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HIGHER ORDER DERIVATIVES
.
1The second order partial derivatives
are found by differentiating the first
order partial derivatives
ii) Notation : 2
2
x
f
y x
f
2
x y
f
22
2
y
f
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y
=
2
2
) x
f
i
x
f Differentiatetwice with respectto x x
y
f Differentiatetwice with respectto y
=
2
2
) y
f ii
y
STEP 2 STEP 1
STEP 2 STEP 1
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=
y x
f i
2
)
y
f Differentiate firstwith respect to yand then withrespect to x x
x
f Differentiate firstwith respect to xand then withrespect to y
=
x y
f ii
2
) y
STEP 2 STEP 1
STEP 2 STEP 1
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EXAMPLE 1 : 2223 33),( y xy y x x y x f ++=Determine :
2
2
) x
f i
=
x f ( )2223 33 y xy y x x ++
Step 1 : find ,let y constant x f
x
y x
f
ii 2
)
Solution: i) To find =
2
2
x
f
x
f
xSTEP 1
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( ) ( ) ( ) (2223 33 y x
xy x
y x x
x x x
f +
+
=
( ) ( ) ( ) ( )2223
33 y x x x y x x y x x
+
+
=
Take out the coefficients
( ) ( )01323322
++= y x y x22 363 y xy x
x
f +=
constant
First order partialderivative is
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Therefore, the second partial derivatives is
= x f
x x f 2
2
( )22 363 y xy x +
( ) ( ) ( )22
363 y x x x y x x
+
=( ) ( ) ( )01623 += y x y x 66 =
Answer from step 1
constant
x
=
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Solution: i) Short-cut
2223
33),(Given y xy y x x y x f ++=1 st order : )0()1(3)2(33 22 ++=
y y x x
x
f
22 363 y xy x +=
2nd order : 0)1(6622
+=
y x x f
y x 66 =
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Solution: ii) To find x y f
2
=
x f
y STEP 1
Therefore, the second order partial derivatives is
=
x
f
y x y
f 2
( )22
363 yxyx +
Answer from step 1 (part i)
y
=
From part (i),first partial derivatives is 22 363 y xy x
x f +=
=
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( ) ( ) ( )22 363 y y
y y
x x y
+
=
( ) ( ) y x 2316)0( +=
yx 66 +=
constant
y x y f
=2
( )22
363 y xy x +Apply sum and different formula and take out the coefficients
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x y
f
y x
f
=
22
Equality of Mixed Second Order Partial Derivative Theorem
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EXAMPLE2:Given
2
),( y xe y x f =Verify (prove/show) that
x y f
y x f
=22
Solution : From the left hand side
= y x f 2
STEP 1 y f
x
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= y f ( )2 xye
Step 1 : find first partial derivativelet x constant
y f
y
( )2 xye=
( ) [ ]
= 22 y y
xe xy
[ ]
2 xy y
Take out the coefficients x
( ) [ ]( ) y xe xy 22=
Recalling that :
[ ] )('.)()( x f eedxd x f x f =
2
2 xy xye=
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= y f x y x f 2
( )2
2 xy xye x=
Answer from step 1
Take out the coefficients 2y
( )22 xy xex
y
= Product Rule
Therefore, the second order partial derivatives is
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( )222
xy xe x
y y x
f
=
[ ]''2 uvvu y +=
Factorizecommon
factor
[ ]))()(())(1(2 222 ye xe y xy xy +=
( )[ ]2122 xye y xy +=
( )2
122
xyye xy
+=
2 xye
Product rule
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SOLUTION : CONTINUE
Find the right hand side
=
x
f
y x y
f 2
STEP 1
x y f
2
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= x f ( )2 xye
Step 1 : find first partial derivativelet x constant
x f
x
( )2 xye=
( ) [ ]
= x x
ye xy 22
[ ]
2 xy x
Take out the coefficients
( )( ))1(22 ye xy= 22 xye y=
[ ] )('.)()( x f eedx
d x f x f =
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=
x
f
y x y
f 2
( )22 xye y y=Answer from step 1
Product Rule
Therefore, the second order partial derivatives is
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( )222
xye y
y x y
f
=
[ ]'' uvvu +=[ ] [ ] + =
22222
)()( xy y
e ye y y
xy xy
[ ]
+=22 22
)())(2( y y xe ye yxy xy
)2())(())(2(22 2 y xe ye y xy xy +=
Product rule
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Factorizecommonfactor ( )212 2 xy ye xy +=
22 322 xy xy e xy ye +=2
2 xy ye
Hence :
x y f
y x f
= 22
( )2122
xy ye xy +=Verify !
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EXAMPLE 3 : Given
= x y
y x f ln),(
, evaluate22
2
2
2
2
y x f
y f
x f at 1,2 == y x
Solution:
= x y
y x f ln),( rewrite x y y x f lnln),( =
The first partial derivatives are :
xx
f 1=
y y
f 1=
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So, the second partial derivatives are :
=
x f
x x f 2
2
=
y f
x y x f 2
= x x1 ( )1
= x x 2
1 x
=
=
y
f
y y
f 2
2
= y y
1 ( )1
= y y
2
1
y
=
Answer of first order partial derivative
Answer of first order partial derivative
= y x1
0=
Answer of first order partial derivative
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Therefore given : 22
2
2
2
2
y x f
y f
x f
( )222 0
11
=
y x
= 22 1
121
1,2 == y x
( )141
=
when
41=
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Solution: Continue.
For this example, in stead of find y x z
2
We will find x y z
2because we can get the same answer quickly
=
x z
y x y z 2
STEP 1
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Take out the coefficients
( )
++
=
12 ye
x xy
x x z y
( )
++
=
12 y
e
x x
x y
y
( ) 01 += y
constant
y x z =
Therefore,
= x
z y x y
z 2
( ) y y=
1=
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Remark : If we differentiate first we respect toy, we have to use quotient rule and
we are for significantly more work !
= y z
++
12 ye
xy y
y
Apply sum and different formula
( )
+
+
= 12 ye
y xy y
y
Quotient rule
Homework !
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More Higher Order Derivatives(3 rd Order, 4 th Order. )
EXAMPLE 5 : Let ye y y x f x += 2),(find
x y f
2
3
Solution: The formula
=
x y
f 2
3 Step 1 :diff. w.r. to x
Step 2 :diff. w.r. to y
Step 3 :diff. w.r. to y again
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=
x
f
ye y y x f x += 2),(Solution : GivenCopy coefficient
02 + xe y
constant
Step 2 : 2 nd order
(diff. w.r.to y)=
x y f 2 xe y2
xe y2=Differentiate w.r.to y
Step 1 : 1 st order
(diff. w.r.to x)
Step 3 : 3 rd order
(diff. w.r.to y again) =
x y f
2
3 xe)1(2 xe2=
Copy coefficient
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Chain Rule for functions of multiple intermediatevariables and 1 independent variable
If ),( y x f w = is differentiable and x and y are
t
dt dy
yw
dt dx
xw
dt dw
+
=
Remark : i) Notice that both partial derivative xw
and single-variable derivatives
dt
dx
are involved.
differentiable functions of t, then w is a differentiablefunction of t and :
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Example 1 :Suppose that y xw 2= where2t x = 3t y =
Use the chain rule to find dt dw
and check the result by expressing w
as a function of t and differentiatingdirectly.
and
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Solution : By using the chainrule
dt dy
yw
dt dx
xw
dt dw
+
=
)()()()(3222
t dt d
y x yt dt d
y x x
+
=
)3)(1.()2)(2( 22 t xt xy +=
)3)(()2)(.2( 2432 t t t t t +=
66
34 t t +=6
7t =
substitut e
y xw 2=2t x =3t y =
Given
Formula :
Answer in termof t
Solution :
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Solution :Continue.
method 2: We can express wdirectly as a function of t
y xw 2=S
o,
Remark : However, this procedure maynot
always be convenient
)()( 322 t t =7t =))(( 34 t t =
67t
dt
dw =
7
t w =substit
ute
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Example 2 : Suppose that y xy z +=
and,cos = x
,
sin= yUse the chain rule to find
d dz when
2
=
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Solution : By using the chain rule
d
dy
y
z
d
dx
x
z
d
dz
+
=
)(sin)()(cos)( dt
d y xy
ydt
d y xy
x+
++
=
substitut e
y xy z += cos= x sin= yGiven
Formula :
( ) ( ) ))(cos11.(21
)sin)(0.1(21
21
21
+++++= x y xy y y xy
Chainrule
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( ) ( ) ))(cos1(21
)sin)((21
21
21
++++=
x y xy y y xyd dz
y xy
x
y xy
y
+
++
+
=2
))(cos1(
2
)sin)((
y xy
x y
+++=
2
))(cos1()sin)((
sin))(sin(cos2
))(cos1(cos)sin)((sin
+++=
Solution : Continue
simplify
Combine the 2 terms
Answer interms of
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When 2
=
, 2sin)
2)(sin
2(cos2
)2)(cos12(cos)2sin)(2(sin
+
++=d dz
1)1)(0(2
)0)(10()1)(1(
+++=
d dz
21
102
01 =
+
+= d
dz
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Method 2 : If we express z directly as a
, we
obtain :
y xy z += cos= x sin= y
sin))(sin(cos += z
To get d
dz , we have to use product rule
function of
and it involve long working step !
substitute
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Case 2 : Given 3 intermediatevariables
),,( z y x f w = is differentiable and
dt dz
z w
dt dy
yw
dt dx
xw
dt dw
++=
x, y, z and 1 independent
variable t .
If x ,y and z are differentiable functions of t,
CHAIN
RULE :
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Example : Given z xyw += where,cos t x = ,sin t y = t z =
Finddt dw when 0=t
Solution:
dt dz
z w
dt dy
yw
dt dx
xw
dt dw
+
+=
)()()(sin)()(cos)( t dt d
z xy z
t dt d
z xy y
t dt d
z xy x
+++
++=
)1)(1())(cos01.()sin)(0.1( ++++= t xt y
substitut
e
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1))(cos()sin)(( ++= t xt y1))(cos(cos)sin)((sin ++= t t t t
1)(cos)(sin22
++= t t So, when 0=t
1)0(cos)0(sin22
++= 1)1()0( 22 ++=2
=
Solution : Continue
simplify
Answer in term of t
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A FORMULA FOR IMPLICITDIFFERENTIATION
),( y x F
and the equation 0),( = y x F
function of x. Then, at any point where0
y F , we obtain that :
y F
x F
dxdy
=
Case 1 :
Supposeis differentiable
defines y implicitly as a differentiable
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Example 1:
722 =+ y xy x, find dx
dy by using formula for implicit
result using implicitdifferentiation.
differentiation and checkthe
Given
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07),(Let22
=+= y xy x y x F Step 2 : By using formula for implicit differentiation,
y F
x F
dxdy
=
Solution : Given
Step 1: 0),( = y x F
722 =+ y xy x
Must bezero
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y F
x F
dxdy
=Formula
7),(Let22
+= y xy x y x F
Solution : Continue
00)1(2 ++= y x y x =2
02)1(0 ++=
y x
y
F y x 2+=
y x y x2
2+=
x y x y
=
22
simplify
x F
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Method 2: differentiating the given equation
722
=+ y xy x ,implicitly)7()( 22
dxd
y xy xdxd =+
x y
x y
dx
dy
=2
2
Remark: The calculation which using formula for implicit differentiation is significantly shorter than thesingle-variable calculation which we learned in Diploma
course.
Solve it by your-self
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Now supposethat z is given implicitly as
),( y x f z = by an equation of the form0),,( = z y x F . Then, at any point where
0 z F , we obtain
that :
z F
x F
x z
=
and z
F y F
y z
=
Case 2: For 3 variables x, y, z
a function of
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Example 2 : Given 223 333 =+++ xyz z y xfind
dx
z and
dy
z
Solution:The given equation can be written as
0223),,( 333 =+++= xyz z y x z y x F By using formula for implicit differentiation,
z F
x F
x z
=
y F
z =
Must bezero
03000009
2
2
++++++
= xy z yz x
xy z yz x
++
= 22
39
xz y += 226