# StudySquare

# Differentiation and graphs for SQA Higher Maths

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.

The second derivative of a function f(x), given by f''(x), and the second derivative of an equation y, given by d²y/dx², can be found by differentiating the function and the equation twice. For graphs, the second derivative represents the rate of change of the gradient.

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.

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.

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.

# 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 = −cosec²x + 18.

# 2

What is the gradient of the given curve?

The curve has equation y = cosx/x². This must be differentiated to find the gradient, which can be done using the Quotient rule. Let u = cosx and v = x². Then, du/dx = −sinx and dv/dx = 2x. So, by the Quotient rule, dy/dx = 2xcosx − x²sinx.

# 3

Find the stationary points of y = x³/3 − x²/2 − 6x and classify them as maxima and minima.

dy/dx = x² − x − 6. At stationary points, dy/dx = 0. So, x² − x − 6 = 0 gives x = 3 or x = −2. To find the y coordinate at at x = 3, y = 3³/3 − 3²/2 − 6(3) = −27/2 and at x = −2, y = (−2)³/3 − (−2)²/2 − 6(−2) = 34/3. Then, d²y/dx² = 2x − 1. At x = 3, d²y/dx² = 2(3) − 1 = 5 > 0, therefore (3, −27/2) is a minimum point. At x = −2, d²y/dx² = 2(−2) − 1 = −5 < 0, therefore (−2, 34/3) is a maximum point.

# 4

Verify that x = 3 is a minimum point of the equation y = x³/3 − 2x² + 4x + 12.

At minimum points, the second derivative is greater than 0. So, dy/dx = x² − 4x + 4 and d²y/dx² = 2x − 4. At x = 3, d²y/dx² = 2(3) − 4 = 2 > 0. Therefore, at x = 2, the graph has a minimum point.

# 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.

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