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