The derivative of $f(x)=mx+b$.

We know that the straight line is defined by the equation <Sign in to see all the formulas> where $m$ is the slope of the function and $b$ is the $y$-intercept. Since $m$ is the slope of the straight line and the derivative is the slope of the tangent line, we therefore expect that <Sign in to see all the formulas>.

Let's construct the function <Sign in to see all the formulas> and then take the limit at $h$ goes to zero. First, <Sign in to see all the formulas>. Now we have
$$\begin{eqnarray*} f'(x)&=&\lim_{h\to 0}\frac{f(x+h)-f(x)}{h},\\ &=&\lim_{h\to 0}\frac{m\,x+m\,h+b-m\,x-b}{h},\\ \mbox{and}&&\\ &=&\lim_{h\to 0}\frac{m\,h}{h},\\ \end{eqnarray*}$$

Now we know that

<Sign in to see all the formulas>

<Sign in to see all the formulas>

Therefore, <Sign in to see all the formulas> and we obtained the known slope of the function.