$(c)'=0$

$f(x)=c$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{c-c}{h}=\lim_{h\longrightarrow 0} 0 = 0$

$(x)'=1$

$f(x)=x$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)-x}{h}=\lim_{h\longrightarrow 0} 1 = 1$

$(x^2)'=2x$

$f(x)=x^2$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)^2-x^2}{h}=\displaystyle\lim_{h\longrightarrow 0} \dfrac{x^2+2xh+h^2-x^2}{h}=\lim_{h\longrightarrow 0} 2x+h = 2x$

$(x^n)'=nx^{n-1}$

$f(x)=x^n$;

According to the binomial expansion

$(x+h)^n=x^n+nx^{n-1}h+\binom{n}{2}x^{n-1}h^2+\cdots+\binom{n}{n-2}x^2h^{n-2}+nxh^{n-1}+h^n$

So

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{(x+h)^n-x^n}{h}=\\ \qquad\displaystyle\lim_{h\longrightarrow 0}\dfrac{x^n+nx^{n-1}h+\binom{n}{2}x^{n-1}h^2+\cdots+\binom{n}{n-2}x^2h^{n-2}+nxh^{n-1}+h^n-x^n}{h}=\\ \qquad\displaystyle\lim_{h\longrightarrow 0} nx^{n-1}+\binom{n}{2}x^{n-1}h+\cdots+\binom{n}{n-2}x^2h^{n-3}+nxh^{n-1}+h^{n-1} =\\ \qquad nx^{n-1}$