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:

$$\frac{\dd }{\dd x}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$$