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:

\[\begin{aligned} y & = & \text{sin} x \\ \frac{d y}{d x} & = & \text{cos} x \\ \frac{d^{2} y}{d x^{2}} & = & − \text{sin} x \\ \frac{d^{3} y}{d x^{3}} & = & − \text{cos} x \\ \frac{d^{4} y}{d x^{4}} & = & \text{sin} x . \end{aligned}\]

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

\[\begin{aligned} \frac{d^{4}}{d x^{4}} \left(\right. \text{sin} x \left.\right) = \frac{d^{8}}{d x^{8}} \left(\right. \text{sin} x \left.\right) = \frac{d^{12}}{d x^{12}} \left(\right. \text{sin} x \left.\right) = \ldots = \frac{d^{4 n}}{d x^{4 n}} \left(\right. \text{sin} x \left.\right) = \text{sin} x \\ \frac{d^{5}}{d x^{5}} \left(\right. \text{sin} x \left.\right) = \frac{d^{9}}{d x^{9}} \left(\right. \text{sin} x \left.\right) = \frac{d^{13}}{d x^{13}} \left(\right. \text{sin} x \left.\right) = \ldots = \frac{d^{4 n + 1}}{d x^{4 n + 1}} \left(\right. \text{sin} x \left.\right) = \text{cos} x . \end{aligned}\]

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

\[\frac{d^{74}}{d x^{74}} \left(\right. \text{sin} x \left.\right) = \frac{d^{72 + 2}}{d x^{72 + 2}} \left(\right. \text{sin} x \left.\right) = \frac{d^{2}}{d x^{2}} \left(\right. \text{sin} x \left.\right) = − \text{sin} x .\]

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 \(v \left(\right. t \left.\right) = s^{'} \left(\right. t \left.\right) :\)

\[v \left(\right. t \left.\right) = s^{'} \left(\right. t \left.\right) = − \text{cos} t .\]

Thus,

\[v \left(\right. \frac{\pi}{4} \left.\right) = - \frac{1}{\sqrt{2}} .\]

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

\[a \left(\right. \frac{\pi}{4} \left.\right) = \frac{1}{\sqrt{2}} .\]

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

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