Post on 28-Mar-2015
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…
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FunctionsIntroduction
Choose h = .
from to :
The straight line has a gradient of :
0.5 1 1.5 2 2.5
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FunctionsIntroduction
))(,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:
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FunctionsIntroduction
))(,( 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...
FunctionsIntroduction
h
xfhxfh
)()(lim 0
dx
df
dx
xd )(sin)(xf
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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.
FunctionsIntroduction
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)(xf )(xf
Differentiating a constant
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FunctionsIntroduction
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
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FunctionsIntroduction
dx
df
nx
Back to summary
Recap Binomial Expansion
Hide Binomial Expansion
12
1
0...
2
)1(lim
n
nn
hnx
xnhnnx
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FunctionsIntroduction
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 :
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FunctionsIntroduction
dx
df
xe
Back to summary
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FunctionsIntroduction
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
FunctionsIntroduction
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
FunctionsIntroduction
So
01...
3211)cos(
32
h
hh
h
h
1...
53)sin(
53
h
hhh
h
h
Recap Maclaurin's Series
Hide Maclaurin's Series
......
)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
FunctionsIntroduction
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
FunctionsIntroduction
So
01...
3211)cos(
32
h
hh
h
h
1...
53)sin(
53
h
hhh
h
h
Recap Maclaurin's Series
Hide Maclaurin's Series
......
)1ln(lim
)ln(lim
ln)ln(lim
00
0
continue
hxh
hxhx
h
xhx
hh
h
Differentiating :
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FunctionsIntroduction
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)
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FunctionsIntroduction
Back to summary
Because after you divide by h, all the other terms have h in them so disappear as h0.
Recap Maclaurin's Series
Hide Maclaurin's Series
xdx
xd
h
xhx
h
xahxa
xadx
d
h
h
1lnln)ln(lim
lnln)ln(lnlim
)ln(ln
0
0
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FunctionsIntroduction
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
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FunctionsIntroduction
Back to summary
Differentiating :)ln( ax
))(ln( axdx
d
xcos
xcos
xtan
Questions
differentiates to:
FunctionsIntroduction
xsin
12 x
22 x
32 x
Questions
differentiates to:
FunctionsIntroduction
2x
xln
1
xln
2
1
x
Questions
differentiates to:
FunctionsIntroduction
x
1
xsec
x2sec
x1tan
Questions
differentiates to:
FunctionsIntroduction
xtan
x
1
xln
1
x
Questions
differentiates to:
FunctionsIntroduction
xln
2
5
5x
2
3
2
5x
2
4
5x
Questions
differentiates to:
FunctionsIntroduction
2
5
x
xxe1xxe
xe
Questions
differentiates to:
FunctionsIntroduction
xe
xtan
xsin
xsin
Questions
differentiates to:
FunctionsIntroduction
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.
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FunctionsIntroduction
h
xfhxfh
)()(lim 0