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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 - y_{1} = m (x - x_{1})

where m = gradient and (x_{1}, y_{1}) is where the tangent meets the curve.

**Example:**

Find the equation of the tangent to the curve y = x^{2} - 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).

**Example:**

Find the equation of the normal to the curve y = x^{3} + 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 x^{2} is positive have a minimum (∪ - shaped); all quadratics where the co-efficient of x^{2} 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 = x^{3} − 12x + 2

Firstly, where are the stationary points?

Find where the gradient = 0.

**Therefore:**

when x = 2 (and y = -14) or when x = -2 (and y = 18).

**Method 1:**

We know that a + x^{3} 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.)

**Method 2:**

**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).

**Method 3:**

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.