2 Finding the gradient at a specific point
In this subsection we shall consider a simple function to illustrate the calculation of a gradient. Look at the graph of the function $y\left(x\right)={x}^{2}$ shown in Figure 4. Notice that the gradient of the graph changes as we move from point to point. In some places the gradient is positive; at others it is negative. The gradient is greater at some points than at others. In fact the gradient changes from point to point as we move along the curve.
Figure 4
Inspect the graph carefully and make the following observations:
 $A$ is the point with coordinates $\left(1,1\right)$ .
 $B$ is the point with coordinates $\left(4,16\right)$ .

We can calculate the gradient of the line
$AB$
from the formula
$$\text{gradient}=\frac{\text{differencebetween}y\text{coordinates}}{\text{differencebetween}x\text{coordinates}}$$Therefore the gradient of chord $AB$ is equal to $\frac{161}{41}=\frac{15}{3}=5$ . The gradient of $AB$ is not the same as the gradient of the graph at $A$ but we can regard it as an approximation, or estimate of the gradient at $A$ . Is it an overestimate or underestimate ?
Task!
Add the point $C$ to the graph in Figure 4 where $C$ has coordinates $\left(3,9\right)$ . Draw the line $AC$ and calculate its gradient.
$\frac{91}{31}=4$ . Would you agree that this is a better estimate of the gradient at $A$ than using $AB$ ?
We now carry the last task further by introducing point $D$ at $\left(2,4\right)$ and point $E$ at $\left(1.5,2.25\right)$ as shown in Figure 5. The gradient of $AD$ is found to be 3 and the gradient of $AE$ is 2.5.
Figure 5
Observe that each time we carry out this procedure, and move the second point closer to $A$ , the gradient of the line drawn is getting closer and closer to the gradient of the tangent at $A$ . If we continue, the value we eventually obtain is the gradient of the tangent at $A$ whose value is 2 as we will see shortly. This procedure illustrates how we define the gradient of the curve at $A$ .