If you already know that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$, then you can proceed to read this short article. If you wish to learn why $\dfrac d{dx}\sin x = \cos x$ and $\dfrac d{dx}\cos x = -\sin x$, you can refer to the articles on how to find the derivative of $\cos x$ and a more detailed article on how to find the derivative of $\sin x$.

Here we will show that taking multiple derivatives of $\sin x$ and $\cos x$ will result in positive or negative values of $\sin x$ and $\cos x$. In fact, you will see that the derivatives of $\sin x$ and $\cos x$ cycle as follows:

\[ \begin{align*} &\dfrac{d}{dx} \sin x = \cos x\\ &\dfrac{d}{dx} \cos x = -\sin x\\ &\dfrac{d}{dx} (-\sin x) = -\cos x\\ &\dfrac{d}{dx} (-\cos x) = \sin x \end{align*} \]

If you think you understand the cyclic nature of the derivative, calculate

$$\dfrac {d^{21}}{dx^{21}}\sin x$$

If you got $\cos x$, then you are correct! How do we do this? First, divide 21 by 4 with remainder:

$$21\div 4 = 5 R 1$$

Therefore, the derivatives of $\sin x$ will cycle 5 times and then there will be one more derivative left over. Therefore,

$$\dfrac {d^{21}}{dx^{21}}\sin x = \dfrac d{dx}\sin x = \cos x$$

Can you find the 102nd derivative of $\cos x$? You will follow the same steps:

$$102\div 4 = 25 R 2$$

$$\dfrac {d^{102}}{dx^{102}}\cos x = \dfrac {d^2}{dx^2}\cos x = -\cos x$$

We hope that this article helped you understand how to take higher derivatives of $\sin x$ and $\cos x$.