Consider the function $y(x) = 5x^2 - 3x + 10$. What is the derivative of this function in $x=1$? In the figure below the graph of the function $y(x)$, as well as the line tangent to the graph at $(x,y)=(1,12)$, is shown. This line is called a tangent line. The derivative of $y(x)$ in $x=1$ is equal to the slope of the tangent line.
To get some idea about the value of the derivative in $x=1$, we first calculate the difference quotient for increasingly smaller values of $\Delta x$; we use $y(1)=5\cdot1^2 -3\cdot1 + 10 = 12$.
$$
\begin{array}{c|c|c}
\Delta x & y(1+\Delta x) & \dfrac{\Delta y}{\Delta x}\\[3mm]
\hline
2 & 46 & \dfrac{46-12}{2} = 17\\[3mm]
1 & 24 & \dfrac{24-12}{1} = 12\\[3mm]
\tfrac{1}{2} & 16\tfrac{3}{4} & \dfrac{16\tfrac{3}{4}-12}{\tfrac{1}{2}} = 9\tfrac{1}{2}\\[3mm]
\tfrac{1}{10} & 12\tfrac{3}{4} & \dfrac{12\tfrac{3}{4}-12}{\tfrac{1}{10}} = 7\tfrac{1}{2}\\[3mm]
\tfrac{1}{100} & 12\tfrac{141}{2000} & \dfrac{12\tfrac{141}{2000}-12}{\tfrac{1}{100}} = 7\tfrac{1}{20}
\end{array}
$$
When $\Delta x$ approaches 0, the difference quotient approaches 7. Does this mean that $y'(1)=7$? Let us show this formally. We calculate the difference quotient in $x=1$ at a change $\Delta x$:
$$
\begin{align}
\dfrac{(y(1+\Delta x)-y(1)}{\Delta x}& = \dfrac{5(1+\Delta x)^2 -3(1+\Delta x)+10-12}{\Delta x} = \dfrac{5(1+2\Delta x + (\Delta x)^2) - 3(1+\Delta x) - 2}{\Delta x} \\
&= \dfrac{5 + 10\Delta x + 5(\Delta x)^2 - 3 - 3\Delta x - 2}{\Delta x} = \dfrac{5(\Delta x)^2 + 7\Delta x}{\Delta x} = 5\Delta x + 7.
\end{align}
$$
If $\Delta x$ becomes 0, then the difference quotient becomes $5\cdot 0 + 7=7$. Hence, the derivative of $y(x)$ in $x=1$ is indeed $y'(1)=7$.