www.le.ac.uk differentiation – product, quotient and chain rules department of mathematics...
Post on 31-Mar-2015
217 Views
Preview:
TRANSCRIPT
www.le.ac.uk
Differentiation – Product, Quotient and Chain Rules
Department of MathematicsUniversity of Leicester
Content
Product Rule
Quotient Rule
Chain Rule
Inversion Rule
Introduction
Introduction• Previously, we differentiated simple
functions using the definition:
• Now, we introduce some rules that allow us to differentiate any complex function just by remembering the derivatives of the simple functions…
Next
h
xfhxf
dx
dfh
)()(lim 0
Product Quotient Chain InversionIntro
• The product rule is used for functions like:
where and are two functions.
• The product rule says:
• Differentiate the 1st term and times it by the 2nd, then differentiate the 2nd term and times it by the 1st.
Product rule
)()( xvxuy
dx
dvu
dx
duv
dx
dy
)(xu )(xv
Product Quotient Chain InversionIntro
Click here for a proof
Next
h
xvxuhxvxu )()()()( ......continue
h
hxvxuhxvhxu
h
xvxuhxvhxu
h
h
)()()()(lim
)()()()(lim
0
0dx
df
)()()( xvxuxf Let . Then:
Go back to Product Rule
dx
dvu
dx
duv
dx
dvxuhxv
dx
du
h
hxvxuhxu
h
h
)()(lim
)()()(lim
0
0
h
xvhxvxuh
)()()(lim
0
Go back to Product Rule
... DEQ
Product rule example
• Find .
Next
)ln( 2 xxdx
d
dx
dvu
dx
duv
dx
dy
2xu xu 2
xv
1xv ln
xxxx
xxx
ln21
)()2)((ln 2
Product Quotient Chain InversionIntro
Next
Differentiate these: dx
dvu
dx
duv
dx
uvd
)(
x
xx
x
ex
xee
xe
2
xx
xxxx
xxxx
22 cos2sin2
)2sin(cos2)2cos(sin2
)2sin(sin2)2cos(cos2
xx
x
x
x
xxx
ln93
9
3ln9
3
2
22
xxe
)2cos(sin xx
xx ln3 3
Product Quotient Chain InversionIntro
Check Answers Clear Answers Show Answers
Hint
Hint
Hint
• The quotient rule is used for functions like:
where and are two functions.
• The quotient rule says:
• This time, it’s a subtraction, and then you divide by .
Quotient rule
Next
)(xu )(xv)(
)(
xv
xuy
2vdxdvu
dxduv
dx
dy
2v
Product Quotient Chain InversionIntro
Click here for a proof
)()(
)()()()(lim
)()(
)()(
lim
0
0
xvhxhv
hxvxuxvhxu
hxvxu
hxvhxu
h
hdx
df
Let . Then:)(
)()(
xv
xuxf
Go back to Quotient Rule
......continue
)()(
)()()()()()(lim
)()(
)()()()()()()()(lim
0
0
xvhxhv
hxvxvxuxvxuhxu
xvhxhv
hxvxuxvxuxvxuxvhxu
h
h
......continue
Go back to Quotient Rule
2
0)(
)()()(
)()(lim
vdxdvu
dxduv
xuh
xvhxvxv
h
xuhxuh
)()(
1
xvhxv
Go back to Quotient Rule
... DEQ
Quotient rule example
• Find .
22 )23(
7
)23(
3646
xx
xx
Next
23
12
x
x
dx
d 12 xu 2u
3v23 xv
22 )23(
3)12(2)23(
x
xx
vdxdvu
dxduv
dx
dy
Product Quotient Chain InversionIntro
(give your answers as decimals)
Next
Differentiate these: 2v
dxdvu
dxduv
dxvu
d
1,1
2
xx
x
8,
2)2sin(
23)2sin(
xx
x
2,1
23
xx
x
Product Quotient Chain InversionIntro
Check Answers Clear Answers Show Answers
Hint
Hint
Hint
• The chain rule is used for functions, , which have one expression inside another expression.
• Let be the inside part, so that now is just a function of .
• Then the chain rule says:
Chain rule
Next
dx
du
du
dy
dx
dy
, which has inside.2)2( xy 2x
2)( xxu 2uy , then
y
)(xuu
y
Product Quotient Chain InversionIntro
Click here for a proof
Instead of writing:
We write:
h
xfhxf
dx
dyh
)()(lim
0
The best way to prove the chain rule is to write the definition of derivative in a different way:
xa
xfaf
dx
dyxa
)()(lim
If we put , we see that these two definitions are the same.
xah
......continue
Go back to Chain Rule
xa
xuau
xuau
xufauf
xa
xufauf
xa
xa
)()(
)()(
))(())((lim
))(())((lim
We have .))(( xufy
dx
dy
......continue
Go back to Chain Rule
dx
du
du
df
dx
du
ub
ufbf
xa
xuau
xuau
xufauf
ub
xaxa
)()(lim
)()(lim
)()(
))(())((lim
u(x) is just u, and u(a) is just a number, so we can call it b.
Then the first term matches the definition of .du
df
Go back to Chain Rule
... DEQ
Chain rule example
• Find .))(sin( 2xdx
d
Next
)sin( 2xy 2xu
)cos(2cos2 2xxuxdx
du
du
dy
dx
dy
uy sin, so
udu
dycosx
dx
du2
Product Quotient Chain InversionIntro
differentiates to
Next
True or False?12xe 12 2
)1( xex
))12sin(( 2x ))12cos(()12(2 2 xx
3)1(sin x 2)1(sincos3 xx
differentiates to
differentiates to
Product Quotient Chain InversionIntro
True
True
True
False
False
False
Check Answers Clear Answers Show Answers
Hint
Hint
Hint
Inversion Rule
• If you have a function that is written in
terms of y, eg.
Then you can use this fact:
• So if , then .
dydxdx
dy 1
)(yfdy
dx
)(
1
yfdx
dy
Next
132 yyx
Product Quotient Chain InversionIntro
Click here for a proof
First note that , because we’re
differentiating the function .
Then:
1dx
dx
dydxdx
dy 1
xxf )(
1dx
dx
dx
dy
dy
dx
This is a function, so we can divide by
it…We get: .
dx
du
du
dy
dx
dy
by the chain rule,
Go back to Inversion Rule
... DEQ
Inversion Rule Example
• A curve has an equation .
Find when .
• , therefore
• Then when ,
142 yyx
dx
dy
42 ydy
dx
42
1
ydx
dy
2
1
dx
dy
Next
1y
1y
Product Quotient Chain InversionIntro
2
2
12
2
dyxddx
yd
Note, it is NOT TRUE that
Next
Product Quotient Chain InversionIntro
Next
Find at the specified values of y:
dydxdx
dy 1
dx
dy
1,4
2
yyy
x
3,sin2
yyx
Product Quotient Chain InversionIntro
Check Answers Clear Answers Show Answers
Hint
Hint
More complicated example
• Find .
• Quotient rule:
• Chain rule on :
– is ‘inside’, so let
– Then , so
1
12
2
x
x
e
e
dx
d
Next
12 xeu 12 xev
x2
1)( 2 xexu
wewuxxw )(,2)(
wewuxw )(,2)(xe
dx
dw
dw
du
dx
du 22
Product Quotient Chain InversionIntro
• Chain rule on also gives .
• Then quotient rule gives:
22
2222
22
2
)1(
2)1()1(2
1
1
x
xxxx
x
x
e
eeee
v
vuvu
e
e
dx
d
22
222222
)1(
2)(22)(2
x
xxxx
e
eeee
Next
1)( 2 xexv xedx
dv 22
22
2
)1(
4
x
x
e
e
Product Quotient Chain InversionIntro
Differentiate
xxxx
xxxx
cossin)cos(2
sin)sin(33
232
xxxxx cossin)sin(6 32
xxxx
xx
cossin)cos(2
sin)13cos(3
22
:sin)cos( 23 xxx
xxxx
xxxxx
cossin)cos(2
sin)cos()(3
233
Product Quotient Chain InversionIntro
Conclusion
• We can differentiate simple functions using the definition:
• We have found rules for differentiating products, quotients, compositions and functions written in terms of x.
• Using these two things we can now differentiate ANY function.
Next
Product Quotient Chain InversionIntro
h
xfhxf
dx
dfh
)()(lim 0
top related