## Sketching derivative graphs (cubics)

The important thing to remember here is that we are doing a sketch. It does not have to be accurate, but id does have to conatian all the key points.

Hints:

- A cubic differentiates to a quadratic (parabola)
- A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
- A point with a positive gradient will be above the x axis in the sketch
- A point with a pnegative gradient will be below the x axis in the sketch

Examples

Describe and sketch the derivative graph for the function below

The critical points are , because this is where the turning points occur.

These points will be on the x axis in the sketch.

For the gradient is positive.

These points will be above the x axis in the sketch.

For the gradient is negative .

These points will be below the x axis in the sketch.

The graph is a cubic so the sketch will be a parabola

### Practice

## Questions: Differentiation

- Find the coordinates of the point where the curve has a gradient of 4.
- Find the gradient of the curve at .
- Find the x coordinate where the curves and have the same gradient.
- Find the gradients at the points of intersection of the two curves and
- Find the coordinates of the points where the curve has a gradient of 0.
- Find the gradient of the curve with equation at the point
- Differentiate and hence show that the curve never has a gradient of zero.
- Show that the curve can be written in the form and find the gradient at
- Find the gradient of the curve at the point
- Find the three gradients where the curve cuts the axes

## Finding the turning points of a polynomial

The turning points are sometimes called the stationary points or maximums and minimums.

Lets look at what happens to the gradient at a turning point

- If we consider the maximum, you can see that before it has a positive gradient and after it has a negative gradient. During this transition it has a gradient of zero. So at a maximum
- If we consider the minimum, you can see that before it has a negative gradient and after it has a positive gradient. During this transition it has a gradient of zero. So at a minimum
- So we can see that at a turning point (maximum or minimum)

Examples

### Find the maximum of the curve

- We need to find the gradient function
- We let , because this is when the maximum will occur
- We solve the equation to find the value of x where the maximum occurs
- We substitute this value in the original equation to get the value for y

Let

If

the maximum value is

### Find the coordinates of the turning points of

- We need to find the gradient function
- We let , because this is when the turning points will occur
- We solve the equation to find the values of x where the turning points occurs
- We substitute the values into the original equation to get the values for y

Let

So of

If then and if then

So the coordinates of the turning points are and

### Given that the curve has a stationary point at find the value of a

- We need to find the gradient function
- We let , because this is when the turning points will occur and we will let
- We will solve the resulting equation to find the value of a

when

### Find the equation of the tangent to the curve at

- We need to find the gradient function , because to get the equation of a line we need the gradient and a point it passes through
- Substitute into the equation for y to get the required point the line passes through
- Substitute into the equation for to get the gradient of the line
- Use to get the equation of the line

At

At

So substituting into gives

## Differentiating Polynomials from first principles

The gradient of a curve is always changing. To calculate the gradient at a point we can consider the gradient of a chord going through that point and gradually make the length of the chord shorter. As the length gets closer to zero the gradient of the chord should get closer to the gradient of the tangent at the point.

When we are finding the gradient function we are differentiating

Consider the diagram

You can see from the graph that as gets smaller so does and that the gradient of the chord gets closer to that of the tangent.

So as , the gradient at x

We refer to the gradient at x as

### Finding the gradient function of

Let

The point that is across from has the coordinates

Substituting into gives

Since we get

If we now divide through by we get

Now if we let we get

### Finding the gradient function of

Note that

Let

The point that is across from has the coordinates

Substituting into gives

So

Since

we get

If we now divide through by

we get

Now if we let we get

You should be able to see that when we differentiate that we multiply by the power and then subtract one from the power.

Examples