Www.le.ac.uk Differentiation Department of Mathematics University of Leicester.

www.le.ac.uk

Differentiation

Department of MathematicsUniversity of Leicester

Content

Differentiation of functions

Introduction

Introduction

•Differentiation is the process of

finding the rate of change of some quantity (eg. a line), at a general point x.

• The rate of change at x is equal to the gradient of the tangent at x.

• We can approximate the gradient of the tangent using a straight line joining 2 points on the graph…

Next

FunctionsIntroduction

Choose h = .

from to :

The straight line has a gradient of :

0.5 1 1.5 2 2.5

Next


))(,5.0( xf ))5.0(,5.0( hfh

Gradient of tangent at is

0.4.5.0x

0

(Tangent Line at 0.5)

)5.10( h

Draw Line

Clear

Gradients• The gradient of the line from to

is .

• ie. it’s the difference between the 2

points.

• As h gets smaller the line gets closer to the tangent, so we let h tend to 0.

• We get:

Next


))(,( xfx

))(,( hxfhx h

xfhxf )()(

h

xfhxfh

)()(lim 0

Differentiation

• When you differentiate f(x), you find

• This is called the derivative , and is

written as or ,

or (for example).

• Each function has its own derivative...


h

xfhxfh

)()(lim 0

dx

df

dx

xd )(sin)(xf

Next

c

nx

xe

xln

)ln(ax

)ln( ax

xsin

xcos

xtan

0

1nnx

xe

x

1

x

axsin

xcos

x2sec

x

1

SummaryClick on the functions to see how they are derived.


Next

)(xf )(xf

Differentiating a constant

Next


0lim0

h

cc

dx

dfh

Back to summary

......

...2)1(

lim

)(lim

221

0

0

continue

h

xxnnh

hnxx

h

xhx

nn

nn

h

nn

h

Differentiating :

This is using the binomial expansion

Next


dx

df

nx

Back to summary

Recap Binomial Expansion

Hide Binomial Expansion

12

1

0...

2

)1(lim

n

nn

hnx

xnhnnx

Next


Back to summary

All these terms contain h, so disappear when we take the limit as h0

......

)1(lim

lim

0

0

continue

h

ee

h

ee

hx

h

xhx

h

Differentiating :

Next


dx

df

xe

Back to summary

Next


Back to summary

x

x

h

x

h

e

hhe

h

hhe

...32

1lim

1...2

1

lim

2

0

2

0This is using the

Maclaurin’s Series for eh.

Recap Maclaurin's Series

Hide Maclaurin's Series

......

)sin(cos

1)cos(sinlim

sincos)sin()cos(sinlim

)sin()sin(lim

0

0

0

continue

h

hx

h

hx

h

xxhhx

h

xhx

h

h

h

Differentiating :

This is using the Trigonometric Identity for sin(a+b)

NextBack to summary


dx

dfxsin Recap Trig Identity

Hide Trig Identity

xh

hx

h

hx

dx

dfh

cos)sin(

cos1)cos(

sinlim0

This is using the Maclaurin’s Series for sin(x) and cos(x)

NextBack to summary


So

01...

3211)cos(

32

h

hh

h

h

1...

53)sin(

53

h

hhh

h

h



......

)sin(sin

1)cos(coslim

cos)sin(sin)cos(coslim

)cos()cos(lim

0

0

0

continue

h

hx

h

hx

h

xhxhx

h

xhx

h

h

h

Differentiating :

This is using the Trigonometric Identity for cos(a+b)

NextBack to summary


dx

df

xcos Recap Trig Identity

Hide Trig Identity

xh

hx

h

hx

dx

dfh

sin)sin(

sin1)cos(

coslim0

This is using the Maclaurin’s Series for sin(x) and cos(x)

NextBack to summary


So

01...

3211)cos(

32

h

hh

h

h

1...

53)sin(

53

h

hhh

h

h



......

)1ln(lim

)ln(lim

ln)ln(lim

00

0

continue

hxh

hxhx

h

xhx

hh

h

Differentiating :

Next


dx

df

xln

Back to summary

`

x

h

xh

xh

xh

xh

hxh

hh

1

...432lim

)1ln(lim

432

00

This is using the Macluarin’s Series for ln(a+1)

Next


Back to summary

Because after you divide by h, all the other terms have h in them so disappear as h0.



xdx

xd

h

xhx

h

xahxa

xadx

d

h

h

1lnln)ln(lim

lnln)ln(lnlim

)ln(ln

0

0

Next


Back to summary

Differentiating :)ln(ax

))(ln(axdx

d

x

a

dx

xda

h

xhxa

h

xahxa

xadx

d

h

h

lnln)ln(lim

ln)ln(lim

)ln(

0

0

Next


Back to summary

Differentiating :)ln( ax

))(ln( axdx

d

xcos

xcos

xtan

Questions

differentiates to:


xsin

12 x

22 x

32 x

Questions

differentiates to:


2x

xln

1

xln

2

1

x

Questions

differentiates to:


x

1

xsec

x2sec

x1tan

Questions

differentiates to:


xtan

x

1

xln

1

x

Questions

differentiates to:


xln

2

5

5x

2

3

2

5x

2

4

5x

Questions

differentiates to:


2

5

x

xxe1xxe

xe

Questions

differentiates to:


xe

xtan

xsin

xsin

Questions

differentiates to:


xcos

Conclusion

• Differentiation is the process of finding a general expression for the rate of change of a function.

• It is defined as

• Differentiation is a process of subtraction.

• Using this official definition, we can derive rules for differentiating any function.

Next


h

xfhxfh

)()(lim 0

Www.le.ac.uk Differentiation Department of Mathematics University of Leicester.

Documents

summary slide

summary functionsintroduction

summary differentiating

maclaurins series

tangent line

macluarins series

cosx nextback

straight line