Basic Differentiation Rules
Solved Problems
Example 7.
Differentiate the function \(y = \frac{{\sin x}}{2} + 3{x^3}.\)
Solution.
Using the sum and constant multiple rules, we obtain:
\[y^\prime = \left( {\frac{{\sin x}}{2} + 3{x^3}} \right)^\prime = \left( {\frac{{\sin x}}{2}} \right)^\prime + \left( {3{x^3}} \right)^\prime = \frac{1}{2}\left( {\sin x} \right)^\prime + 3\left( {{x^3}} \right) = \frac{1}{2} \cdot \cos x + 3 \cdot 3{x^2} = \frac{{\cos x}}{2} + 9{x^2}.\]
Example 8.
Find the derivative of the function \(y = \sqrt[3]{x} + 8x.\)
Solution.
Using the differentiation formulas above, we have
\[y'\left( x \right) = \left( {\sqrt[3]{x} + 8x} \right)^\prime = \left( {\sqrt[3]{x}} \right)^\prime + \left( {8x} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime + 8\left( x \right)^\prime .\]
Here the first term is a power function with the exponent \(1/3.\) Then the derivative is given by the following expression:
\[y'\left( x \right) = \left( {{x^{\frac{1}{3}}}} \right)^\prime + 8\left( x \right)^\prime = \frac{1}{3}{x^{\frac{1}{3} - 1}} + 8 \cdot 1 = \frac{1}{3}{x^{ - \frac{2}{3}}} + 8 = \frac{1}{{3\sqrt[3]{{{x^2}}}}} + 8.\]
Example 9.
Calculate the derivative of the function \(y = x\left( {2 + 3x} \right)\) without using the product rule.
Solution.
We can write
\[y = x\left( {2 + 3x} \right) = 2x + 3{x^2}.\]
Using the basic differentiation rules, we have
\[y^\prime = \left( {2x + 3{x^2}} \right)^\prime = \left( {2x} \right)^\prime + \left( {3{x^3}} \right)^\prime = 2x^\prime + 3\left( {{x^2}} \right)^\prime = 2 \cdot 1 + 3 \cdot 2x = 2 + 6x.\]
Example 10.
Find the derivative of the function \({y = 3{x^2} - 2\sqrt x + 1.}\)
Solution.
We first apply the sum and difference rules:
\[y^\prime = \left( {3{x^2} - 2\sqrt x + 1} \right)^\prime = \left( {3{x^2}} \right)^\prime - \left( {2\sqrt x } \right)^\prime + 1^\prime.\]
For the first two terms, we can use the constant multiple rule:
\[y^\prime = 3\left( {{x^2}} \right)^\prime - 2\left( {\sqrt x } \right)^\prime + 1^\prime.\]
The derivatives of \({{x^2}}\) and \({\sqrt x }\) are
\[\left( {{x^2}} \right)^\prime = 2x,\;\;\left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }}.\]
The derivatives of the constant number \(1\) is zero. Hence
\[y^\prime = 3 \cdot 2x - 2 \cdot \frac{1}{{2\sqrt x }} + 0 = 6x - \frac{1}{{\sqrt x }}.\]
Example 11.
Calculate the derivative of the function \(y = \left( {2 - {\frac{x}{3}}} \right)\left( {{\frac{1}{3}} + {x^2}} \right).\)
Solution.
To solve this example using the above differentiation rules, we multiply the expressions in the brackets and write the function in the form
\[y\left( x \right) = \left( {2 - \frac{x}{3}} \right)\left( {\frac{1}{3} + {x^2}} \right) = \frac{2}{3} - \frac{x}{9} + 2{x^2} - \frac{{{x^3}}}{3}.\]
Now it is easy to find the derivative:
\[y'\left( x \right) = \left( {\frac{2}{3} - \frac{x}{9} + 2{x^2} - \frac{{{x^3}}}{3}} \right)^\prime = - \frac{1}{9} + 4x - {x^2} = 4x - {x^2} - \frac{1}{9}.\]
Example 12.
Calculate the derivative of the function \(y = \left( {x - 1} \right){\left( {x - 2} \right)^2}\) without using the product rule for the derivative.
Solution.
By expanding the brackets, we can write the given function as
\[y\left( x \right) = \left( {x - 1} \right){\left( {x - 2} \right)^2} = \left( {x - 1} \right)\left( {{x^2} - 4x + 4} \right) = {x^3} - {5{x^2}} + {8x} - 4.\]
The derivative of this function is
\[y'\left( x \right) = \left( {{x^3} - 5{x^2} + 8x - 4} \right)^\prime = 3{x^2} - 10x + 8.\]
Example 13.
Calculate the derivative of the function \(y = \frac{{{x^2} - x - 1}}{x}\) without using the quotient rule.
Solution.
We first rewrite the function in the form
\[\require{cancel} y = \frac{{{x^2} - x - 1}}{x} = {\frac{{{x^\cancel{2}}}}{\cancel{x}}} - \frac{\cancel{x}}{\cancel{x}} - {\frac{1}{x}} = x - 1 - {\frac{1}{x}}.\]
Differentiate this expression using the difference and constant rules:
\[y^\prime = \left( {x - 1 - \frac{1}{x}} \right)^\prime = x^\prime - 1^\prime - \left( {\frac{1}{x}} \right)^\prime = 1 - 0 - \left( { - \frac{1}{{{x^2}}}} \right) = 1 + \frac{1}{{{x^2}}}.\]
Example 14.
Find the derivative of the function \(y = {\frac{{{x^2} + 3x + 1}}{x}}\) without using the quotient rule for the derivative.
Solution.
By dividing the numerator by the denominator term by term, we write the function as
\[y\left( x \right) = \frac{{{x^2} + 3x + 1}}{x} = \frac{{{x^2}}}{x} + \frac{{3x}}{x} + \frac{1}{x} = x + 3 + \frac{1}{x}.\]
Next, using the linear properties of the derivative, we have
\[y'\left( x \right) = \left( {x + 3 + \frac{1}{x}} \right)^\prime = x^\prime + 3' + \left( {\frac{1}{x}} \right)^\prime = 1 + 0 + \left( { - \frac{1}{{{x^2}}}} \right) = 1 - \frac{1}{{{x^2}}} = \frac{{{x^2} - 1}}{{{x^2}}}.\]
Example 15.
Find the derivative of the irrational function \(y = 2\sqrt x - 3\sqrt[3]{x}.\)
Solution.
By the difference and constant multiple rule,
\[y^\prime = \left( {2\sqrt x - 3\sqrt[3]{x}} \right)^\prime = \left( {2\sqrt x } \right)^\prime - \left( {3\sqrt[3]{x}} \right)^\prime = 2\left( {\sqrt x } \right)^\prime - 3\left( {\sqrt[3]{x}} \right)^\prime.\]
As \(\left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }},\) \(\left( {\sqrt[3]{x}} \right)^\prime = \frac{1}{{3\sqrt[3]{{{x^2}}}}},\) we have
\[y^\prime = 2 \cdot \frac{1}{{2\sqrt x }} - 3 \cdot \frac{1}{{3\sqrt[3]{{{x^2}}}}} = \frac{1}{{\sqrt x }} - \frac{1}{{\sqrt[3]{{{x^2}}}}}.\]
Example 16.
Find the derivative of the function \(y = {\left( {x + 4} \right)^3}\) without using the power rule.
Solution.
Using the perfect cube identity
\[{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3},\]
we can write:
\[y = {\left( {x + 4} \right)^3} = {x^3} + 3{x^2} \cdot 4 + 3x \cdot {4^2} + {4^3} = {x^3} + 12{x^2} + 48x + 64.\]
Apply the basic differentiation properties:
\[y^\prime = \left( {{x^3} + 12{x^2} + 48x + 64} \right)^\prime = \left( {{x^3}} \right)^\prime + \left( {12{x^2}} \right)^\prime + \left( {48x} \right)^\prime + 64^\prime = \left( {{x^3}} \right)^\prime + 12\left( {{x^2}} \right)^\prime + 48x^\prime + 64^\prime = 3{x^2} + 12 \cdot 2x + 48 \cdot 1 + 0 = 3{x^2} + 24x + 48 = 3\left( {{x^2} + 8x + 16} \right) = 3{\left( {x + 4} \right)^2}.\]
Example 17.
Determine the derivative of \(y = {\left( {x - 1} \right)^4}\) without using the power rule.
Solution.
We apply the product identity to expand the given expression:
\[{\left( {a - b} \right)^4} = {a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}.\]
Hence
\[y = {\left( {x - 1} \right)^4} = {x^4} - 4{x^3} + 6{x^2} - 4x + 1.\]
Take the derivative using the sum, difference and constant multiple rules.
\[y^\prime = \left( {{{\left( {x - 1} \right)}^4}} \right)^\prime = \left( {{x^4} - 4{x^3} + 6{x^2} - 4x + 1} \right)^\prime = \left( {{x^4}} \right)^\prime - \left( {4{x^3}} \right)^\prime + \left( {6{x^2}} \right)^\prime - \left( {4x} \right)^\prime + 1^\prime = \left( {{x^4}} \right)^\prime - 4\left( {{x^3}} \right)^\prime + 6\left( {{x^2}} \right)^\prime - 4x^\prime + 1^\prime = 4{x^3} - 4 \cdot 3{x^2} + 6 \cdot 2x - 4 \cdot 1 + 0 = 4{x^3} - 12{x^2} + 12x - 4 = 4\left( {{x^3} - 3{x^2} + 3x - 1} \right) = 4{\left( {x - 1} \right)^3}.\]
