This App will help you to avoid any unwanted slip-ups in the exam. Although most of the reminders are common sense, but from the evidence students still need reminding of them. Read through the tips and take note of the most relevant ones before tackling your exam.
Uses of differentiation
As mentioned before, the main use for differentiation is to find the gradient of a function at any point on the graph. Having found the gradient at a specific point we can use our coordinate geometry skills to find the equation of the tangent to the curve.
To do this we:
1. Differentiate the function.
2. Put in the x-value into
to find the gradient of the tangent.
3. Put in the x-value into the function (y = ...) to find the coordinates of the point where the tangent touches the curve.
4. Put these values into the formula for a straight line:
y - y1 = m (x - x1)
where m = gradient and (x1, y1) is where the tangent meets the curve.
Find the equation of the tangent to the curve y = x2 - 2x - 3, when x = -1:
so when x = -1,
and y = 1 + 2 - 3 = 0
Therefore the equation of the tangent is y − 0 = -4(x + 1)
So now we know that y = -4x - 4 is the equation of the tangent at (-1, 0).
The normal to a curve is the line at right angles to the curve at a particular point.
This means that the normal is perpendicular to the tangent and therefore the gradient of the normal is -1 × the gradient of the tangent.
To find the equation of the normal, follow the same procedures as before, (remembering to multiply the gradient of the tangent by -1 to calculate the gradient of the normal).
Find the equation of the normal to the curve y = x3 + x − 10 when x = 2.
and y = 8 + 2 − 10 = 0
As the gradient of the tangent = 13, the gradient of the normal = -1/13
Therefore the equation of the normal is:
y − 0 = -(x - 2) / 13
Therefore: 13y = 2 − x is the equation of the normal at (2, 0).
As mentioned before, the main use for differentiation is to find the gradient of a function at any point on the graph. In particular differentiation is useful to find one of the main features of the graph - the Stationary Points.
These are points where the gradient = 0.
There are three types of stationary point:
There are three possible ways to determine the nature of a stationary point.
1. From experience - if you know the shape of the graph, then you know if it is a max/min. All quadratics where the co-efficient of x2 is positive have a minimum (∪ - shaped); all quadratics where the co-efficient of x2 is negative have a maximum (∩ - shaped).
2. By looking at the gradient either side of the stationary point.
3. By using the second derivative,
which often shown as
For a particular value for x, when
An example using all three methods:
Find the coordinates and nature of the turning points of the curve y = x3 − 12x + 2
Firstly, where are the stationary points?
Find where the gradient = 0.
when x = 2 (and y = -14) or when x = -2 (and y = 18).
We know that a + x3 graph has a maximum followed by a minimum, so (-2, 18) must be a maximum, and (2, -14) must be a minimum. (Also the value of the y-coordinates confirms that this must be true.)
For this graph the gradient = 0 when x = -2 and x = 2.
We can use the fact that the gradient is a multiple of (x + 2)(x - 2) to determine the sign of the gradient either side of these values. This is best illustrated in a table.
At x = -2, the gradient goes from positive to negative. This is a ∩-shape, and means that there is a maximum at (-2, 18).
At x = 2, the gradient goes from negative to positive. This is a ∪-shape, and means that there is a minimum at (2, -14).
When x = -2,
Therefore there is a maximum at (-2, 18).
When x = 2,
Therefore there is a minimum at (2, -14).
As a guideline - if you know the graph use method 1. Otherwise use method 3 unless the second derivative is hard to find.
If you do not know the shape of the graph, or you cannot differentiate twice or
then use the method 2.