Finding Derivatives Made Easy! Product Rule of Differentiation

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?

The product rule tells you how to differentiate the product of two functions: \((fg)’ = f’g + fg’\). Each function gets multiplied by the other function’s derivative, and you add the two results. It’s required any time you take the derivative of two functions multiplied together.

Why isn’t (fg)’ just f’g’?

Because the derivative measures how the product changes, and changes happen for two reasons: \(f\) changing (with \(g\) constant), and \(g\) changing (with \(f\) constant). \(f’g\) captures the first; \(fg’\) captures the second. Adding them gives the total rate of change. Just multiplying the derivatives loses both of those pieces.

When do I use the product rule?

Use it any time you have two functions multiplied: \(x \ln x\), \(\sin x \cos x\), \(e^x x^2\), \((x^2+1)(x^3-2)\), etc. If the expression is a constant times a function (like \(5x^2\)), the product rule isn’t needed — the constant just pulls out front by the constant multiple rule.

How does the product rule work for three functions?

\((fgh)’ = f’gh + fg’h + fgh’\). Each function gets its turn being differentiated while the other two stay as-is. The pattern extends — for four functions, you’d get four terms, each with one of the four factors differentiated.

What’s the most common product rule mistake?

Writing \((fg)’ = f’g’\). This is wrong — and very tempting because it feels symmetric with the sum rule. Always write out both terms: \(f’g + fg’\). Another common mistake is forgetting the sign when one of the derivatives is negative.

Can I use the product rule instead of expanding?

Yes, but expanding is sometimes faster. For \((x^2+1)(x^3-2)\), you can either apply the product rule directly or expand to \(x^5 + x^3 – 2x^2 – 2\) and use the power rule term by term. Both give the same answer; expanding is faster when the polynomials are simple.

How does the product rule relate to the quotient rule?

The quotient rule \((f/g)’ = (f’g – fg’)/g^2\) can be derived from the product rule by writing \(f/g = fg^{-1}\) and applying the product rule plus the chain rule on \(g^{-1}\). Many calc students prefer to memorize quotient as a separate rule, but they’re related.

What if one factor is a constant?

If \(f(x) = k\) (a constant), then \(f'(x) = 0\), so \((fg)’ = 0 \cdot g + k \cdot g’ = kg’\). That’s the constant multiple rule. The product rule still works — it just collapses because the constant’s derivative is zero.

Does the product rule work for vector functions?

Yes, with appropriate definitions of multiplication. For a scalar function \(f\) and vector \(\vec{v}\), \((f\vec{v})’ = f’\vec{v} + f\vec{v}’\). For two vectors with a dot product, \((\vec{u}\cdot\vec{v})’ = \vec{u}’\cdot\vec{v} + \vec{u}\cdot\vec{v}’\). Cross products and matrix products follow analogous rules.

Where does the product rule show up on tests?

Every calculus test from AP Calc AB to college finals. Expect at least 3-5 problems requiring the product rule on any derivatives unit test. It also feeds into implicit differentiation, related rates, optimization, and integration by parts (which is the product rule run backwards).

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