The higher-order derivatives of [latex]\sin x[/latex] and [latex]\cos x[/latex] follow a repeating pattern. By following the pattern, we can find any higher-order derivative of [latex]\sin x[/latex] and [latex]\cos x[/latex].
Example: Finding Higher-Order Derivatives of [latex]y= \sin x[/latex]
Find the first four derivatives of [latex]y= \sin x[/latex].
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Each step in the chain is straightforward:
[latex]\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}[/latex]
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 [latex]\sin x[/latex] equals [latex]\sin x[/latex], so
For [latex]y= \cos x[/latex], find [latex]\dfrac{d^4 y}{dx^4}[/latex].
Show Solution
[latex]\cos x[/latex]
Hint
See the previous example.
Watch the following video to see the worked solution to Example: Finding Higher-Order Derivatives of [latex]y= \sin x[/latex] and the above Try It.
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A particle moves along a coordinate axis in such a way that its position at time [latex]t[/latex] is given by [latex]s(t)=2- \sin t[/latex].
Find [latex]v\left(\frac{\pi}{4}\right)[/latex] and [latex]a\left(\frac{\pi}{4}\right)[/latex]. Compare these values and decide whether the particle is speeding up or slowing down.
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First find [latex]v(t)=s^{\prime}(t)[/latex]: [latex]v(t)=s^{\prime}(t)=−\cos t[/latex]. Thus, [latex]v\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}[/latex].
Next, find [latex]a(t)=v^{\prime}(t)[/latex].
Thus, [latex]a(t)=v^{\prime}(t)= \sin t[/latex] and we have [latex]a\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}[/latex].
Since [latex]v\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}<0[/latex] and [latex]a\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}>0[/latex], 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.
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A block attached to a spring is moving vertically. Its position at time [latex]t[/latex] is given by [latex]s(t)=2 \sin t[/latex].
Find [latex]v\left(\frac{5\pi}{6}\right)[/latex] and [latex]a\left(\frac{5\pi}{6}\right)[/latex]. Compare these values and decide whether the block is speeding up or slowing down.
Show Solution
[latex]v\left(\frac{5\pi}{6}\right)=−\sqrt{3}<0[/latex] and [latex]a\left(\frac{5\pi}{6}\right)=-1<0[/latex]. The block is speeding up.
Hint
Use the last example as a guide.
Licenses and Attributions
CC licensed content, Original
3.5 Derivatives of Trigonometric Functions (edited). Authored by: Ryan Melton. License: CC BY: Attribution
