The product rule is one of the basic rules of differential calculus. It is used when a function is given as the product of two other functions.
At first glance, it may seem that the derivative of such an expression can be found very easily: differentiate each factor separately and then multiply the results. But this raises an important question: does this approach take into account the change of the entire product? In fact, it does not, because when the argument changes, not just one factor changes. Both parts of the product change at the same time.
That is why the product of functions needs a separate rule. This rule correctly takes into account the contribution of each factor to the overall change of the function.
Product Rule: The Basic Formula for Two Functions
Suppose we have two differentiable functions \( u(x) \) and \( v(x) \). Let us consider a function that is their product:
\[
y=u(x)\cdot v(x).
\]
Then the derivative of this product is found by the formula
\[
y’=u'(x)\cdot v(x)+u(x)\cdot v'(x).
\]
This is the basic formula of the product rule. It is also often written using the differentiation operator:
\[
\frac{d}{dx}\bigl(u(x)\cdot v(x)\bigr)=u'(x)\cdot v(x)+u(x)\cdot v'(x).
\]
This notation is especially convenient because it immediately shows that we are finding the derivative of the entire product \( u(x)\cdot v(x) \), not only one of its factors. That is exactly why the formula contains two terms.
The first term \( u'(x)\cdot v(x) \) accounts for the change of the first factor \( u(x) \), while the second factor \( v(x) \) remains unchanged at that moment. The second term \( u(x)\cdot v'(x) \) accounts for the change of the second factor \( v(x) \), while the first factor \( u(x) \) remains unchanged.
So, the derivative of a product is not equal to the product of the derivatives. Instead, we take the derivative of the first function, multiply it by the second function, and then add the product of the first function and the derivative of the second. This is how the formula takes into account the change of both factors.
Step by Step: Deriving the Formula from the Definition of the Derivative
Now let us look closely at where the product rule formula comes from. To do this, we will use the definition of the derivative using a limit. Suppose the function \( y \) is given as the product of two differentiable functions:
\[
y=u(x)\cdot v(x).
\]
By the definition of the derivative, we have
\[
y’=\lim_{h\to 0}\frac{y(x+h)-y(x)}{h}.
\]
Since \( y(x)=u(x)\cdot v(x) \), the value of this function at the point \( x+h \) is
\[
y(x+h)=u(x+h)\cdot v(x+h).
\]
Now substitute these expressions into the definition of the derivative:
\[
y’=\lim_{h\to 0}\frac{u(x+h)\cdot v(x+h)-u(x)\cdot v(x)}{h}.
\]
At this stage, we have obtained the correct expression for the derivative, but it does not yet look like the product rule formula. What should we do next? We need to transform the numerator so that the change of the function \( u(x) \) appears separately, and the change of the function \( v(x) \) also appears separately.
To do this, we add and subtract the same expression:
\[
u(x+h)\cdot v(x).
\]
This does not change the value of the numerator, because we add and subtract the same term at the same time. However, this step helps us split the change of the product into two parts. In one part, we will see the change of the function \( v(x) \), and in the other part, the change of the function \( u(x) \).
So, we write:
\[
y’=\lim_{h\to 0}
\frac{
u(x+h)\cdot v(x+h)-u(x+h)\cdot v(x)+u(x+h)\cdot v(x)-u(x)\cdot v(x)
}{h}.
\]
Now let us group the terms so that the first group has the common factor \( u(x+h) \), and the second group has the common factor \( v(x) \):
\[
y’=\lim_{h\to 0}
\frac{
u(x+h)\cdot \bigl(v(x+h)-v(x)\bigr)+v(x)\cdot \bigl(u(x+h)-u(x)\bigr)
}{h}.
\]
In this form, it is already clear why we added and subtracted the intermediate expression. The first difference \( v(x+h)-v(x) \) shows the change of the function \( v(x) \), while the second difference \( u(x+h)-u(x) \) shows the change of the function \( u(x) \). In other words, we transformed the numerator on purpose so that we could get two differences, which will later give us two derivatives.
Next, we write this fraction as the sum of two fractions:
\[
y’=\lim_{h\to 0}
\left(
u(x+h)\cdot \frac{v(x+h)-v(x)}{h}
+
v(x)\cdot \frac{u(x+h)-u(x)}{h}
\right).
\]
Now let us look at each factor separately. The fraction
\[
\frac{v(x+h)-v(x)}{h}
\]
is the difference quotient for the function \( v(x) \). Therefore, when \( h\to 0 \), this fraction approaches the derivative \( v'(x) \):
\[
\lim_{h\to 0}\frac{v(x+h)-v(x)}{h}=v'(x).
\]
Similarly, the fraction
\[
\frac{u(x+h)-u(x)}{h}
\]
is the difference quotient for the function \( u(x) \). Therefore, as \( h\to 0 \), we have
\[
\lim_{h\to 0}\frac{u(x+h)-u(x)}{h}=u'(x).
\]
It remains to pay attention to the factor \( u(x+h) \) in the first term. Since the function \( u(x) \) is differentiable, it is continuous at the point \( x \). Therefore, as \( h\to 0 \), the value \( u(x+h) \) approaches \( u(x) \):
\[
u(x+h)\to u(x).
\]
Now we can take the limit of the whole expression:
\[
y’=u(x)\cdot v'(x)+v(x)\cdot u'(x).
\]
Since the order of terms in a sum can be changed, we write the result in the usual form:
\[
y’=u'(x)\cdot v(x)+u(x)\cdot v'(x).
\]
So, we finally get the product rule formula:
\[
\frac{d}{dx}\bigl(u(x)\cdot v(x)\bigr)=u'(x)\cdot v(x)+u(x)\cdot v'(x).
\]
Thus, the product rule follows directly from the definition of the derivative using a limit. The main step in the derivation is to transform the numerator correctly and split the change of the product into two parts. That is why the final formula contains two terms: one accounts for the change of the first factor, and the other accounts for the change of the second factor.
Product Rule: Practical Application of the Formula
Now let us move on to the practical part and see how the product rule works in typical problems. Here it is important not to rush. First, we need to correctly identify the first and second factors, and only then apply the formula step by step. In each example, we will find the derivatives of both factors and substitute them into the product rule.
Example 1. Find the derivative of the function \( y=(x^2+1)\cdot (3\cdot x-5) \)
In this example, we have the product of two functions. The first factor is the expression \( x^2+1 \), and the second factor is the expression \( 3\cdot x-5 \). For convenience, let us denote
\[
u(x)=x^2+1,\qquad v(x)=3\cdot x-5.
\]
Then the function can be written as
\[
y=u(x)\cdot v(x).
\]
By the product rule, we have
\[
y’=u'(x)\cdot v(x)+u(x)\cdot v'(x).
\]
Now let us find the derivative of the first factor:
\[
u'(x)=(x^2+1)’=2\cdot x.
\]
Next, let us find the derivative of the second factor:
\[
v'(x)=(3\cdot x-5)’=3.
\]
Now substitute the derivatives we found into the product rule formula:
\[
y’=2\cdot x\cdot (3\cdot x-5)+(x^2+1)\cdot 3.
\]
Now expand the brackets:
\[
y’=6\cdot x^2-10\cdot x+3\cdot x^2+3.
\]
Combine like terms, and we get the final answer:
\[
y’=9\cdot x^2-10\cdot x+3.
\]
Example 2. Find the derivative of the function \( y=(x^3-2\cdot x)\cdot \sqrt{x} \)
Here the function is also given as the product of two expressions. The first factor is the polynomial \( x^3-2\cdot x \), and the second factor is the square root function \( \sqrt{x} \). Let us denote
\[
u(x)=x^3-2\cdot x,\qquad v(x)=\sqrt{x}.
\]
First, let us find the derivative of the first factor:
\[
u'(x)=(x^3-2\cdot x)’=3\cdot x^2-2.
\]
Now let us find the derivative of the second factor. Since \( v(x)=\sqrt{x} \), we have
\[
v'(x)=\frac{1}{2\cdot \sqrt{x}}.
\]
Substitute these expressions into the product rule formula:
\[
y’=(3\cdot x^2-2)\cdot \sqrt{x}+(x^3-2\cdot x)\cdot \frac{1}{2\cdot \sqrt{x}}.
\]
If needed, this expression can be brought to a common denominator. Let us rewrite the first term with the denominator \( 2\cdot \sqrt{x} \):
\[
(3\cdot x^2-2)\cdot \sqrt{x}
=
\frac{2\cdot x\cdot (3\cdot x^2-2)}{2\cdot \sqrt{x}}.
\]
Then we have
\[
y’=
\frac{2\cdot x\cdot (3\cdot x^2-2)}{2\cdot \sqrt{x}}
+
\frac{x^3-2\cdot x}{2\cdot \sqrt{x}}.
\]
Now combine the fractions:
\[
y’=\frac{2\cdot x\cdot (3\cdot x^2-2)+x^3-2\cdot x}{2\cdot \sqrt{x}}.
\]
Expand the brackets in the numerator:
\[
y’=\frac{6\cdot x^3-4\cdot x+x^3-2\cdot x}{2\cdot \sqrt{x}}.
\]
Combine like terms, and we get the final answer:
\[
y’=\frac{7\cdot x^3-6\cdot x}{2\cdot \sqrt{x}}.
\]
In this example, it is important to remember that the function contains \( \sqrt{x} \). Therefore, if we work with real numbers, the derivative of this function is valid only for \( x>0 \).
Example 3. Find the derivative of the function \( y=x^2\cdot \sin(x) \)
In this case, the first factor is the power function \( x^2 \), and the second factor is the trigonometric function \( \sin(x) \). So, let us denote
\[
u(x)=x^2,\qquad v(x)=\sin(x).
\]
Find the derivative of the first factor:
\[
u'(x)=(x^2)’=2\cdot x.
\]
Now find the derivative of the second factor:
\[
v'(x)=(\sin(x))’=\cos(x).
\]
Substitute these results into the product rule formula, and we get the answer:
\[
y’=2\cdot x\cdot \sin(x)+x^2\cdot \cos(x).
\]
In this example, nothing else needs to be simplified. The answer is already written in a convenient form. The first term accounts for the change of the factor \( x^2 \), and the second accounts for the change of the factor \( \sin(x) \).
Example 4. Find the derivative of the function \( y=e^x\cdot \ln(x) \)
Here we have the product of an exponential function and the natural logarithm. The first factor is \( e^x \), and the second is \( \ln(x) \). Let us denote
\[
u(x)=e^x,\qquad v(x)=\ln(x).
\]
Find the derivative of the first factor. Since the derivative of \( e^x \) is the function itself, we have
\[
u'(x)=(e^x)’=e^x.
\]
Now find the derivative of the second factor:
\[
v'(x)=(\ln(x))’=\frac{1}{x}.
\]
Substitute these derivatives into the product rule formula:
\[
y’=e^x\cdot \ln(x)+e^x\cdot \frac{1}{x}.
\]
We get
\[
y’=e^x\cdot \ln(x)+\frac{e^x}{x}.
\]
We can factor out the common factor \( e^x \) and write the final answer:
\[
y’=e^x\cdot \left(\ln(x)+\frac{1}{x}\right).
\]
Here it is worth remembering that the function contains \( \ln(x) \). Therefore, if we work with real numbers, we must consider only \( x>0 \). This is important not only for the function itself, but also for the correct form of its derivative.
Example 5. Find the derivative of the function \( y=(x^2+1)\cdot \cos(2\cdot x) \)
In this example, we have the product of two functions, but the second factor is a composite function. The first factor is \( x^2+1 \), and the second is \( \cos(2\cdot x) \). Let us denote
\[
u(x)=x^2+1,\qquad v(x)=\cos(2\cdot x).
\]
First, let us find the derivative of the first factor:
\[
u'(x)=(x^2+1)’=2\cdot x.
\]
Now let us find the derivative of the second factor:
\[
v'(x)=(\cos(2\cdot x))’.
\]
Here we need to be careful. The product rule helps us find the derivative of the whole product, but to find the derivative of the second factor, we also need to use the rule for differentiating a composite function. The outer function is cosine, and the inner function is the expression \( 2\cdot x \). Therefore,
\[
(\cos(2\cdot x))’=-\sin(2\cdot x)\cdot (2\cdot x)’.
\]
Since \( (2\cdot x)’=2 \), we get
\[
v'(x)=-2\cdot \sin(2\cdot x).
\]
Now substitute everything into the product rule formula:
\[
y’=2\cdot x\cdot \cos(2\cdot x)+(x^2+1)\cdot \bigl(-2\cdot \sin(2\cdot x)\bigr).
\]
Expand the brackets while taking the minus sign into account, and we get the final answer:
\[
y’=2\cdot x\cdot \cos(2\cdot x)-2\cdot (x^2+1)\cdot \sin(2\cdot x).
\]
This example shows an important point: sometimes, when applying the product rule, we also need to use another differentiation rule inside one of the factors. So first we identify the factors, and then carefully find the derivative of each of them.
What to Read Next: Useful Topics for Further Study
After the product rule, it is worth moving on to other basic rules of differentiation. These rules often work together, so knowing each of them helps you solve more complex examples with greater confidence.
- Power Rule: Formula, Proof, and Examples — This article will explain how to find derivatives of power functions and how to apply this rule in typical problems.
- Chain Rule: Formula, Proof, and Examples — This material will help you understand how to differentiate composite functions and correctly identify the inner and outer parts.
- Quotient Rule: Formula, Proof, and Examples — This article will discuss the derivative of the quotient of two functions and the step-by-step application of this rule in calculations.
Product Rule: An Idea for a Programming Experiment
If you are interested not only in solving examples by hand, but also in seeing how a mathematical formula works in a program, try implementing the algorithm from the flowchart in your favorite programming language.
In this algorithm, the derivative of a function is calculated in two ways: analytically, using the product rule, and approximately, using a small increment of the argument. And what is the most interesting part? You will be able to compare both results and see how close the numerical method comes to the exact formula.
