$\tan x=\frac{\sin x}{\cos x}$.
This means that when we find the derivative of $\tan x$, we would need to have the derivative of $\sin x$ and $\cos x$, which are $\cos x$ and $-\sin x$ respectively. However, I would like to know how to find $\tan x$ can be found without using the derivative of $\sin x$ and $\cos x$. Alternatively, a proof of how the derivative of $\sin x$ and $\cos x$ equal to $\cos x$ and $-\sin x$ respectively, would help too.
Do note that the answer is a positive one.
For all $\cos{x}\neq0$ we have: $$(\tan{x})'=\lim_{h\rightarrow0}\frac{\tan(x+h)-\tan{x}}{h}=\lim_{h\rightarrow0}\frac{\sin{h}}{h\cos{x}\cos(x+h)}=\frac{1}{\cos^2x}$$ because $\lim\limits_{h\rightarrow0}\frac{\sin{h}}{h}=1$ and $\cos$ is a continuous function.