The quotient rule is one of the basic rules of differential calculus. It is used when a function is written as the quotient of two other functions. In other words, we have an expression where one function is in the numerator and another function is in the denominator.
At first, it may seem that finding the derivative of such a function is very simple: differentiate the numerator separately, differentiate the denominator separately, and then divide the results. But here an important question arises: does this approach really describe the change of the whole fraction correctly? In fact, no. When the argument changes, not only the numerator changes, but the denominator changes too. That is why a separate rule is needed for the quotient of functions. This rule correctly takes into account the change of both parts of the fraction.
Quotient Rule: How to Find the Derivative of a Fraction
Suppose we have two differentiable functions \( u(x) \) and \( v(x) \). Let us consider a function that is their quotient:
\[
y=\frac{u(x)}{v(x)}.
\]
Here it is important to remember one condition: the denominator cannot be equal to zero. So the formula makes sense only at the points where
\[
v(x)\ne 0.
\]
Why is this important? Because if the denominator is equal to zero, the fraction itself is not defined. Therefore, we cannot talk about the derivative of this fraction at that point.
Then the derivative of the quotient is calculated using the formula
\[
y’=\frac{u'(x)\cdot v(x)-u(x)\cdot v'(x)}{\bigl(v(x)\bigr)^2}.
\]
This is the main formula of the quotient rule. It is also often written using the differentiation operator:
\[
\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=
\frac{u'(x)\cdot v(x)-u(x)\cdot v'(x)}{\bigl(v(x)\bigr)^2}.
\]
Pay attention to the order of operations in the numerator. First, we take the derivative of the numerator \( u'(x) \) and multiply it by the denominator \( v(x) \). Then we subtract the numerator \( u(x) \), multiplied by the derivative of the denominator \( v'(x) \). In the denominator, we get the square of the original denominator.
The minus sign in the numerator is exactly where mistakes often happen. So it is better not to memorize the formula mechanically, but to understand its structure: the change of the numerator is represented by the first term, and the change of the denominator is represented by the second term, which is subtracted.
Deriving the Formula: Explanation Through the Definition of the Derivative
Now let us look closely at where the quotient rule formula comes from. To do this, we will use the definition of the derivative through a limit. Suppose the function \( y \) is given as the quotient of two differentiable functions:
\[
y=\frac{u(x)}{v(x)}.
\]
By the definition of the derivative, we have
\[
y’=\lim_{h\to 0}\frac{y(x+h)-y(x)}{h}.
\]
Since \( y(x)=\frac{u(x)}{v(x)} \), the value of this function at the point \( x+h \) is
\[
y(x+h)=\frac{u(x+h)}{v(x+h)}.
\]
Now substitute these expressions into the definition of the derivative:
\[
y’=\lim_{h\to 0}
\frac{
\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}
}{h}.
\]
At this stage, we have obtained the correct expression for the derivative, but it does not yet look like the quotient rule formula. What should we do next? We need to transform the difference of fractions in the numerator. To do this, we bring them to a common denominator:
\[
y’=\lim_{h\to 0}
\frac{
\frac{u(x+h)\cdot v(x)-u(x)\cdot v(x+h)}
{v(x+h)\cdot v(x)}
}{h}.
\]
Now the division by \( h \) can be written so that \( h \) appears in the denominator of the whole fraction:
\[
y’=\lim_{h\to 0}
\frac{
u(x+h)\cdot v(x)-u(x)\cdot v(x+h)
}
{
v(x+h)\cdot v(x)\cdot h
}.
\]
At first glance, this expression may look more complicated. But right now, we can prepare it for the appearance of the derivatives \( u'(x) \) and \( v'(x) \). To do this, we need to get the differences \( u(x+h)-u(x) \) and \( v(x+h)-v(x) \), because these are exactly the differences that appear in the definition of the derivative.
To make this happen, we add and subtract the same expression:
\[
u(x)\cdot v(x).
\]
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 divide the change of the fraction into two parts. In one part, we will see the change of the function \( u(x) \), and in the other part, the change of the function \( v(x) \).
So we write:
\[
y’=\lim_{h\to 0}
\frac{
u(x+h)\cdot v(x)-u(x)\cdot v(x)+u(x)\cdot v(x)-u(x)\cdot v(x+h)
}
{
v(x+h)\cdot v(x)\cdot h
}.
\]
Now let us group the terms so that the first group has the common factor \( v(x) \), and the second group has the common factor \( u(x) \):
\[
y’=\lim_{h\to 0}
\frac{
v(x)\cdot \bigl(u(x+h)-u(x)\bigr)
–
u(x)\cdot \bigl(v(x+h)-v(x)\bigr)
}
{
v(x+h)\cdot v(x)\cdot h
}.
\]
In this form, it is already clear why we added and subtracted the intermediate expression. The first difference \( u(x+h)-u(x) \) shows the change of the function \( u(x) \), and the second difference \( v(x+h)-v(x) \) shows the change of the function \( v(x) \). In other words, we intentionally transformed the numerator so that we could get two differences, which will later give us two derivatives.
Next, we need to distribute the factor \( h \) carefully. It is in the common denominator, so we can move it to each difference separately. In other words, we rewrite the expression so that two difference quotients appear: one for the function \( u(x) \), and the other for the function \( v(x) \).
Therefore, we get:
\[
y’=\lim_{h\to 0}
\frac{1}{v(x+h)\cdot v(x)}\cdot
\left(
v(x)\cdot \frac{u(x+h)-u(x)}{h}
–
u(x)\cdot \frac{v(x+h)-v(x)}{h}
\right).
\]
Now let us consider each element separately. The fraction
\[
\frac{u(x+h)-u(x)}{h}
\]
is the difference quotient for the function \( u(x) \). Therefore, when \( h\to 0 \), this fraction approaches the derivative \( u'(x) \):
\[
\lim_{h\to 0}\frac{u(x+h)-u(x)}{h}=u'(x).
\]
Similarly, the fraction
\[
\frac{v(x+h)-v(x)}{h}
\]
is the difference quotient for the function \( v(x) \). Therefore, as \( h\to 0 \), we have
\[
\lim_{h\to 0}\frac{v(x+h)-v(x)}{h}=v'(x).
\]
It remains to pay attention to the factor \( v(x+h) \) in the denominator. Since the function \( v(x) \) is differentiable, it is continuous at the point \( x \). Therefore, as \( h\to 0 \), the value \( v(x+h) \) approaches \( v(x) \):
\[
v(x+h)\to v(x).
\]
Now we can take the limit of the whole expression:
\[
y’=
\frac{1}{v(x)\cdot v(x)}\cdot
\left(
v(x)\cdot u'(x)-u(x)\cdot v'(x)
\right).
\]
Since \( v(x)\cdot v(x)=\bigl(v(x)\bigr)^2 \), we can write the result more briefly:
\[
y’=
\frac{
v(x)\cdot u'(x)-u(x)\cdot v'(x)
}
{
\bigl(v(x)\bigr)^2
}.
\]
Now we use the usual property of multiplication: \( v(x)\cdot u'(x)=u'(x)\cdot v(x) \). So the same formula can be written in its standard form:
\[
y’=
\frac{
u'(x)\cdot v(x)-u(x)\cdot v'(x)
}
{
\bigl(v(x)\bigr)^2
}.
\]
Therefore, we finally obtain the quotient rule formula:
\[
\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=
\frac{u'(x)\cdot v(x)-u(x)\cdot v'(x)}{\bigl(v(x)\bigr)^2}.
\]
So, the quotient rule follows directly from the definition of the derivative through a limit. The main step in the derivation is to transform the numerator correctly and obtain two differences: one for the function \( u(x) \), and the other for the function \( v(x) \). That is exactly why the final formula contains two terms in the numerator with a minus sign between them: one accounts for the change of the numerator, and the other accounts for the change of the denominator.
Quotient Rule: Practice Finding Derivatives
After the formula and its derivation, it is important to see how the quotient rule works in real calculations. After all, the formula becomes much easier to understand when we apply it several times step by step. Next, let us look at typical examples where a function is given as the quotient of two other functions.
Example 1. Find the derivative of the function \( y=\frac{x^2+1}{3\cdot x-5} \)
In this example, we have the quotient of two functions. The numerator contains the expression \( x^2+1 \), and the denominator contains 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
\[
y=\frac{u(x)}{v(x)}.
\]
By the quotient rule, we have
\[
y’=
\frac{u'(x)\cdot v(x)-u(x)\cdot v'(x)}
{\bigl(v(x)\bigr)^2}.
\]
First, let us find the derivative of the numerator:
\[
u'(x)=(x^2+1)’=2\cdot x.
\]
Next, let us find the derivative of the denominator:
\[
v'(x)=(3\cdot x-5)’=3.
\]
Now substitute the found derivatives into the quotient rule formula:
\[
y’=
\frac{
2\cdot x\cdot (3\cdot x-5)-(x^2+1)\cdot 3
}
{(3\cdot x-5)^2}.
\]
Expand the brackets in the numerator:
\[
y’=
\frac{
6\cdot x^2-10\cdot x-3\cdot x^2-3
}
{(3\cdot x-5)^2}.
\]
Combine like terms and get the final answer:
\[
y’=
\frac{
3\cdot x^2-10\cdot x-3
}
{(3\cdot x-5)^2}.
\]
Example 2. Find the derivative of the function \( y=\frac{x^3-2\cdot x}{\sqrt{x}} \)
Here the function is given as the quotient of a polynomial and a square root function. In the numerator, we have the expression \( x^3-2\cdot x \), and in the denominator, we have \( \sqrt{x} \). Let us denote
\[
u(x)=x^3-2\cdot x,\qquad v(x)=\sqrt{x}.
\]
Then
\[
y=\frac{u(x)}{v(x)}.
\]
Find the derivative of the numerator:
\[
u'(x)=(x^3-2\cdot x)’=3\cdot x^2-2.
\]
Now find the derivative of the denominator. Since \( v(x)=\sqrt{x} \), we have
\[
v'(x)=\frac{1}{2\cdot \sqrt{x}}.
\]
By the quotient rule, we have
\[
y’=
\frac{
u'(x)\cdot v(x)-u(x)\cdot v'(x)
}
{\bigl(v(x)\bigr)^2}.
\]
Substitute the found expressions:
\[
y’=
\frac{
(3\cdot x^2-2)\cdot \sqrt{x}
–
(x^3-2\cdot x)\cdot \frac{1}{2\cdot \sqrt{x}}
}
{(\sqrt{x})^2}.
\]
Since \( (\sqrt{x})^2=x \), we get
\[
y’=
\frac{
(3\cdot x^2-2)\cdot \sqrt{x}
–
\frac{x^3-2\cdot x}{2\cdot \sqrt{x}}
}
{x}.
\]
Next, we work only with algebraic simplification of the expression we obtained. The quotient rule has already been applied, and now we need to carefully bring the fractions into a convenient form.
Let us bring the first term in the numerator of the large fraction to 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
\[
y’=
\frac{
\frac{2\cdot x\cdot (3\cdot x^2-2)}{2\cdot \sqrt{x}}
–
\frac{x^3-2\cdot x}{2\cdot \sqrt{x}}
}
{x}.
\]
Combine the fractions in the numerator:
\[
y’=
\frac{
\frac{
2\cdot x\cdot (3\cdot x^2-2)-(x^3-2\cdot x)
}
{2\cdot \sqrt{x}}
}
{x}.
\]
Expand the brackets:
\[
y’=
\frac{
\frac{
6\cdot x^3-4\cdot x-x^3+2\cdot x
}
{2\cdot \sqrt{x}}
}
{x}.
\]
Combine like terms:
\[
y’=
\frac{
\frac{
5\cdot x^3-2\cdot x
}
{2\cdot \sqrt{x}}
}
{x}.
\]
Dividing by \( x \) gives
\[
y’=
\frac{5\cdot x^3-2\cdot x}{2\cdot x\cdot \sqrt{x}}.
\]
Now cancel \( x \) in the numerator and denominator:
\[
y’=
\frac{5\cdot x^2-2}{2\cdot \sqrt{x}}.
\]
So, the final answer is
\[
y’=
\frac{5\cdot x^2-2}{2\cdot \sqrt{x}}.
\]
In this example, it is important to remember that the function contains \( \sqrt{x} \). Therefore, if we work with real numbers, it is correct to work with this function only when \( x>0 \).
Example 3. Find the derivative of the function \( y=\frac{x^2}{\sin(x)} \)
In this case, the numerator is the power function \( x^2 \), and the denominator is the trigonometric function \( \sin(x) \). Let us denote
\[
u(x)=x^2,\qquad v(x)=\sin(x).
\]
By the quotient rule, we have
\[
y’=
\frac{u'(x)\cdot v(x)-u(x)\cdot v'(x)}
{\bigl(v(x)\bigr)^2}.
\]
Find the derivative of the numerator:
\[
u'(x)=(x^2)’=2\cdot x.
\]
Now find the derivative of the denominator:
\[
v'(x)=(\sin(x))’=\cos(x).
\]
Substitute these derivatives into the formula:
\[
y’=
\frac{
2\cdot x\cdot \sin(x)-x^2\cdot \cos(x)
}
{(\sin(x))^2}.
\]
So, finally, we have
\[
y’=
\frac{
2\cdot x\cdot \sin(x)-x^2\cdot \cos(x)
}
{\sin^2(x)}.
\]
Example 4. Find the derivative of the function \( y=\frac{e^x}{\ln(x)} \)
Here we have the quotient of an exponential function and the natural logarithm. The numerator contains \( e^x \), and the denominator contains \( \ln(x) \). Let us denote
\[
u(x)=e^x,\qquad v(x)=\ln(x).
\]
First, find the derivative of the numerator. Since the derivative of \( e^x \) is equal to the function itself, we have
\[
u'(x)=(e^x)’=e^x.
\]
Now find the derivative of the denominator:
\[
v'(x)=(\ln(x))’=\frac{1}{x}.
\]
By the quotient rule, we have
\[
y’=
\frac{
u'(x)\cdot v(x)-u(x)\cdot v'(x)
}
{\bigl(v(x)\bigr)^2}.
\]
Substitute the found derivatives:
\[
y’=
\frac{
e^x\cdot \ln(x)-e^x\cdot \frac{1}{x}
}
{(\ln(x))^2}.
\]
In the numerator, we can factor out the common factor \( e^x \):
\[
y’=
\frac{
e^x\cdot \left(\ln(x)-\frac{1}{x}\right)
}
{(\ln(x))^2}.
\]
This is already a correct answer. But sometimes it is convenient to write it without a fraction inside the brackets. To do this, let us bring the expression in the brackets to a common denominator:
\[
\ln(x)-\frac{1}{x}
=
\frac{x\cdot \ln(x)-1}{x}.
\]
Then we get
\[
y’=
\frac{
e^x\cdot (x\cdot \ln(x)-1)
}
{
x\cdot (\ln(x))^2
}.
\]
So, the final answer can be written as
\[
y’=
\frac{
e^x\cdot (x\cdot \ln(x)-1)
}
{
x\cdot (\ln(x))^2
}.
\]
Here it is worth remembering that the natural logarithm \( \ln(x) \) is defined for positive values of the argument. Therefore, when working with real numbers, we need to take into account the condition \( x>0 \). Since \( \ln(x) \) is in the denominator, we also need \( \ln(x)\ne 0 \), so \( x\ne 1 \).
Example 5. Find the derivative of the function \( y=\frac{x^2+1}{\cos(2\cdot x)} \)
In this example, the function is also given as the quotient of two functions. But the denominator is a composite function, so we need to be especially careful when finding its derivative. Let us denote
\[
u(x)=x^2+1,\qquad v(x)=\cos(2\cdot x).
\]
Then
\[
y=\frac{u(x)}{v(x)}.
\]
Find the derivative of the numerator:
\[
u'(x)=(x^2+1)’=2\cdot x.
\]
Now find the derivative of the denominator:
\[
v'(x)=(\cos(2\cdot x))’.
\]
Here we 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 apply the quotient rule:
\[
y’=
\frac{
u'(x)\cdot v(x)-u(x)\cdot v'(x)
}
{\bigl(v(x)\bigr)^2}.
\]
Substitute all the found expressions:
\[
y’=
\frac{
2\cdot x\cdot \cos(2\cdot x)
–
(x^2+1)\cdot \bigl(-2\cdot \sin(2\cdot x)\bigr)
}
{(\cos(2\cdot x))^2}.
\]
Now notice that there is a minus sign before the second product, and the derivative of the denominator also has a minus sign. So, as a result, we get a plus:
\[
y’=
\frac{
2\cdot x\cdot \cos(2\cdot x)
+
2\cdot (x^2+1)\cdot \sin(2\cdot x)
}
{(\cos(2\cdot x))^2}.
\]
Finally, we can write
\[
y’=
\frac{
2\cdot x\cdot \cos(2\cdot x)
+
2\cdot (x^2+1)\cdot \sin(2\cdot x)
}
{\cos^2(2\cdot x)}.
\]
This example shows an important point: the quotient rule helps us find the derivative of the whole fraction, but to find the derivative of the numerator or the denominator, we sometimes need to use other differentiation rules as well. Here, that rule is the rule for a composite function.
What to Read Next: Useful Topics for Further Study
After the quotient rule, it is worth moving on to other differentiation rules. They are often combined in the same problem, so studying these topics step by step will help you understand the whole process of differentiation much better.
- Power Rule: Formula, Proof, 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, Examples — This material will help you understand how to find derivatives of composite functions step by step.
- Product Rule: Formula, Proof, Examples — This article will discuss the derivative of the product of two functions and the correct order of applying the formula.
Quotient Rule: Checking the Formula Through Programming
If you like programming, try turning the quotient rule you have learned into a small computational experiment. The flowchart below shows an algorithm that calculates the derivative of a given function at a selected point in two ways: analytically, using the ready-made formula, and numerically, using the central difference method.
And then everything depends on you: you can implement this algorithm in Pascal, Python, C++, JavaScript, or any other language you like. Isn’t it interesting to see how a formula from mathematical analysis turns into a working program and gives a result that can be checked by computation?
