How to differentiate using the product rule

In Add Math you often meet expressions where two functions are multiplied together, such as $y=(2x-1)(x^{2}+1)$, $y=x^{3}(2x+5)$ or $y=x\sqrt{4x+1}$. The product rule is the tool for differentiating this kind of expression when the two factors are not simply combined into one term.

It is tempting to differentiate each factor and multiply the derivatives, but that is wrong: $\frac{d}{dx}(uv)$ is not $\frac{du}{dx}\times\frac{dv}{dx}$. Some simple products, like $(2x-1)(x^{2}+1)$, can be expanded first and differentiated term by term, but many cannot be expanded neatly, especially when a root or an awkward power is involved.

The product rule gives a reliable, exam-marked method that always works, and it appears throughout the Differentiation chapter in tangents, rates of change and stationary-point problems.

Reach for the product rule whenever you see two functions of $x$ multiplied together and neither is a constant. The clearest signal is a genuine product such as $x^{2}(3x-1)$, $(x+2)(2x-5)$ or $x\sqrt{2x+1}$, where both factors change as $x$ changes.

Be careful to tell the three rules apart. Use the product rule when two functions are multiplied; use the chain rule when one function is wrapped inside another, like $(3x-1)^{4}$; and use the quotient rule when one function is divided by another.

A quick test: if you can circle two separate factors sitting side by side, and both contain $x$, the product rule is the right choice. When a factor is itself a bracket to a power, you will use the chain rule inside the product rule.

1. 1

   Split into two factors

   Write the expression as a product and name the factors: let $u$ be the first and $v$ be the second, so $y=uv$.

2. 2

   Differentiate each factor

   Find $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately.

3. 3

   Apply the formula

   Substitute into $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Keep each factor with the derivative of the other one.

4. 4

   Expand each product

   Multiply out the two brackets carefully, watching signs.

5. 5

   Collect like terms

   Add the results and simplify to a single tidy expression.

6. 6

   Use it if a value is asked

   Substitute the given $x$-value to find a gradient, or set $\frac{dy}{dx}=0$ for a stationary point.

• Writing $\frac{dy}{dx}=\frac{du}{dx}\times\frac{dv}{dx}$, multiplying the two derivatives. This is the single biggest error; the rule adds two terms, it does not multiply.
• Pairing each factor with its own derivative. The rule keeps the first factor with the derivative of the second, and the second factor with the derivative of the first.
• Sign slips when expanding brackets, especially with a negative term such as $(2x-1)(2x)$.
• Forgetting the chain rule for a factor that is a bracket to a power, e.g. differentiating $(3x-1)^{4}$ inside the product.
• Leaving the answer unsimplified when the question asks for a single expression or a value.

The step students most often get wrong is the pairing: they write $u'\times v'$ instead of $u\,v'+v\,u'$, and the whole answer collapses. In a one-to-one lesson our teachers watch you set out $u$, $v$, $\frac{du}{dx}$ and $\frac{dv}{dx}$ in a neat little table before you touch the formula, so the terms can never be crossed.

Because your working is shown line by line, the two terms of the rule stay visible and every method mark is easy to award. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To see how we teach the product rule, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.