Differentiation Lesson 1

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Differentiation Lesson 1. Chapter 7. Gradient =. We need to be able to find the gradient of a straight line joining two points:. Find the gradient of the line joining (4, -11) and (-2, 7). Describe the way the gradient is changing on these graphs:. Finding the gradient of a curve. - PowerPoint PPT Presentation

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DifferentiationLesson 1

Chapter 7

We need to be able to find the gradient of a straight line joining two points:

x

y

Gradient =

Find the gradient of the line joining (4, -11) and (-2, 7)

Describe the way the gradient is changing on these graphs:

Finding the gradient of a curve

(differentiation)

How can you find the gradient of a curve if it keeps changing??

x

y

x

y

E.g. the function y = x2

x

y

Go to GSP file

The ideal way to find a gradient of a curve is to find the gradient of the tangent at the point we are interested in

x

y

Finding the gradient

The process of finding the gradient of a curve is called “differentiation”

You can differentiate any function to find its gradient

In general it can be shown that

If f(x) = xn where n is a real number

Then

f ’(x) = nxn-1

The function

The derivative (or differential)of the function

f(x) = xn

In general it can be shown that

The function

The derivative (or differential)of the function.This is the gradient function

f(x) = xnf(x) = xn

f ’(x) = nxn-1

In other words…

3xxf

23xxf '

1) Multiply function by power of x

2) Subtract 1 from the power

E.g. 1

f(x) = xn

f ’(x) = nxn-1

f(x) = xn

f ’(x) = nxn-1

-1

In other words…

5xxf

65 xxf '

1) Multiply function by power of x

2) Subtract 1 from the power

E.g. 1

f(x) = xn

f ’(x) = nxn-1

f(x) = xn

f ’(x) = nxn-1

Notation

xfdxdy

xfy

'

x

yz1

Now do Ex 7B on page 109Now do Ex 7B on page 109

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