application of differentiation

To find the equation of a tangent to a curve:
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) =