Product Rule of Differentiation: Finding Derivatives Made Easy

Finding the derivative of product of two functions:

As mentioned earlier, we can’t find the derivative of the product of two function by multiplying the derivative of each individual function. This only works when dealing with addition/subtraction. Instead, we need to use the formula:

\( (f(x) \times g(x))’ = f'(x) \cdot g(x) + g'(x) \cdot f(x) \)

So basically, finding the derivative of first function and multiplying it by the second, and adding it to the derivative of second function multiplied by the first function.

You can sometimes avoid using product rule:

Sometimes it can be easier to simplify the given functions by multiplying them first, and taking the derivative of the resulting expression, meaning to avoid using product rule. Both methods yield the same answer, yet sometimes one method can be easier than the other.

For instance, the derivative of \( 2x(x^2) \)

First, using the product rule:

\( f(x) = 2x \cdot x^2 \Rightarrow f'(x) = 2x \cdot 2x + 2 \cdot x^2 = 4x^2 + 2x^2 = 6x^2 \)

Now, if simplifying first:

\( f(x) = 2x \cdot x^2 = 2x^3 \Rightarrow f'(x) = 3 \cdot 2x^{2} = 6x^2 \)

Product rule involving other functions:

Multiplication of other functions, like radicals and trigonometric functions, follows the same principle. For example:

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\( f(x) = x \sin x \Rightarrow f'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x \)

Here is another example:

\( f(x) = \cos x \cdot x^{1/3} \Rightarrow f'(x) = -\sin x \cdot x^{1/3} + \frac{1}{3} \cos x \cdot x^{-2/3} \)

Finding the derivative of product of three functions:

Here is a more complex example, involving chain rule:

\( f(x) = \cos 2x \cdot (3x)^{1/3} \Rightarrow f'(x) = -2 \sin 2x \cdot (3x)^{1/3} + \cos 2x \cdot \frac{1}{3} \cdot (3x)^{-2/3} \cdot 3 \)

If you have three functions and want to apply the product rule for differentiation, the process involves a bit more complexity than with two functions.

1. Differentiate the first function and multiply it by the unchanged second and third functions.
2. Keep the first function unchanged, differentiate the second function, and multiply by the unchanged third function.
3. Keep the first and second functions unchanged and differentiate the third function.

For the functions \( f(x), g(x), h(x) \), the mathematical formula for the derivative of product of these functions, is:

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

Here’s an example:

\( f(x) = x^2, \ g(x) = x, \ h(x) = x \)

 \( f'(x) = 2x, \ g'(x) = 1, \ h'(x) = 1 \)

\(\Rightarrow f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) = 2x \cdot x \cdot x + x^2 \cdot 1 \cdot x + x^2 \cdot x \cdot 1 \)

\(= 2x^3 + x^3 + x^3 \)

\(= 4x^3 \)

Recommended EffortlessMath Books

For a calculus prep workbook that builds derivative rules into a complete Calc I toolkit, the Pre-Calculus for Beginners covers the algebra and trig foundation derivatives depend on. For deeper derivative work including chain rule, implicit differentiation, and optimization, the Algebra II for Beginners nails the algebra skills you’ll lean on in every calc problem.

Frequently Asked Questions

What is the product rule of differentiation formula?

The product rule says that the derivative of f(x)·g(x) is f′(x)·g(x) + f(x)·g′(x). In words, differentiate the first function and multiply by the second, then add the first function times the derivative of the second. It is not f′(x)·g′(x), which is the most common mistake students make when they first learn the rule.

What is an example of the product rule of differentiation?

Take y = x² sin x. Let f = x² and g = sin x, so f′ = 2x and g′ = cos x. The product rule gives y′ = 2x·sin x + x²·cos x. For a quick check on a simpler case, 2x·x² gives 2·x² + 2x·2x = 6x², which matches differentiating 2x³ directly.

How do you use the product rule of differentiation for 3 functions?

For three factors, (fgh)′ = f′gh + fg′h + fgh′. Each function takes one turn being differentiated while the other two stay the same, and you add the three terms. For example, the derivative of x·eˣ·sin x is eˣ sin x + x eˣ sin x + x eˣ cos x.

What is the product rule of differentiation in terms of u and v?

If u and v are both functions of x, then d(uv)/dx = u·dv/dx + v·du/dx, often written (uv)′ = u′v + uv′. For example, with u = x³ and v = ln x, the derivative is 3x²·ln x + x³·(1/x) = 3x² ln x + x².

What is the proof of the product rule of differentiation?

Start from the limit definition: the derivative of f·g is the limit as h → 0 of [f(x+h)g(x+h) − f(x)g(x)]/h. Add and subtract f(x+h)g(x) in the numerator, then split the fraction into f(x+h)·[g(x+h) − g(x)]/h plus g(x)·[f(x+h) − f(x)]/h. Taking the limit gives f(x)g′(x) + g(x)f′(x).

What are the main methods for finding derivatives?

Most derivatives come from a few rules: the power rule, the constant multiple and sum rules, the product rule, the quotient rule and the chain rule. Sometimes you can skip the product rule by multiplying out first. For example, 2x·x² simplifies to 2x³, whose derivative is 6x², the same answer the product rule gives.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• Derivative of logarithmic functions
• Second derivatives
• Sketching curves with derivatives
• Vector function derivatives
• Derivatives of polar coordinates

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