Uses of differentiation
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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.
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