Select Page

# Calculus

Differentiation of Functions

# The Product and Quotient Rules

Page 1
Problems 1-3
Page 2
Problems 4-18

### The Product Rule

Let $$u\left( x \right)$$ and $$v\left( x \right)$$ be differentiable functions. Then the product of the functions $$u\left( x \right)v\left( x \right)$$ is also differentiable and

${\left( {uv} \right)^\prime } = u’v + uv’.$

We prove the above formula using the definition of the derivative. For this we find the increment of the functions $${uv}$$ assuming that the argument changes by $$\Delta x$$:

${\Delta \left( {uv} \right) }={ u\left( {x + \Delta x} \right)v\left( {x + \Delta x} \right) }-{ u\left( x \right)v\left( x \right).}$

Take into account that

${u\left( {x + \Delta x} \right) = u\left( x \right) + \Delta u,\;\;\;}\kern0pt {v\left( {x + \Delta x} \right) = v\left( x \right) + \Delta v,}$

where $$\Delta u$$ and $$\Delta v$$ are the increments, respectively, of the functions $$u$$ and $$v$$. Omitting for brevity the argument $$x$$ of the functions $$u$$ and $$v$$, we can write the increment $$\Delta \left( {uv} \right)$$ in the following form:

$\require{cancel} {\Delta \left( {uv} \right) = \left( {u + \Delta u} \right)\left( {v + \Delta v} \right) – uv } = {\cancel{uv} + u\Delta v + v\Delta u + \Delta u\Delta v – \cancel{uv} } = {u\Delta v + v\Delta u + \Delta u\Delta v.}$

We proceed to calculate the derivative of the product using the properties of limits:

${{\left( {uv} \right)^\prime } = {\lim\limits_{\Delta x \to 0} \frac{{\Delta \left( {uv} \right)}}{{\Delta x}}} } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v + v\Delta u + \Delta u\Delta v}}{{\Delta x}} } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v}}{{\Delta x}} + \lim\limits_{\Delta x \to 0} \frac{{v\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v.}$

In the first limit the function $$u$$ does not depend on the increment $$\Delta x$$. Therefore, it can be taken outside the limit sign. The same applies to the function $$v$$ in the second term. We calculate separately the limit $$\lim\limits_{\Delta x \to 0} \Delta v:$$

${\lim\limits_{\Delta x \to 0} \Delta v }={ \lim\limits_{\Delta x \to 0} \left( {\frac{{\Delta v}}{{\Delta x}} \cdot \Delta x} \right) } = {\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta x }={ v’ \cdot 0 = 0.}$

Thus, the derivative of the product is given by

${{\left( {uv} \right)^\prime } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v}}{{\Delta x}} + \lim\limits_{\Delta x \to 0} \frac{{v\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v }} = {u\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}} + v\lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v } = {uv’ + vu’ + u’ \cdot 0 } = {u’v + uv’.}$

Important: The derivative of the product is NOT equal to the product of the derivatives!

From this formula, it is easy to obtain an expression for the derivative of the function $$kf\left( x \right)$$, where $$k$$ is a constant:

${{\left( {kf\left( x \right)} \right)^\prime } = {k’f\left( x \right) + kf’\left( x \right) }} = {0 \cdot f\left( x \right) + kf’\left( x \right) } = {kf’\left( x \right),}$

so that a constant factor can be taken out of the sign of derivative.

### The Quotient Rule

Let $$u\left( x \right)$$ and $$v\left( x \right)$$ be again differentiable functions. Then, if $$v\left( x \right) \ne 0$$, the derivative of the quotient of these functions is calculated by the formula

${\left( {\frac{u}{v}} \right)^\prime } = {\frac{{u’v – uv’}}{{{v^2}}}.}$

This formula is proved similarly. The increment of the quotient can be written as

${\Delta \left( {\frac{u}{v}} \right) }={ \frac{{u + \Delta u}}{{v + \Delta v}} – \frac{u}{v} } = {\frac{{\left( {u + \Delta u} \right)v – u\left( {v + \Delta v} \right)}}{{v\left( {v + \Delta v} \right)}} } = {\frac{{\cancel{uv} + v\Delta u – \cancel{uv} – u\Delta v}}{{v\left( {v + \Delta v} \right)}} } = {\frac{{v\Delta u – u\Delta v}}{{v\left( {v + \Delta v} \right)}}.}$

The derivative of the quotient is expressed as follows:

${{\left( {\frac{u}{v}} \right)^\prime } }={ \lim\limits_{\Delta x \to 0} \frac{{\Delta \left( {\frac{u}{v}} \right)}}{{\Delta x}} } = {\lim\limits_{\Delta x \to 0} \frac{{\frac{{v\Delta u – u\Delta v}}{{{v^2} + v\Delta v}}}}{{\Delta x}} } = {\lim\limits_{\Delta x \to 0} \frac{{v\frac{{\Delta u}}{{\Delta x}} – u\frac{{\Delta v}}{{\Delta x}}}}{{{v^2} + v\Delta v}}.}$

Further, using the properties of limits, we find:

${{\left( {\frac{u}{v}} \right)^\prime } }={ \lim\limits_{\Delta x \to 0} \frac{{v\frac{{\Delta u}}{{\Delta x}} – u\frac{{\Delta v}}{{\Delta x}}}}{{{v^2} + v\Delta v}} } = {\frac{{v\lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} – u\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}}}}{{\lim\limits_{\Delta x \to 0} {v^2} + \lim\limits_{\Delta x \to 0} \left( {v\Delta v} \right)}} } = {\frac{{u’v – uv’}}{{{v^2} + v\lim\limits_{\Delta x \to 0} \Delta v}}.}$

Taking into account that $${\lim\limits_{\Delta x \to 0} \Delta v} = 0$$ we obtain the final expression for the derivative of the quotient of two functions:

${\left( {\frac{u}{v}} \right)^\prime } = \frac{{u’v – uv’}}{{{v^2}}}.$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Find the derivative of the function $${y = {\large\frac{2}{x}\normalsize}}$$.

### ✓Example 2

Find the derivative of a power function with the negative exponent $$y = {x^{ – n}}$$.

### ✓Example 3

Calculate the derivative of $$y = \tan x$$ using the quotient rule.

### ✓Example 4

Let $$y = {\sin ^2}x$$. Differentiate this function without using the chain rule.

### ✓Example 5

Find the derivative of the secant function $$y = \sec x$$.

### ✓Example 6

Find the derivative of the function $$y = {e^x}\cos x$$.

### ✓Example 7

Find the derivative of the function $$y = {\large\frac{{{e^x} – 1}}{{{e^x} + 1}}\normalsize}$$ and calculate its value at $$x = 0$$.

### ✓Example 8

Find the derivative of the function $$y = {x^2}\sin x$$.

### ✓Example 9

Given the function $$z = \left( {{x^2} + 1} \right)\left( {x – 1} \right)$$. Find the value of its derivative at $$x = -1$$.

### ✓Example 10

Derive a formula for the derivative of the product of three functions.

### ✓Example 11

Find the derivative of the function $$y = {\large\frac{{2x}}{{1 – {x^2}}}\normalsize}$$.

### ✓Example 12

Calculate the derivative of the function $$y = {\large\frac{{1 + \cos x}}{{\sin x}}\normalsize}$$.

### ✓Example 13

Differentiate the function $$y = \left( {3 – 2x} \right)\left( {2 – 3x} \right)$$.

### ✓Example 14

Differentiate the function $$y = {x^3}{2^x}$$.

### ✓Example 15

Find the derivative of the linear fractional function $$y = {\large\frac{{ax + b}}{{cx + d}}\normalsize}$$.

### ✓Example 16

Calculate the derivative of the function $$y = {\large\frac{{3x – 1}}{{{x^4}}}\normalsize}$$.

### ✓Example 17

Calculate the derivative of the following function: $$y = {\large\frac{{\sqrt x – 1}}{{\sqrt x + 1}}\normalsize}$$.

### ✓Example 18

Derive a formula for the derivative of the function $$f\left( x \right) = {\large\frac{{u\left( x \right)v\left( x \right)}}{{w\left( x \right)}}\normalsize}$$.

### Example 1.

Find the derivative of the function $${y = {\large\frac{2}{x}\normalsize}}$$.

#### Solution.

Using the quotient rule, we have

${y’\left( x \right) = {\left( {\frac{2}{x}} \right)^\prime } } = {\frac{{2′ \cdot x – 2 \cdot x’}}{{{x^2}}} } = {\frac{{0 \cdot x – 2 \cdot 1}}{{{x^2}}} } = { – \frac{2}{{{x^2}}}.}$

### Example 2.

Find the derivative of a power function with the negative exponent $$y = {x^{ – n}}$$.

#### Solution.

We write the function in the form $$y\left( x \right) = {\large\frac{1}{{{x^n}}}\normalsize}$$ and use the quotient rule.

${y’\left( x \right) = {\left( {\frac{1}{{{x^n}}}} \right)^\prime } } = {\frac{{1′ \cdot {x^n} – 1 \cdot {{\left( {{x^n}} \right)}^\prime }}}{{{{\left( {{x^n}} \right)}^2}}} } = {\frac{{0 \cdot {x^n} – n{x^{n – 1}}}}{{{x^{2n}}}} } = { – \frac{n}{{{x^{2n – n + 1}}}} } = { – \frac{n}{{{x^{n + 1}}}}.}$

### Example 3.

Calculate the derivative of $$y = \tan x$$ using the quotient rule.

#### Solution.

We can write the tangent function as $$\tan x = {\large\frac{{\sin x}}{{\cos x}}\normalsize}$$. Then

${y’\left( x \right) }={ {\left( {\tan x} \right)^\prime } = {\left( {\frac{{\sin x}}{{\cos x}}} \right)^\prime } } = {\frac{{{{\left( {\sin x} \right)}^\prime }\cos x – \sin x{{\left( {\cos x} \right)}^\prime }}}{{{{\left( {\cos x} \right)}^2}}}.}$

As $${\left( {\sin x} \right)^\prime } = \cos x$$, $${\left( {\cos x} \right)^\prime } = -\sin x,$$ the derivative is given by

${{\left( {\tan x} \right)^\prime } }={ \frac{{\cos x \cdot \cos x – \sin x \cdot \left( { – \sin x} \right)}}{{{{\cos }^2}x}} } = {\frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} }={ \frac{1}{{{{\cos }^2}x}}.}$
Page 1
Problems 1-3
Page 2
Problems 4-18