Derivatives of Trigonometric Functions: Learn It 3

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Derivatives of Trigonometric Functions: Learn It 3

Higher-Order Derivatives of Trig Functions

The higher-order derivatives of $\sin x$ and $\cos x$ exhibit a cyclical pattern, making it possible to predict any higher-order derivative of these functions. By understanding this repeating sequence, you can easily compute derivatives beyond the first order.

To illustrate, let’s calculate the first four derivatives of $y= \sin x$ .

$\begin{array}{ll} \text{Start with the function itself:} & y = \sin x \\ \text{The first derivative of } \sin x \text{ is:} & \dfrac{dy}{dx} = \cos x \\ \text{The second derivative becomes:} & \dfrac{d^2y}{dx^2} = -\sin x \\ \text{Continuing, the third derivative is:} & \dfrac{d^3y}{dx^3} = -\cos x \\ \text{The fourth derivative brings us back to the starting function:} & \dfrac{d^4y}{dx^4} = \sin x \end{array}$

This sequence of derivatives demonstrates a pattern that repeats every four derivatives.

$\sin x$ leads to $\cos x$
$\cos x$ leads to $- \sin x$
$- \sin x$ leads to $- \cos x$
$- \cos x$ leads back to $\sin x$

Understanding this cyclical pattern not only simplifies calculations but also equips us with a systematic approach for determining any higher-order derivative.

How to: Use the Cyclic Pattern to Determine Higher-Order Derivatives of Sine and Cosine Functions

Determine the order of the derivative you need (let’s call it $n$ ).
Calculate the remainder when $n$ is divided by $4$ . The remainder determines the position in the cycle:
- Remainder 0: The derivative returns to the original function.
- Remainder 1: The derivative progresses to the next function in the cycle.
- Remainder 2: The derivative is the negative of the original function.
- Remainder 3: The derivative is the negative of the next function in the cycle.

For $sin x$ :
- If $n$ is divisible by $4$ (remainder $0$ ), the $n$ -th derivative is $\sin x$ .
- If $n$ divided by $4$ gives a remainder of $1$ , it is $\cos x$ .
- If the remainder is $2$ , it is $- \sin x$ .
- If the remainder is $3$ , it is $- \cos x$ .
For $cos x$ :
- Remainder $0$ : $\cos x$
- Remainder $1$ : $- \sin x$
- Remainder $2$ : $- \cos x$
- Remainder $3$ : $\sin x$

Find $\dfrac{d^{74}}{dx^{74}}(\sin x)$ .

We can see right away that for the $74$ th derivative of $\sin x, \, 74=4(18)+2$ , so,

\frac{d^{74}}{d x^{74}} (\sin x) = \frac{d^{72 + 2}}{d x^{72 + 2}} (\sin x) = \frac{d^{2}}{d x^{2}} (\sin x) = - \sin x

$\dfrac{d^{74}}{dx^{74}}(\sin x)=\dfrac{d^{72+2}}{dx^{72+2}}(\sin x)=\dfrac{d^2}{dx^2}(\sin x)=−\sin x$ .

A particle moves along a coordinate axis in such a way that its position at time $t$ is given by $s(t)=2- \sin t$ .

Find $v\left(\dfrac{\pi}{4}\right)$ and $a\left(\dfrac{\pi}{4}\right)$ . Compare these values and decide whether the particle is speeding up or slowing down.

First find $v(t)=s^{\prime}(t)$ : $v(t)=s^{\prime}(t)=−\cos t$ . Thus, $v\left(\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}$ .

Next, find $a(t)=v^{\prime}(t)$ .

Thus, $a(t)=v^{\prime}(t)= \sin t$ and we have $a\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$ .

Since $v\left(\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}<0$ and $a\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}>0$ , 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 traveling.

Consequently, the particle is slowing down.

Watch the following video to see the worked solution to this example.

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