) = ln(x)

Now that we’ve defined the components of f(x), we can plug in g(x) and h(x) into the product rule, using their respective derivatives:

g'(x) = 3x^{2}

h'(x) = 1/x

f'(x) = 3x^{2}ln(x) + x^{3}(1/x)

We can reduce and combine terms further to get the final answer:

f'(x) = 3x^{2}ln(x) + x^{2}

f'(x) = x^{2}(3ln(x) + 1)

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