Calculus

Differentiation of Functions

Differentiation of Functions Logo

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