Everything Maths Grade 12

CHAPTER 7. DIFFERENTIAL CALCULUS 7.2

We know how to calculate the average gradient between two points on a curve, but we needtwo
points. The problem now is that we only haveone point, namely P. To get around the problem we
first consider a secant to the curve that passesthrough point P and another point onthe curve Q,
where Q is an arbitrary distance(h) from P , as shown in the figure. We can now find the average
gradient of the curve between points P and Q.

f (x)

�

�

P

Q

a a +h

f (a +h)

f (a)

secant

x

y

If the x-coordinate of P is a +h, then the y-coordinate is f (a +h). Similarly, if the x-coordinate of Q
is a, then the y-coordinate is f (a). If we choose a +h as x 2 and a as x 1 , then:

y 1 = f (a)

y 2 = f (a +h).

We can now calculate the average gradient as:

y 2 −y 1

x 2 −x 1

=

f (a +h)−f (a)

(a +h)−a

(7.12)

=

f (a +h)−f (a)

h

(7.13)

Now imagine that Q moves along the curvetoward P. The secant line approaches the tangent line as
its limiting position. This means that the averagegradient of the secant approaches the gradient of the
tangent to the curve at P. In (??) we see that as point Q approaches point P , h gets closer to 0. When
h = 0, points P and Q are equal. We can nowuse our knowledge of limits to write this as:

gradient at P = lim

h→ 0

f (a +h)−f (h)

h

. (7.14)

and we say that the gradient at point P is the limit of the average gradient as Q approaches P along
the curve.

See video: VMgye at http://www.everythingmaths.co.za

Example 5: Limits

QUESTION

For the function f (x) = 2x^2 − 5 x, determine the gradient of the tangent to the curve at the