# Very accurate method as its diﬃcult to draw a tangent

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very accurate method, as it’s diﬃcult to draw a tangent accurately by eye.The next subsection shows you a better method. It considers theparticular example of the graph ofy=x2.217
Unit 6Differentiation1.3Gradients of the graph ofy=x2The graph ofy=x2has a gradient ateverypoint, because it has atangent, which is not vertical, at every point.Let’s try to find the gradient of this graph at the point (1,1), which isshown in Figure 10. That is, we want to find the gradient of the tangent tothe graph at this point, which is also shown in Figure 10.xy¡3¡2¡1123¡11234Figure 10The point (1,1) on the graph ofy=x2and the tangent atthis pointThe way to find the gradient at (1,1) is to begin by thinking about howyou could find anapproximatevalue for this gradient. Here’s how you cando that. You choose a second point on the graph ofy=x2, fairly close to(1,1), as illustrated in blue in Figure 11. The straight line that passesthrough both (1,1) and the second point is an approximation for thetangent to the graph at (1,1). So the gradient of this line, which you cancalculate using the two points on the line in the usual way, is anapproximation for the gradient of the graph at (1,1).xy¡3¡2¡1123¡11234Figure 11An approximation to the tangent to the graph ofy=x2at(1,1)The closer to (1,1) you choose the second point to be, the better theapproximation will be. For example, the second point in Figure 12 willgive a better approximation than the second point in Figure 11.218
1What is differentiation?xy¡3¡2¡1123¡11234Figure 12Another approximation to the tangent to the graph ofy=x2at (1,1)As an example, let’s calculate the approximation to the gradient at (1,1)that you get by choosing the second point to be the point withx-coordinate 1.1, which is the point shown in blue in Figure 12. Since theequation of the graph isy=x2, they-coordinate of this second point is1.12= 1.21. So the second point is (1.1,1.21). The gradient of the linethrough (1,1) and (1.1,1.21) isriserun=1.2111.11=0.210.1= 2.1.So an approximate value for the gradient of the graph at (1,1) is 2.1.The second point that you choose on the graph can lie either to the left orto the right of (1,1). In the next activity, you’re asked to calculate theapproximation to the gradient that you get by choosing a second point onthe graph that lies to the left of (1,1), as illustrated in Figure 13.xy¡3¡2¡1123¡11234Figure 13Yet another approximation to the tangent to the graph ofy=x2at (1,1)219
Unit 6DifferentiationActivity 3Calculating an approximation to the gradient at a pointCalculate the approximation to the gradient of the graph ofy=x2at thepoint withx-coordinate 1 that you get by taking the second point on thegraph to be the point withx-coordinate 0.9.

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Term
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Gottfried Leibniz, Gottfried Wilhelm
• • • 