The Product Rule for Derivatives (Including Proof)

The product rule is a formal rule for differentiating problems where one function is multiplied by another. The rule follows from the limit definition of the derivative and states that the derivative of $f(x) \times g(x)$ will be $f'(x) \times g(x) + f(x) \times g'(x)$ . This rule only works if f and g are both differentiable functions. This pattern works for any number of functions as I will demonstrate below.

With a Single Function

With just one function, say $f(x)$ , we can arbitrarily define a second function $g(x) = 1$ and multiply the two together to create a new function we will call $h(x) = f(x) \times g(x) = f(x) \times 1 = f(x)$ .

Can you see how $h(x)$ is the same function as $f(x)$ ?

Now if we take the derivative of $h(x)$ we get $h'(x) = f'(x) \times g(x) + f(x) \times g'(x)$ .

Now if we recall our constant rule, we know that the derivative of a constant (in this case 1) will be zero. So if we substitute in zero we get $h'(x) = f'(x) \times g(x) + f(x) \times 0$ .

Now if we substitute in $g(x)$ we get $h'(x) = f'(x) \times 1 = f'(x)$ .

We have now shown that we can use our product rule for derivatives on a single function by defining a second function equal to one.

With Three Functions

Let $k(x) = f(x) \times g(x) \times h(x)$ , assuming that f, g, h are differentiable functions.

We can arbitrarily define a new function $j(x) = f(x) \times g(x)$ and we can rewrite $k(x)$ as $k(x) = j(x) \times h(x)$ .

We can now use our two-function definition of the product rule to find the derivative:

k'(x) = j'(x) \times h(x) + j(x) \times h'(x)

We can now independently find the derivative of $j(x)$ using the same method:

j'(x) = f'(x) \times g(x) + f(x) \times g'(x)

We can now plug $j'(x)$ back into our derivative of $k'(x)$ to get:

k'(x) = f'(x) \times g(x) \times h(x) + f(x) \times h'(x) \times g(x) + f(x) \times h(x) \times g'(x)

This pattern is infinitely long, and if you are wondering why, you can read about it in the proof below.

Proof

To start off our proof we will write out the definition of the derivative:

(fg)'(x) = \displaystyle\lim_{h \to 0} \frac{(fg)(x+h) - (fg)(x)}{h}

(f(x) \times g(x))' = \displaystyle\lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}

We can now add and subtract $f(x+h)g(x)$ from the numerator, which is allowed because by adding then subtracting the same thing we will not change our end result:

= \displaystyle\lim_{h \to 0} \frac{f(x+h)g(x+h) + f(x+h)g(x) - f(x+h)g(x) - f(x)g(x)}{h}

We will then factor $f(x+h)$ from the first and second-last terms:

= \displaystyle\lim_{h \to 0} \frac{f(x+h)[g(x+h) - g(x)] + f(x+h)g(x) - f(x)g(x)}{h}

We will then factor $g(x)$ from our remaining 2 terms in the numerator:

= \displaystyle\lim_{h \to 0} \frac{f(x+h)[g(x+h) - g(x)] + g(x)[f(x+h) - f(x)]}{h}

Now if we recall the addition property of limits:

If a is a real number and both $\displaystyle\lim_{x \to a} f(x)$ and $\displaystyle\lim_{x \to a} g(x)$ exist, then $\displaystyle\lim_{x \to a} [f(x) + g(x)] = \displaystyle\lim_{x \to a} f(x) + \displaystyle\lim_{x \to a} g(x)$ .

Since we have already stated that $f(x)$ and $g(x)$ must be differentiable functions, we can apply this rule here to expand our equation:

= \displaystyle\lim_{h \to 0} \frac{f(x+h)[g(x+h) - g(x)]}{h} + \displaystyle\lim_{h \to 0} \frac{g(x)[f(x+h) - f(x)]}{h}

Now if we recall the multiplicative property of limits:

As long as both $\displaystyle\lim_{x \to a} f(x)$ and $\displaystyle\lim_{x \to a} g(x)$ exist, then $\displaystyle\lim_{x \to a} [f(x)g(x)] = \displaystyle\lim_{x \to a} f(x) \times \displaystyle\lim_{x \to a} g(x)$ .

Since we have already stated that $f(x)$ and $g(x)$ must be differentiable functions, we can apply this rule here to expand our equation:

= \displaystyle\lim_{h \to 0} f(x+h) \times \displaystyle\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} + \displaystyle\lim_{h \to 0} \frac{g(x)[f(x+h) - f(x)]}{h}

We can now simplify the leftmost limit. Since $f(x)$ is differentiable and since all differentiable functions are continuous, we can conclude that $f(x)$ must be continuous, allowing us to substitute what h is approaching directly into $f(x+h)$ , giving us $f(x+0) = f(x)$ .

We can now substitute that into our equation:

= f(x) \displaystyle\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} + \displaystyle\lim_{h \to 0} \frac{g(x)[f(x+h) - f(x)]}{h}

We can then apply the multiplicative property of limits again on the right limit to get:

= f(x) \displaystyle\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} + \displaystyle\lim_{h \to 0} g(x) \times \displaystyle\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Since our limit variable h is not involved in g(x) in any way, we can simplify it further to:

= f(x) \displaystyle\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} + g(x) \displaystyle\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Now if we recall the definition of the derivative:

f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

g'(x) = \displaystyle\lim_{h \to 0} \frac{g(x+h) - g(x)}{h}

We can substitute those into our equation:

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

We have now completed our proof.

Example 1: $x^2$

$f(x) = x^2$

Now you may be wondering how this rule is applicable if we only have one function, but what we are able to do here is define two functions that when multiplied together equate to $f(x)$ . In this case we can let $g(x) = h(x) = x$ since $f(x) = g(x) \times h(x)$ .

We can now solve for the derivative:

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

Since the derivative of x is just one (as it is a linear line with a slope of 1), we can substitute $h'(x)$ and $g'(x)$ with 1. To get:

f'(x) = 1 \times g(x) + h(x) \times 1

We can substitute in the functions and simplify to get:

f'(x) = x + x = 2x

Which is exactly what the power rule would have given you. This example was purely for demonstration. I would recommend you read up on the power rule to take derivatives in the form of $x^n$ as it is much faster than this.

Example 2: $(20x+1) \times x^3$

$f(x) = (20x+1) \times x^3$

Let’s start by defining our functions: let $h(x) = (20x+1)$ and $g(x) = x^3$ since $f(x) = g(x) \times h(x)$ .

We can write:

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

The derivative of $h(x)$ is $h'(x) = 20$ .

The derivative of $g(x)$ is $g'(x) = 3x^2$ (this is done with power rule).

Now we can substitute into $f'(x)$ to get:

f'(x) = 20 \times x^3 + (20x+1) \times 3x^2

Now we can simplify:

f'(x) = 80x^3 + 3x^2

Example 3: $(x^2+x+6)^2$

$f(x) = (x^2+x+6)^2$

Without the product rule we would be stuck using either the chain rule or expanding out, but with the product rule we can define $g(x) = h(x) = (x^2+x+6)$ .

Then $f(x) = g(x) \times h(x)$ and since $f'(x) = g'(x)h(x) + g(x)h'(x)$ we can differentiate:

g'(x) = h'(x) = 2x + 1

Now we can substitute into $f'(x)$ to get:

(2x+1)(x^2+x+6) + (x^2+x+6)(2x+1)

We can now simplify to get:

2 \times (2x+1)(x^2+x+6)

We can validate our result using chain rule or by expanding before taking the derivative.

Example 4: $x|x|$

$f(x) = x|x|$

Now, you may be thinking that we should let $h(x) = x$ and $g(x) = |x|$ , but that is not allowed. Because by the definition of the product rule, both functions must be differentiable, and $|x|$ is not a differentiable function (it has a corner at x=0).

So we cannot directly apply the product rule here. What we would have to do is create a piecewise function where it becomes $-x^2$ when $x \leq 0$ or $x^2$ when $x > 0$ :

f(x) = \begin{cases} -x^2 & x \leq 0 \\ x^2 & x > 0 \end{cases}

Then we can differentiate the two segments independently and use the definition of the derivative to find the derivative at their point of intersection.

Example 5: $x \times \frac{1}{x}$

$f(x) = x \times \frac{1}{x}$

This one is a little bit strange because using the product rule you will end up with:

f'(x) = \frac{1}{x} - \frac{x}{x^2}

But if you leave it like this you will end up with a derivative that is not defined at x=0. In reality you need to simplify to:

f'(x) = 0

For your solution to be valid.

Pro Tips

All functions must be differentiable for you to be able to use the product rule
Can be used with any number of functions from 1 to infinity
Can be used in place of the chain rule (with limited success)
Can be used in place of the power rule (with limited success)

THANKS FOR READING

With a Single Function

With Three Functions

Proof

Example 1: x2x^2

Example 2: (20x+1)×x3(20x+1) \times x^3

Example 3: (x2+x+6)2(x^2+x+6)^2