The Chain and Power Rules Combined
We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form \(h \left(\right. x \left.\right) = \left(\left(\right. g \left(\right. x \left.\right) \left.\right)\right)^{n} ,\) we need to use the chain rule combined with the power rule. To do so, we can think of \(h \left(\right. x \left.\right) = \left(\right. g \left(\right. x \left.\right) \left.\right)^{n}\) as \(f \left(\right. g \left(\right. x \left.\right) \left.\right)\) where \(f \left(\right. x \left.\right) = x^{n} .\) Then \(f^{'} \left(\right. x \left.\right) = n x^{n - 1} .\) Thus, \(f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) = n \left(\right. g \left(\right. x \left.\right) \left.\right)^{n - 1} .\) This leads us to the derivative of a power function using the chain rule,
Rule: Power Rule for Composition of Functions
For all values of x for which the derivative is defined, if
Then
Example 3.48
Using the Chain and Power Rules
Find the derivative of \(h \left(\right. x \left.\right) = \frac{1}{\left(\right. 3 x^{2} + 1 \left.\right)^{2}} .\)
Solution
First, rewrite \(h \left(\right. x \left.\right) = \frac{1}{\left(\right. 3 x^{2} + 1 \left.\right)^{2}} = \left(\right. 3 x^{2} + 1 \left.\right)^{−2} .\)
Applying the power rule with \(g \left(\right. x \left.\right) = 3 x^{2} + 1 ,\) we have
Rewriting back to the original form gives us
Checkpoint 3.34
Find the derivative of \(h \left(\right. x \left.\right) = \left(\right. 2 x^{3} + 2 x - 1 \left.\right)^{4} .\)
Example 3.49
Using the Chain and Power Rules with a Trigonometric Function
Find the derivative of \(h \left(\right. x \left.\right) = \text{sin}^{3} x .\)
Solution
First recall that \(\text{sin}^{3} x = \left(\right. \text{sin} x \left.\right)^{3} ,\) so we can rewrite \(h \left(\right. x \left.\right) = \text{sin}^{3} x\) as \(h \left(\right. x \left.\right) = \left(\right. \text{sin} x \left.\right)^{3} .\)
Applying the power rule with \(g \left(\right. x \left.\right) = \text{sin} x ,\) we obtain
Example 3.50
Finding the Equation of a Tangent Line
Find the equation of a line tangent to the graph of \(h \left(\right. x \left.\right) = \frac{1}{\left(\left(\right. 3 x - 5 \left.\right)\right)^{2}}\) at \(x = 2 .\)
Solution
Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the y-coordinate, substitute 2 into \(h \left(\right. x \left.\right) .\) Since \(h \left(\right. 2 \left.\right) = \frac{1}{\left(\right. 3 \left(\right. 2 \left.\right) - 5 \left.\right)^{2}} = 1 ,\) the point is \(\left(\right. 2 , 1 \left.\right) .\)
For the slope, we need \(h^{'} \left(\right. 2 \left.\right) .\) To find \(h^{'} \left(\right. x \left.\right) ,\) first we rewrite \(h \left(\right. x \left.\right) = \left(\right. 3 x - 5 \left.\right)^{−2}\) and apply the power rule to obtain
By substituting, we have \(h^{'} \left(\right. 2 \left.\right) = −6 \left(\right. 3 \left(\right. 2 \left.\right) - 5 \left.\right)^{−3} = −6 .\) Therefore, the line has equation \(y - 1 = −6 \left(\right. x - 2 \left.\right) .\) Rewriting, the equation of the line is \(y = −6 x + 13 .\)
Checkpoint 3.35
Find the equation of the line tangent to the graph of \(f \left(\right. x \left.\right) = \left(\right. x^{2} - 2 \left.\right)^{3}\) at \(x = −2 .\)
This lesson is part of:
Derivatives