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

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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

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