Quotient Rule
The quotient rule is a formal rule for differentiating of a quotient of functions.
Let u (x) and v (x) be again differentiable functions. Then, if v (x) ≠ 0, the derivative of the quotient of these functions is calculated by the formula
\[\left( {\frac{u}{v}} \right)^\prime = \frac{{u'v - uv'}}{{{v^2}}}.\]
To prove this formula, consider the increment of the quotient:
\[\require{cancel} \Delta \left( {\frac{u}{v}} \right) = \frac{{u + \Delta u}}{{v + \Delta v}} - \frac{u}{v} = \frac{{\left( {u + \Delta u} \right)v - u\left( {v + \Delta v} \right)}}{{v\left( {v + \Delta v} \right)}} = \frac{{\cancel{uv} + v\Delta u - \cancel{uv} - u\Delta v}}{{v\left( {v + \Delta v} \right)}} = \frac{{v\Delta u - u\Delta v}}{{v\left( {v + \Delta v} \right)}}.\]
The derivative of the quotient is expressed as follows:
\[\left( {\frac{u}{v}} \right)^\prime = \lim\limits_{\Delta x \to 0} \frac{{\Delta \left( {\frac{u}{v}} \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\frac{{v\Delta u - u\Delta v}}{{{v^2} + v\Delta v}}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{v\frac{{\Delta u}}{{\Delta x}} - u\frac{{\Delta v}}{{\Delta x}}}}{{{v^2} + v\Delta v}}.\]
Next, 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
Example 1.
Find the derivative of the function \({y = {\frac{2}{x}}}.\)
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 \(y\left( x \right) = {\frac{1}{{{x^n}}}}\) 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 \({y = \frac{{x + 1}}{{x - 1}}.}\)
Solution.
Let \(u = x + 1,\) \(v = x - 1.\)
By the quotient rule \(\left( {\frac{u}{v}} \right)^\prime = \frac{{u^\prime v - uv^\prime}}{{{v^2}}},\) 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 \(y = \frac{3}{{{x^3}}}.\)
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 \(y = \frac{{2x + 1}}{{2x - 1}}\) 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 \(\tan x = {\frac{{\sin x}}{{\cos x}}}\). 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 \(\cot x = \frac{{\cos x}}{{\sin x}},\) we can apply the quotient rule:
\[y^\prime = \left( {\cot x} \right)^\prime = \left( {\frac{{\cos x}}{{\sin x}}} \right)^\prime = \frac{{\left( {\cos x} \right)^\prime\sin x - \cos x\left( {\sin x} \right)^\prime}}{{{{\sin }^2}x}} = \frac{{\left( { - \sin x} \right) \cdot \sin x - \cos x \cdot \cos x}}{{{{\sin }^2}x}} = \frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}} = - \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} = - \frac{1}{{{{\sin }^2}x}} = - {\csc ^2}x.\]
Example 8.
Find the derivative of the secant function \(y = \sec x.\)
Solution.
The derivative of secant can be calculated using the quotient rule:
\[y'\left( x \right) = {\left( {\sec x} \right)^\prime } = {\left( {\frac{1}{{\cos x}}} \right)^\prime } = \frac{{1' \cdot \cos x - 1 \cdot {{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^2}x}} = \frac{{0 \cdot \cos x - 1 \cdot \left( { - \sin x} \right)}}{{{{\cos }^2}x}} = \frac{{\sin x}}{{{{\cos }^2}x}} = \frac{{\sin x}}{{\cos x}} \cdot \frac{1}{{\cos x}} = \tan x\sec x.\]
See more problems on Page 2.
