Product Rule
Solved Problems
Example 7.
Given the function \(z = \left( {{x^2} + 1} \right)\left( {x - 1} \right)\). Find the value of its derivative at \(x = -1.\)
Solution.
We determine the derivative by the product rule:
\[z'\left( x \right) = \left[ {\left( {{x^2} + 1} \right)\left( {x - 1} \right)} \right]^\prime = 2x \cdot \left( {x - 1} \right) + \left( {{x^2} + 1} \right) \cdot 1 = \color{blue}{2{x^2}} - \color{black}{2x} + \color{blue}{x^2} + \color{black}{1} = \color{blue}{3{x^2}} - \color{black}{2x} + \color{black}{1}.\]
At the point \(x = -1,\) the derivative is equal
\[z'\left( { - 1} \right) = 3 \cdot {\left( { - 1} \right)^2} - 2 \cdot \left( { - 1} \right) + 1 = 3 + 2 + 1 = 6.\]
Example 8.
Determine the derivative of the function \(y = x\ln x - x.\)
Solution.
Given that \(\left( {\ln x} \right)^\prime = \frac{1}{x}\) and using the product rule, we get
\[\require{cancel} y^\prime = \left( {x\ln x - x} \right)^\prime = \left( {x\ln x} \right)^\prime - x^\prime = x^\prime\ln x + x\left( {\ln x} \right)^\prime - x^\prime = 1 \cdot \ln x + x \cdot \frac{1}{x} - 1 = \ln x + \cancel{1} - \cancel{1} = \ln x.\]
Example 9.
Differentiate the function \(y = \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right).\)
Solution.
We use the product rule:
\[y^\prime = \left[ {\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)} \right]^\prime = \left( {\sqrt x - 1} \right)^\prime\left( {\sqrt x + 1} \right) + \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)^\prime = \frac{1}{{2\sqrt x }} \cdot \left( {\sqrt x + 1} \right) + \left( {\sqrt x - 1} \right) \cdot \frac{1}{{2\sqrt x }} = \frac{\cancel{\sqrt x }}{{2\cancel{\sqrt x }}} + \cancel{\frac{1}{{2\sqrt x }}} + \frac{\cancel{\sqrt x }}{{2\cancel{\sqrt x }}} - \cancel{\frac{1}{{2\sqrt x }}} = \frac{1}{2} + \frac{1}{2} = 1.\]
Example 10.
Find the derivative of the fractional function \(y = \frac{{2x + 1}}{{{x^3}}}.\)
Solution.
We write the given function as the product of two functions:
\[y = \frac{{2x + 1}}{{{x^3}}} = \left( {2x + 1} \right){x^{ - 3}}.\]
Then
\[y^\prime = \left( {\left( {2x + 1} \right){x^{ - 3}}} \right)^\prime = \left( {2x + 1} \right)^\prime{x^{ - 3}} + \left( {2x + 1} \right)\left( {{x^{ - 3}}} \right)^\prime = 2 \cdot {x^{ - 3}} + \left( {2x + 1} \right) \cdot \left( { - 3{x^{ - 4}}} \right) = \frac{2}{{{x^3}}} - \frac{{3\left( {2x + 1} \right)}}{{{x^4}}} = \frac{{2x - 6x - 3}}{{{x^4}}} = - \frac{{4x + 3}}{{{x^4}}}.\]
Example 11.
Differentiate the function \(y = {x^3}{2^x}.\)
Solution.
By the product rule, we obtain:
\[y'\left( x \right) = \left( {{x^3}{2^x}} \right)^\prime = {\left( {{x^3}} \right)^\prime }{2^x} + {x^3}{\left( {{2^x}} \right)^\prime } = 3{x^2} \cdot {2^x} + {x^3} \cdot {2^x}\ln 2 = {x^2}{2^x}\left( {3 + x\ln 2} \right).\]
Example 12.
Find the derivative of the function \(y = \left( {{e^x} - 1} \right)\left( {{e^x} + 2} \right)\) and calculate its value at \(x = 0.\)
Solution.
By the product rule,
\[y^\prime = \left[ {\left( {{e^x} - 1} \right)\left( {{e^x} + 2} \right)} \right]^\prime = \left( {{e^x} - 1} \right)^\prime\left( {{e^x} + 2} \right) + \left( {{e^x} - 1} \right)\left( {{e^x} + 2} \right)^\prime = {e^x}\left( {{e^x} + 2} \right) + \left( {{e^x} - 1} \right){e^x} = \color{blue}{e^{2x}} + \color{red}{2{e^x}} + \color{blue}{e^{2x}} - \color{red}{e^x} = \color{blue}{2{e^{2x}}} + \color{red}{e^x}.\]
The derivative at \(x = 0\) is equal to
\[y^\prime\left( 0 \right) = 2{e^{2 \cdot 0}} + {e^0} = 2 \cdot 1 + 1 = 3.\]
Example 13.
Find the derivative of the function \(y = {e^x}\sin x\) at \(x = 0.\)
Solution.
By the product rule,
\[y^\prime\left( x \right) = \left( {{e^x}\sin x} \right)^\prime = \left( {{e^x}} \right)^\prime\sin x + {e^x}\left( {\sin x} \right)^\prime = {e^x} \cdot \sin x + {e^x} \cdot \cos x = {e^x}\left( {\sin x + \cos x} \right).\]
At \(x = 0\) the derivative has the following value:
\[y^\prime\left( 0 \right) = {e^0}\left( {\sin 0 + \cos 0} \right) = 1 \cdot \left( {0 + 1} \right) = 1.\]
Example 14.
Differentiate the function \(y = {e^x}\left( {\sin x - \cos x} \right).\)
Solution.
Using the product rule, we get:
\[y^\prime = \left[ {{e^x}\left( {\sin x - \cos x} \right)} \right]^\prime = \left( {{e^x}} \right)^\prime\left( {\sin x - \cos x} \right) + {e^x}\left( {\sin x - \cos x} \right)^\prime = {e^x}\left( {\sin x - \cos x} \right) + {e^x}\left( {\cos x + \sin x} \right) = {e^x}\left( {\color{blue}{\sin x} - \cancel{\color{red}{\cos x}} + \cancel{\color{red}{\cos x}} + \color{blue}{\sin x}} \right) = 2{e^x}\sin x.\]
Example 15.
Derive a formula for the derivative of the product of three functions.
Solution.
Let \(f\left( x \right) = u\left( x \right)v\left( x \right)w\left( x \right)\). Apply the product rule twice:
\[f'\left( x \right) = \left[ {u\left( x \right)v\left( x \right)w\left( x \right)} \right]^\prime = {\left[ {u\left( x \right)v\left( x \right)} \right]^\prime }w\left( x \right) + \left[ {u\left( x \right)v\left( x \right)} \right]w'\left( x \right). \]
Since \({\left[ {u\left( x \right)v\left( x \right)} \right]^\prime } = u'v + uv',\) we obtain the following formula:
\[f'\left( x \right) = {\left( {uvw} \right)^\prime } = \left( {u'v + uv'} \right)w + uvw' = u'vw + uv'w + uvw'. \]
Example 16.
Find the derivative of the function \(y = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\) at \(x = 0.\)
Solution.
The derivative of the product of three functions is given by the formula
\[\left( {uvw} \right)^\prime = u^\prime vw + uv^\prime w + uvw^\prime .\]
Hence
\[y^\prime = \left[ {\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)} \right]^\prime = \left( {x - 1} \right)^\prime\left( {x - 2} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\left( {x - 2} \right)^\prime\left( {x - 3} \right) + \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)^\prime = \left( {x - 2} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\left( {x - 3} \right) + \left( {x - 1} \right)\left( {x - 2} \right) = \left( {{x^2} - 2x - 3x + 6} \right) + \left( {{x^2} - x - 3x + 3} \right) + \left( {{x^2} - x - 2x + 2} \right) = \color{darkgreen}{x^2} - \color{red}{5x} + \color{blue}{6} + \color{darkgreen}{x^2} - \color{red}{4x} + \color{blue}{3} + \color{darkgreen}{x^2} - \color{red}{3x} + \color{blue}{2} = \color{darkgreen}{3{x^2}} - \color{red}{12x} + \color{blue}{11}.\]
Substituting \(x = 0,\) we have
\[y^\prime\left( 0 \right) = 3 \cdot {0^2} - 12 \cdot 0 + 11 = 11.\]
