The table below lists the derivatives of the trigonometric functions and their inverses.

\[\begin{array}{|c|c|} \hline

\mathit{function} & \mathit{derivative} \\ \hline\\

\sin(x) & \cos(x) \\ \\

\cos(x) & -\sin(x) \\ \\

\tan(x) & \dfrac{1}{\cos(x)^2}\\ \\ \hline\\

\arcsin(x) & \dfrac{1}{\sqrt{1-x^2}} \\ \\

\arccos(x) & -\dfrac{1}{\sqrt{1-x^2}} \\ \\

\arctan(x) & \dfrac{1}{1+x^2}\\ \\ \hline

\end{array}\] What is immediately striking is that the derivative of a trigonometric function again is a trigonometric function. More generally, the derivative of a periodic function is also itself again a periodic function. Also notable is that the derivative of an inverse trigonometric function is a function which contains no trigonometric function anymore.

For math enthusiasts we present the proofs of some formulas. Let us first determine the derivative of the sine function assuming that \(\displaystyle \lim_{u\to 0}\frac{\sin(u)}{u}=0\) and common calculation rules for limits can be applied: \[\begin{aligned}\frac{\dd}{\dd x}\sin(x)&= \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}\\[0.25cm] &=\lim_{h\to 0} \frac{2\cos(x+\frac{h}{2})\sin(\frac{h}{2})}{h}\\[0.25cm] &=\lim_{u\to 0} \frac{2\cos(x+u)\sin(u)}{2u}\\[0.25cm] &=\lim_{u\to 0}\cos(x+u)\cdot \lim_{u\to 0}\frac{\sin(u)}{u}\\[0.25cm] &= \cos(x)\cdot 1\\[0.25cm] &= \cos(x)\end{aligned}\] Now the derivative of \(\cos x\) can also be calulated via the chain rule: \[\begin{aligned}\frac{\dd }{\dd x}\cos(x)&=\frac{\dd }{\dd x}\sin(x+\tfrac{1}{2}\pi)\\[0.25cm]&=\cos(x+\tfrac{1}{2}\pi)\cdot \frac{\dd }{\dd x}(x+\tfrac{1}{2}\pi)\\[0.25cm] &=-\sin(x)\cdot 1\\[0.25cm] &=-\sin(x)\end{aligned}\] The derivative of the tangent function can then be calculated using the quotient rule:: \[\begin{aligned}\frac{\dd}{\dd x}\tan(x)&=\frac{\dd}{\dd x}\left(\frac{\sin(x)}{\cos{x}}\right)\\[0.25cm] &=\frac{\sin'(x)\cos(x)-\sin(x)\cos'(x)}{\cos^2(x)}\\[0.25cm] &=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}\\[0.25cm]&= \frac{1}{\cos^2(x)}\end{aligned}\]

We give one example of the proof of the formula for the derivative of an inverse trigonometric function, namely of \(\arcsin\).

Suppose \(y=\arcsin(x)\), then \(x=\sin(y)\) and \(-\tfrac{1}{2}\pi\le y \le \tfrac{1}{2}\pi\).

With implicit differentiation we find \(1=\cos(y)\cdot \frac{\dd y}{\dd x}\), that is \(\frac{\dd y}{\dd x}=\frac{1}{\cos(y)}\).

Because \(-\tfrac{1}{2}\pi\le y \le \tfrac{1}{2}\pi\) we know that \(\cos(y)\ge 0\) and therefore that \(\cos(y)=\sqrt{1-\sin^2(y)}=\sqrt{1-x^2}\).

So: \[\frac{\dd }{\dd x}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}\]

Differentiation of \(\sin x\) and \(\cos x\) (15:08)