Higher-Order Derivatives
The higher-order derivatives of \(\text{sin} x\) and \(\text{cos} x\) follow a repeating pattern. By following the pattern, we can find any higher-order derivative of \(\text{sin} x\) and \(\text{cos} x .\)
Example 3.45
Finding Higher-Order Derivatives of \(y = \text{sin} x\)
Find the first four derivatives of \(y = \text{sin} x .\)
Solution
Each step in the chain is straightforward:
Analysis
Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x, so
Checkpoint 3.31
For \(y = \text{cos} x ,\) find \(\frac{d^{4} y}{d x^{4}} .\)
Example 3.46
Using the Pattern for Higher-Order Derivatives of \(y = \text{sin} x\)
Find \(\frac{d^{74}}{d x^{74}} \left(\right. \text{sin} x \left.\right) .\)
Solution
We can see right away that for the 74th derivative of \(\text{sin} x , 74 = 4 \left(\right. 18 \left.\right) + 2 ,\) so
Checkpoint 3.32
For \(y = \text{sin} x ,\) find \(\frac{d^{59}}{d x^{59}} \left(\right. \text{sin} x \left.\right) .\)
Example 3.47
An Application to Acceleration
A particle moves along a coordinate axis in such a way that its position at time \(t\) is given by \(s \left(\right. t \left.\right) = 2 - \text{sin} t .\) Find \(v \left(\right. \pi / 4 \left.\right)\) and \(a \left(\right. \pi / 4 \left.\right) .\) Compare these values and decide whether the particle is speeding up or slowing down.
Solution
First find \(v \left(\right. t \left.\right) = s^{'} \left(\right. t \left.\right) :\)
Thus,
Next, find \(a \left(\right. t \left.\right) = v^{'} \left(\right. t \left.\right) .\) Thus, \(a \left(\right. t \left.\right) = v^{'} \left(\right. t \left.\right) = \text{sin} t\) and we have
Since \(v \left(\right. \frac{\pi}{4} \left.\right) = - \frac{1}{\sqrt{2}} < 0\) and \(a \left(\right. \frac{\pi}{4} \left.\right) = \frac{1}{\sqrt{2}} > 0 ,\) we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down.
Checkpoint 3.33
A block attached to a spring is moving vertically. Its position at time \(t\) is given by \(s \left(\right. t \left.\right) = 2 \text{sin} t .\) Find \(v \left(\right. \frac{5 \pi}{6} \left.\right)\) and \(a \left(\right. \frac{5 \pi}{6} \left.\right) .\) Compare these values and decide whether the block is speeding up or slowing down.
This lesson is part of:
Derivatives