differentiation - amazon s3 · 2019-01-10 · introduction •differentiation is the process of...
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Differentiation
D STAVROVA
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
Functions Introduction
Choose h = .
from to :
The straight line has a gradient of :
0.5 1 1.5 2 2.5
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Functions Introduction
))(,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
4
7
287
0.5
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
Functions Introduction
))(,( 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...
Functions Introduction
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
Summary
Click on the functions to see how they are derived.
Functions Introduction
Next
)(xf )(xf
Differentiating a constant
Next
Functions Introduction
0lim0
h
cc
dx
df
h
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
Functions Introduction
dx
df
nx
Back to summary
The Binomial expansion gives a general formula for (x+y)n.
It says:
nyx )(
nnn
nnn
ynxyyxnnn
yxnnynxx
133
221
...3
)2)(1(
2
)1(
9
Recap Binomial Expansion
Hide Binomial Expansion
12
1
0...
2
)1(lim
n
nn
hnx
xnhnnx
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Functions Introduction
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
Functions Introduction
dx
df
xe
Back to summary
The Maclaurin’s Series gives an expansion for ex.
It says:
...
54321
5432 xxxxxex
11
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Functions Introduction
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.
The Maclaurin’s Series gives an expansion for ex.
It says:
...
54321
5432 xxxxxex
Recap Maclaurin's Series
Hide Maclaurin's Series
12
......
)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)
Next Back to summary
Functions Introduction
dx
df
xsin
The Trig Identity says:
abbaba cossincossin)sin(
Recap Trig Identity
Hide Trig Identity
13
xh
hx
h
hx
dx
df
hcos
)sin(cos
1)cos(sinlim
0
This is using the
Maclaurin’s Series
for sin(x) and cos(x)
Next Back to summary
Functions Introduction
So
01...
321
1)cos(
32
h
hh
h
h
1...
53)sin(
53
h
hhh
h
h The Maclaurin’s Series gives expansions for sinx and cosx
It says:
...642
1cos
...752
sin
642
753
xxxx
xxxxx
Recap Maclaurin's Series
Hide Maclaurin's Series
14
......
)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)
Next Back to summary
Functions Introduction
dx
df
xcos
The Trig Identity says:
bababa sinsincoscos)cos(
Recap Trig Identity
Hide Trig Identity
15
xh
hx
h
hx
dx
df
hsin
)sin(sin
1)cos(coslim
0
This is using the
Maclaurin’s Series
for sin(x) and cos(x)
Next Back to summary
Functions Introduction
So
01...
321
1)cos(
32
h
hh
h
h
1...
53)sin(
53
h
hhh
h
h The Maclaurin’s Series gives expansions for sinx and cosx
It says:
...642
1cos
...752
sin
642
753
xxxx
xxxxx
Recap Maclaurin's Series
Hide Maclaurin's Series
16
......
)1ln(lim
)ln(lim
ln)ln(lim
00
0
continue
h
x
h
h
x
hx
h
xhx
hh
h
Differentiating :
Next
Functions Introduction
dx
df
xln
Back to summary
`
13
x
h
x
h
x
h
x
h
x
h
h
x
h
hh
1
...432lim
)1ln(lim
432
00
This is using the Macluarin’s
Series for ln(a+1)
Next
Functions Introduction
Back to summary
Because after you divide by h, all the other
terms have h in them so disappear as h0.
The Maclaurin’s Series gives an expansion for ln(a + 1).
It says:
...5432
)1ln(5432 aaaa
aa
Recap Maclaurin's Series
Hide Maclaurin's Series
18
xdx
xd
h
xhx
h
xahxa
xadx
d
h
h
1lnln)ln(lim
lnln)ln(lnlim
)ln(ln
0
0
Next
Functions Introduction
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
Functions Introduction
Back to summary
Differentiating : )ln( ax
))(ln( axdx
d
xcos
xcos
xtan
Questions
differentiates to:
Functions Introduction
xsin
12 x
22 x
32 x
Questions
differentiates to:
Functions Introduction
2x
xln
1
xln
2
1
x
Questions
differentiates to:
Functions Introduction
x
1
xsec
x2sec
x1tan
Questions
differentiates to:
Functions Introduction
xtan
x
1
xln
1
x
Questions
differentiates to:
Functions Introduction
xln
2
5
5x
2
3
2
5x
2
4
5x
Questions
differentiates to:
Functions Introduction
2
5
x
xxe
1xxe
xe
Questions
differentiates to:
Functions Introduction
xe
xtan
xsin
xsin
Questions
differentiates to:
Functions Introduction
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
Functions Introduction
h
xfhxfh
)()(lim 0
