Quotient Rule
Solved Problems
Example 9.
Find the derivative of the function \(y = {\frac{{{e^x} - 1}}{{{e^x} + 1}}}\) and calculate its value at \(x = 0.\)
Solution.
By the quotient rule,
\[\require{cancel} y'\left( x \right) = {\left( {\frac{{{e^x} - 1}}{{{e^x} + 1}}} \right)^\prime } = \frac{{{e^x}\left( {{e^x} + 1} \right) - \left( {{e^x} - 1} \right){e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}} = \frac{{\cancel{e^{2x}} + {e^x} - \cancel{e^{2x}} + {e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}} = \frac{{2{e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}}.\]
The derivative at \(x = 0\) is equal to
\[y'\left( 0 \right) = \frac{{2 \cdot {e^0}}}{{{{\left( {{e^0} + 1} \right)}^2}}} = \frac{{2 \cdot 1}}{{{{\left( {1 + 1} \right)}^2}}} = \frac{1}{2}.\]
Example 10.
Find the derivative of the function \(y = \frac{{{e^x} + 2}}{{{e^x}}}\) at \(x = 0.\)
Solution.
By the quotient rule,
\[y^\prime = \left( {\frac{{{e^x} + 2}}{{{e^x}}}} \right)^\prime = \frac{{\left( {{e^x} + 2} \right)^\prime{e^x} - \left( {{e^x} + 2} \right)\left( {{e^x}} \right)^\prime}}{{{{\left( {{e^x}} \right)}^2}}} = \frac{{{e^x} \cdot {e^x} - \left( {{e^x} + 2} \right) \cdot {e^x}}}{{{e^{2x}}}} = \frac{{\cancel{e^{2x}} - \cancel{e^{2x}} + 2{e^x}}}{{{e^{2x}}}} = \frac{2}{{{e^x}}}.\]
At \(x = 0,\) we have
\[y^\prime\left( 0 \right) = \frac{2}{{{e^0}}} = \frac{2}{1} = 2.\]
Example 11.
Find the derivative of the function \(y = {\frac{{2x}}{{1 - {x^2}}}}.\)
Solution.
Using the quotient rule, we have
\[y'\left( x \right) = \left( {\frac{{2x}}{{1 - {x^2}}}} \right)^\prime = \frac{{{{\left( {2x} \right)}^\prime }\left( {1 - {x^2}} \right) - 2x{{\left( {1 - {x^2}} \right)}^\prime }}}{{{{\left( {1 - {x^2}} \right)}^2}}} = \frac{{2\left( {1 - {x^2}} \right) - 2x \cdot \left( { - 2x} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}} = \frac{{2 - \color{blue}{2{x^2}} + \color{blue}{4{x^2}}}}{{{{\left( {1 - {x^2}} \right)}^2}}} = \frac{{2 + \color{blue}{2{x^2}}}}{{{{\left( {1 - {x^2}} \right)}^2}}} = \frac{{2\left( {1 + {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}.\]
Example 12.
Find the derivative of the function \(y = \frac{{2x - 1}}{{{x^3}}}.\)
Solution.
By the quotient rule,
\[y^\prime = \left( {\frac{{2x - 1}}{{{x^3}}}} \right)^\prime = \frac{{\left( {2x - 1} \right)^\prime \cdot {x^3} - \left( {2x - 1} \right) \cdot \left( {{x^3}} \right)^\prime}}{{{{\left( {{x^3}} \right)}^2}}} = \frac{{2 \cdot {x^3} - \left( {2x - 1} \right) \cdot 3{x^2}}}{{{x^6}}} = \frac{{\color{blue}{2{x^3}} - \color{blue}{6{x^3}} + \color{red}{3{x^2}}}}{{{x^6}}} = \frac{{\color{red}{3{x^2}} - \color{blue}{4{x^3}}}}{{{x^6}}} = \frac{{{x^2}\left( {3 - 4x} \right)}}{{{x^6}}} = \frac{{3 - 4x}}{{{x^4}}}.\]
Example 13.
Differentiate the function \(y = \frac{1}{{\ln x}}.\)
Solution.
We take the derivative using the quotient rule. Given that \(1^\prime = 0,\) and \(\left( {\ln x} \right)^\prime = \frac{1}{x},\) we get:
\[y^\prime = \left( {\frac{1}{{\ln x}}} \right)^\prime = \frac{{1^\prime \cdot \ln x - 1 \cdot \left( {\ln x} \right)^\prime}}{{{{\ln }^2}x}} = \frac{{0 \cdot \ln x - 1 \cdot \frac{1}{x}}}{{{{\ln }^2}x}} = - \frac{1}{{x{{\ln }^2}x}}.\]
Example 14.
Differentiate the rational function \(y = \frac{{3{x^2} + 1}}{{x - 1}}.\)
Solution.
Using the quotient and power rules, we have
\[y^\prime = \left( {\frac{{3{x^2} + 1}}{{x - 1}}} \right)^\prime = \frac{{6x \cdot \left( {x - 1} \right) - \left( {3{x^2} + 1} \right) \cdot 1}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{\color{darkgreen}{6{x^2}} - \color{red}{6x} - \color{darkgreen}{3{x^2}} - \color{blue}{1}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{\color{darkgreen}{3{x^2}} - \color{red}{6x} - \color{blue}{1}}}{{{{\left( {x - 1} \right)}^2}}}.\]
Example 15.
Differentiate the function \(z = \frac{{1 + {x^2}}}{{1 - {x^2}}}.\)
Solution.
Using the quotient rule, we have
\[z^\prime = \left( {\frac{{1 + {x^2}}}{{1 - {x^2}}}} \right)^\prime = \frac{{2x \cdot \left( {1 - {x^2}} \right) - \left( {1 + {x^2}} \right) \cdot \left( { - 2x} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}} = \frac{{\color{blue}{2x} - \cancel{\color{red}{2{x^3}}} + \color{blue}{2x} + \cancel{\color{red}{2{x^3}}}}}{{{{\left( {1 - {x^2}} \right)}^2}}} = \frac{{4x}}{{{{\left( {1 - {x^2}} \right)}^2}}}.\]
Example 16.
Calculate the derivative of the function \(y = {\frac{{1 + \cos x}}{{\sin x}}}.\)
Solution.
By the quotient rule, we can write
\[y'\left( x \right) = \left( {\frac{{1 + \cos x}}{{\sin x}}} \right)^\prime = \frac{{\left( { - \sin x} \right)\sin x - \left( {1 + \cos x} \right)\cos x}}{{{{\sin }^2}x}} = \frac{{ - {{\sin }^2}x - \cos x - {{\cos }^2}x}}{{{{\sin }^2}x}} = \frac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - \cos x}}{{{{\sin }^2}x}} = \frac{{ - 1 - \cos x}}{{{{\sin }^2}x}} = \frac{{ - 1 - \cos x}}{{1 - {{\cos}^2}x}} = \frac{{ - \cancel{\left( {1 + \cos x} \right)}}}{{\left( {1 - \cos x} \right)\cancel{\left( {1 + \cos x} \right)}}} = \frac{{ - 1}}{{1 - \cos x}} = \frac{1}{{\cos x - 1}}.\]
Note that the final expression for the derivative has a domain different from the domain of the original function. This is caused by the loss of the root when the numerator and denominator are reduced by \({\left( {1 + \cos x} \right)}\). In fact, the domain of the original function and its derivative is the whole set of real numbers \(\mathbb{R}\), except for those ones in which
\[\sin x = 0\;\;\text{or}\;\;x = \pi n,\;n \in \mathbb{Z}.\]
Example 17.
Find the derivative of the cosecant function \(y = \csc x.\)
Solution.
We can calculate the derivative of \(y = \csc x\) with the help of the quotient rule.
\[y^\prime = \left( {\csc x} \right)^\prime = \left( {\frac{1}{{\sin x}}} \right)^\prime = \frac{{1^\prime \cdot \sin x - 1 \cdot \left( {\sin x} \right)^\prime}}{{{{\sin }^2}x}} = \frac{{0 \cdot \sin x - 1 \cdot \cos x}}{{{{\sin }^2}x}} = - \frac{{\cos x}}{{{{\sin }^2}x}} = - \frac{{\cos x}}{{\sin x}} \cdot \frac{1}{{\sin x}} = - \cot x\csc x.\]
Example 18.
Find the derivative of the linear fractional function \(y = {\frac{{ax + b}}{{cx + d}}}.\)
Solution.
By the quotient rule, we get:
\[y'\left( x \right) = \left( {\frac{{ax + b}}{{cx + d}}} \right)^\prime = \frac{{a\left( {cx + d} \right) - c\left( {ax + b} \right)}}{{{{\left( {cx + d} \right)}^2}}} = \frac{{\cancel{\color{blue}{acx}} + ad - \cancel{\color{blue}{acx}} - bc}}{{{{\left( {cx + d} \right)}^2}}} = \frac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}.\]
Note that the numerator can be written in terms of the determinant:
\[ad - bc = \left| {\begin{array}{*{20}{c}} a & b\\ c & d \end{array}} \right|.\]
The final expression for the derivative is as follows:
\[y'\left( x \right) = \frac{{\left| {\begin{array}{*{20}{c}} a & b\\ c & d \end{array}} \right|}}{{{{\left( {cx + d} \right)}^2}}}.\]
Example 19.
Calculate the derivative of the function \(y = {\frac{{3x - 1}}{{{x^4}}}}.\)
\[y'\left( x \right) = \left( {\frac{{3x - 1}}{{{x^4}}}} \right)^\prime = \frac{{3 \cdot {x^4} - \left( {3x - 1} \right) \cdot 4{x^3}}}{{{x^8}}} = \frac{{\color{blue}{3{x^4}} - \color{blue}{12{x^4}} + \color{black}{4{x^3}}}}{{{x^8}}} = \frac{{4{x^3} - \color{blue}{9{x^4}}}}{{{x^8}}} = \frac{{{x^3}\left( {4 - 9x} \right)}}{{{x^8}}} = \frac{{4 - 9x}}{{{x^5}}}.\]
Example 20.
Find the derivative of the function \(y = \frac{3}{{5 - x}} + \frac{{{x^2}}}{3}\) at \(x = 0.\)
Solution.
The function is the sum of two fractions. We apply the quotient rule to the first fraction:
\[\left( {\frac{3}{{5 - x}}} \right)^\prime = \frac{{3^\prime \cdot \left( {5 - x} \right) - 3 \cdot \left( {5 - x} \right)^\prime}}{{{{\left( {5 - x} \right)}^2}}} = \frac{{0 \cdot \left( {5 - x} \right) - 3 \cdot \left( { - 1} \right)}}{{{{\left( {5 - x} \right)}^2}}} = \frac{3}{{{{\left( {5 - x} \right)}^2}}}.\]
Using the constant multiple rule and the power rule, we find the derivative of the second fraction:
\[\left( {\frac{{{x^2}}}{3}} \right)^\prime = \frac{1}{3}\left( {{x^2}} \right)^\prime = \frac{1}{3} \cdot 2x = \frac{{2x}}{3}.\]
Hence, the derivative of the original function is written in the form
\[y^\prime = \frac{3}{{{{\left( {5 - x} \right)}^2}}} + \frac{{2x}}{3}.\]
At \(x = 0,\) we have
\[y^\prime\left( 0 \right) = \frac{3}{{{{\left( {5 - 0} \right)}^2}}} + \frac{{2 \cdot 0}}{3} = \frac{3}{{25}}.\]
Example 21.
Calculate the derivative of the following function: \(y = {\frac{{\sqrt x - 1}}{{\sqrt x + 1}}}.\)
Solution.
Using the quotient rule we have
\[y'\left( x \right) = \left( {\frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right)^\prime = \frac{{\frac{1}{{2\sqrt x }}\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 1} \right)\frac{1}{{2\sqrt x }}}}{{{{\left( {\sqrt x + 1} \right)}^2}}} = \frac{{\frac{1}{{2\sqrt x }}\left( {\cancel{\color{blue}{\sqrt x}} + \color{red}{1} - \cancel{\color{blue}{\sqrt x}} + \color{red}{1}} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}} = \frac{{\frac{1}{{2\sqrt x }} \cdot \color{red}{2}}}{{{{\left( {\sqrt x + 1} \right)}^2}}} = \frac{1}{{\sqrt x {{\left( {\sqrt x + 1} \right)}^2}}}.\]
Example 22.
Derive a formula for the derivative of the function \(f\left( x \right) = {\frac{{u\left( x \right)v\left( x \right)}}{{w\left( x \right)}}}.\)
Solution.
First we differentiate this function by the quotient rule:
\[f'\left( x \right) = \left( {\frac{{uv}}{w}} \right)^\prime = \frac{{{{\left( {uv} \right)}^\prime } \cdot w - uv \cdot w'}}{{{w^2}}}.\]
Next, using the product rule, we get:
\[f'\left( x \right) = \frac{{{{\left( {uv} \right)}^\prime } \cdot w - uv \cdot w'}}{{{w^2}}} = \frac{{\left( {u'v + uv'} \right)w - uvw'}}{{{w^2}}} = \frac{{u'vw + uv'w - uvw'}}{{{w^2}}}.\]
