1) find the derivative,

2) find the gradient, m, of the tangent by substituting in the x-ccordinate of the point;

3) use one of the following formulae to get the equation of the tangent:

EITHER y = mx + c

OR

To find the equation of a normal to a curve:

1) find the derivative ;

2) Substitute in the x-coordinate of the point to find the value of the gradient there.

3) the gradient of the normal is .

4) Use one of the following formulae to get the equation of the normal:

EITHER y = mx + c OR

To find the coordinates of maximum/ minimum points on a curve:

1) differentiate to get

2) solve the equation

3) find the y-coordinates of the points

4) determine whether the points are a maximum or minimum EITHER using the second derivative OR by considering the gradient either side of the point.

Example:

Find the equation of the tangent to the curve at the point where the curve crosses the y-axis.

Solution:

Expanding the brackets:

Differentiate:

The curve crosses the y-axis at the point (0, -2).

The gradient of the tangent at x = 0 is:

To find the equation of the tangent:

EITHER: y = mx + c y = 3x + c

Substitute x = 0, y = -2: -2 = 3(0) + c i.e. c = -2.

So equation is y = 3x - 2

OR: where m = 3

Substitute x = 0, y = -2:

i.e. y = 3x - 2

Applications of Differentiation

Recall: A turning point is a maximum if

A turning point is a minimum if .

Example:

Find the equation of the normal to the curve at the point where x = 4.

Solution:

The curve can be written as

Therefore,

When x = 4, and So the gradient of the normal is .

To find the equation of the tangent: y = mx + c y = 4x + c

Substitute x = 4, y = 2: 2 = 4(4) + c i.e. c = -14.

So equation is y = 4x – 14.

Example:

Find the coordinates of the stationary points on the curve .

Solution:

At a stationary point, .

Therefore, or .

Factorising gives (x + 1)(x – 4) =