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Differentiation and graphs for Edexcel A-level Maths

Differentiation and graphs

This page covers the following topics:

1. Gradient of a curve
2. Second derivative
3. Stationary points and points of inflection
4. Maxima and minima
5. Increasing and decreasing functions

The gradient of a curve is given by the derivative of the equation of the curve.

Gradient of a curve

The second derivative of a function f(x), given by f''(x), and the second derivative of an equation y, given by dยฒy/dxยฒ, can be found by differentiating the function and the equation twice. For graphs, the second derivative represents the rate of change of the gradient.

Second derivative

At the stationary points and points of inflection of a graph, the derivative of the function of the equation of the graph is 0. Therefore, these points can be found by finding the derivative of the function, equating it to 0 and solving for x.

Stationary points and points of inflection

The maxima of a graph are the high points of the graph at which the slope is zero. At maxima, the second derivative of the function is less than 0. The minima of a graph are the low points of the graph at which the slope is zero. At minima, the second derivative of the function is more than 0.

Maxima and minima

A function f(x) is said to be increasing at point x if f'(x) > 0 or decreasing at point x if f'(x) < 0.

Increasing and decreasing functions

1

Find the gradient of the curve given by the equation y = cotx + 18x.

The gradient of the curve is given by the derivative of the equation. So, dy/dx = โˆ’cosecยฒx + 18.

Find the gradient of the curve given by the equation y = cotx + 18x.

2

What is the gradient of the given curve?

The curve has equation y = cosx/xยฒ. This must be differentiated to find the gradient, which can be done using the Quotient rule. Let u = cosx and v = xยฒ. Then, du/dx = โˆ’sinx and dv/dx = 2x. So, by the Quotient rule, dy/dx = 2xcosx โˆ’ xยฒsinx.

What is the gradient of the given curve?

3

Find the stationary points of y = xยณ/3 โˆ’ xยฒ/2 โˆ’ 6x and classify them as maxima and minima.

dy/dx = xยฒ โˆ’ x โˆ’ 6. At stationary points, dy/dx = 0. So, xยฒ โˆ’ x โˆ’ 6 = 0 gives x = 3 or x = โˆ’2. To find the y coordinate at at x = 3, y = 3ยณ/3 โˆ’ 3ยฒ/2 โˆ’ 6(3) = โˆ’27/2 and at x = โˆ’2, y = (โˆ’2)ยณ/3 โˆ’ (โˆ’2)ยฒ/2 โˆ’ 6(โˆ’2) = 34/3. Then, dยฒy/dxยฒ = 2x โˆ’ 1. At x = 3, dยฒy/dxยฒ = 2(3) โˆ’ 1 = 5 > 0, therefore (3, โˆ’27/2) is a minimum point. At x = โˆ’2, dยฒy/dxยฒ = 2(โˆ’2) โˆ’ 1 = โˆ’5 < 0, therefore (โˆ’2, 34/3) is a maximum point.

Find the stationary points of y = xยณ/3 โˆ’ xยฒ/2 โˆ’ 6x and classify them as maxima and minima.

4

Verify that x = 3 is a minimum point of the equation y = xยณ/3 โˆ’ 2xยฒ + 4x + 12.

At minimum points, the second derivative is greater than 0. So, dy/dx = xยฒ โˆ’ 4x + 4 and dยฒy/dxยฒ = 2x โˆ’ 4. At x = 3, dยฒy/dxยฒ = 2(3) โˆ’ 4 = 2 > 0. Therefore, at x = 2, the graph has a minimum point.

Verify that x = 3 is a minimum point of the equation y = xยณ/3 โˆ’ 2xยฒ + 4x + 12.

5

The second derivative of a graph at x = 5 is found to be 27. Explain what this means.

This means that at x = 5, the rate of change of the gradient of the graph is 27.

The second derivative of a graph at x = 5 is found to be 27. Explain what this means.

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