Higher-Order Derivatives of Trig Functions

The higher-order derivatives of $\sin x$ and $\cos x$ follow a repeating pattern. By following the pattern, we can find any higher-order derivative of $\sin x$ and $\cos x$ .

Example: Finding Higher-Order Derivatives of $y= \sin x$

Find the first four derivatives of $y= \sin x$ .

Show Solution

Each step in the chain is straightforward:

\begin{matrix} y & = & sin x \frac{d y}{d x} & = & cos x \frac{d^{2} y}{d x^{2}} & = & - sin x \frac{d^{3} y}{d x^{3}} & = & - cos x \frac{d^{4} y}{d x^{4}} & = & sin x \end{matrix}

$\begin{array}{lll} y & = & \sin x \\ \frac{dy}{dx} & = & \cos x \\ \frac{d^2 y}{dx^2} & = & −\sin x \\ \frac{d^3 y}{dx^3} & = & −\cos x \\ \frac{d^4 y}{dx^4} & = & \sin x \end{array}$

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{matrix} \frac{d^{4}}{d x^{4}} (sin x) = \frac{d^{8}}{d x^{8}} (sin x) = \frac{d^{12}}{d x^{12}} (sin x) = \dots = \frac{d^{4 n}}{d x^{4 n}} (sin x) = sin x \frac{d^{5}}{d x^{5}} (sin x) = \frac{d^{9}}{d x^{9}} (sin x) = \frac{d^{13}}{d x^{13}} (sin x) = \dots = \frac{d^{4 n + 1}}{d x^{4 n + 1}} (sin x) = cos x \end{matrix}

$\begin{array}{l}\frac{d^4}{dx^4}(\sin x)=\frac{d^8}{dx^8}(\sin x)=\frac{d^{12}}{dx^{12}}(\sin x)=\cdots =\frac{d^{4n}}{dx^{4n}}(\sin x)= \sin x \\ \frac{d^5}{dx^5}(\sin x)=\frac{d^9}{dx^9}(\sin x)=\frac{d^{13}}{dx^{13}}(\sin x)=\cdots =\frac{d^{4n+1}}{dx^{4n+1}}(\sin x)= \cos x\end{array}$

Try It

For $y= \cos x$ , find $\dfrac{d^4 y}{dx^4}$ .

Show Solution

$\cos x$

Hint

See the previous example.

Watch the following video to see the worked solution to Example: Finding Higher-Order Derivatives of $y= \sin x$ and the above Try It.

Closed Captioning and Transcript Information for Video

Example: Using the Pattern for Higher-Order Derivatives of $y= \sin x$

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

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We can see right away that for the 74th 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

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

Try It

For $y= \sin x$ , find $\frac{d^{59}}{dx^{59}}(\sin x)$ .

Show Solution

$−\cos x$

Hint

$\frac{d^{59}}{dx^{59}}(\sin x)=\frac{d^{4(14)+3}}{dx^{4(14)+3}}(\sin x)$

Try It

Example: An Application to Acceleration

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

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

Show Solution

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

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

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

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

Consequently, the particle is slowing down.

Watch the following video to see the worked solution to Example: An Application to Acceleration.

Closed Captioning and Transcript Information for Video

Try It

A block attached to a spring is moving vertically. Its position at time $t$ is given by $s(t)=2 \sin t$ .

Find $v\left(\frac{5\pi}{6}\right)$ and $a\left(\frac{5\pi}{6}\right)$ . Compare these values and decide whether the block is speeding up or slowing down.

Show Solution

$v\left(\frac{5\pi}{6}\right)=−\sqrt{3}<0$ and $a\left(\frac{5\pi}{6}\right)=-1<0$ . The block is speeding up.

Hint

Use the last example as a guide.

Module 3: Derivatives