# Quotient Rule

• The quotient rule is a formal rule for differentiating of a quotient of functions.

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

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

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:

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

Important: The derivative of a quotient is NOT the quotient of the derivatives!

• ## Solved Problems

Click a problem to see the 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

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

### Example 4

Find the derivative of the function $$y = \large{\frac{3}{{{x^3}}}}\normalsize.$$

### Example 5

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

### Example 6

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

### Example 7

Find the derivative of the cotangent function $$y = \cot x.$$

### Example 8

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

### Example 9

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

### Example 10

Find the derivative of the function $$y = \large{\frac{{{e^x} + 2}}{{{e^x}}}}\normalsize$$ at $$x = 0.$$

### Example 11

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

### Example 12

Find the derivative of the function $$y = \large{\frac{{2x – 1}}{{{x^3}}}}\normalsize.$$

### Example 13

Differentiate the function $$y = \large{\frac{1}{{\ln x}}}\normalsize.$$

### Example 14

Differentiate the rational function $$y = \large{\frac{{3{x^2} + 1}}{{x – 1}}}\normalsize.$$

### Example 15

Differentiate the function $$z = \large{\frac{{1 + {x^2}}}{{1 – {x^2}}}}\normalsize.$$

### Example 16

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

### Example 17

Find the derivative of the cosecant function $$y = \csc x.$$

### Example 18

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

### Example 19

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

### Example 20

Find the derivative of the function $$y = \large{\frac{3}{{5 – x}}}\normalsize + \large{\frac{{{x^2}}}{3}}\normalsize$$ at $$x = 0.$$

### Example 21

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

### Example 22

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.

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

Solution.

Let $$u = x + 1,$$ $$v = x – 1.$$

By the quotient rule $$\left( {\large{\frac{u}{v}}\normalsize} \right)^\prime = \large{\frac{{u^\prime v – uv^\prime}}{{{v^2}}}}\normalsize,$$ we can write

${y^\prime = \left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime }={ \frac{{\left( {x + 1} \right)^\prime\left( {x – 1} \right) – \left( {x + 1} \right)\left( {x – 1} \right)^\prime}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{1 \cdot \left( {x – 1} \right) – \left( {x + 1} \right) \cdot 1}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{\cancel{\color{red}{x}} – \color{blue}{1} – \cancel{\color{red}{x}} – \color{blue}{1}}}{{{{\left( {x – 1} \right)}^2}}} }={ – \frac{\color{blue}{2}}{{{{\left( {x – 1} \right)}^2}}}.}$

### Example 4.

Find the derivative of the function $$y = \large{\frac{3}{{{x^3}}}}\normalsize.$$

Solution.

Using the quotient rule, we have

${y^\prime = \left( {\frac{3}{{{x^3}}}} \right)^\prime }={ \frac{{3^\prime \cdot {x^3} – 3 \cdot \left( {{x^3}} \right)^\prime}}{{{{\left( {{x^3}} \right)}^2}}} }={ \frac{{0 \cdot {x^3} – 3 \cdot 3{x^2}}}{{{x^6}}} }={ – \frac{{9{x^2}}}{{{x^6}}} }={ – \frac{9}{{{x^4}}}.}$

### Example 5.

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

Solution.

By the quotient rule,

${y^\prime = \left( {\frac{{2x + 1}}{{2x – 1}}} \right)^\prime }={ \frac{{2 \cdot \left( {2x – 1} \right) – \left( {2x + 1} \right) \cdot 2}}{{{{\left( {2x – 1} \right)}^2}}} }={ \frac{{\cancel{\color{red}{4x}} – \color{blue}{2} – \cancel{\color{red}{4x}} – \color{blue}{2}}}{{{{\left( {2x – 1} \right)}^2}}} }={ – \frac{\color{blue}{4}}{{{{\left( {2x – 1} \right)}^2}}}.}$

At $$x = 1,$$ the derivative is equal

${y^\prime\left( 1 \right) }={ – \frac{4}{{{{\left( {2 \cdot 1 – 1} \right)}^2}}} }={ – \frac{4}{{{1^2}}} }={ – 4.}$

### Example 6.

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

${{\left( {\tan x} \right)^\prime } }={ \frac{{\cos x \cdot \cos x – \sin x \cdot \left( { – \sin x} \right)}}{{{{\cos }^2}x}} } = {\frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}} }={ \frac{1}{{{{\cos }^2}x}}.}$

### Example 7.

Find the derivative of the cotangent function $$y = \cot x.$$

Solution.

As $$\cot x = \large{\frac{{\cos x}}{{\sin x}}}\normalsize,$$ we can apply the quotient rule:

${y^\prime = \left( {\cot x} \right)^\prime = \left( {\frac{{\cos x}}{{\sin x}}} \right)^\prime }={ \frac{{\left( {\cos x} \right)^\prime\sin x – \cos x\left( {\sin x} \right)^\prime}}{{{{\sin }^2}x}} }={ \frac{{\left( { – \sin x} \right) \cdot \sin x – \cos x \cdot \cos x}}{{{{\sin }^2}x}} }={ \frac{{ – {{\sin }^2}x – {{\cos }^2}x}}{{{{\sin }^2}x}} }={ – \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} }={ – \frac{1}{{{{\sin }^2}x}} }={ – {\csc ^2}x.}$

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Problems 1-7
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Problems 8-22