multivariable chain rules: 2nd-order derivatives ... chain rules: 2nd-order derivatives calculus iii...
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Multivariable Chain Rules: 2nd-order DerivativesCalculus III
Josh Engwer
TTU
10 December 2014
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 1 / 44
2nd-order Derivatives using Multivariable Chain Rules(Toolkit)
RESULT SAMPLE STATEMENT(D2D) Definition of 2nd-order Derivative d2w
dt2 = ddt
[ dwdt
](D2P) Definition of 2nd-order Partial ∂2w
∂t2 = ∂∂t
[∂w∂t
], ∂2w
∂s∂t =∂∂s
[∂w∂t
](EMP) Equality of Mixed 2nd-order Partials ∂2w
∂u∂v = ∂2w∂v∂u
(PR) Product Rule for Derivatives ddt
[f (t)g(t)
]= f ′(t)g(t) + f (t)g′(t)
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 2 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
”1-2” means 1 intermediate variable (x) and 2 independent var’s (s, t).
How to compute the 2nd-order partial:∂2z∂t2 ??
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 3 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 4 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂t2 = ∂
∂t
[∂z∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 5 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂t2 = ∂
∂t
[∂z∂t
]= ∂
∂t
[ dzdx ·
∂x∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 6 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂t2 = ∂
∂t
[∂z∂t
]= ∂
∂t
[ dzdx ·
∂x∂t
]PR= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂t
[∂x∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 7 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂t2 = ∂
∂t
[∂z∂t
]= ∂
∂t
[ dzdx ·
∂x∂t
]PR= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂t
[∂x∂t
]D2P= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂2x∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 8 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂t2 = ∂
∂t
[∂z∂t
]= ∂
∂t
[ dzdx ·
∂x∂t
]PR= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂t
[∂x∂t
]D2P= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂2x∂t2
1−2=
( ddx
[ dzdx
]∂x∂t
)· ∂x∂t +
dzdx
∂2x∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 9 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂t2 = ∂
∂t
[∂z∂t
]= ∂
∂t
[ dzdx ·
∂x∂t
]PR= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂t
[∂x∂t
]D2P= ∂
∂t
[ dzdx
]· ∂x∂t +
dzdx ·
∂2x∂t2
1−2=
( ddx
[ dzdx
]∂x∂t
)· ∂x∂t +
dzdx
∂2x∂t2
D2D= d2z
dx2
(∂x∂t
)2+ dz
dx∂2x∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 10 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
∴∂2z∂t2 =
d2zdx2
(∂x∂t
)2
+dzdx
∂2x∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 11 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
”1-2” means 1 intermediate variable (x) and 2 independent var’s (s, t).
How to compute the 2nd-order partial:∂2z∂s∂t
??
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 12 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 13 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂s∂t = ∂
∂s
[∂z∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 14 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂s∂t = ∂
∂s
[∂z∂t
]= ∂
∂s
[ dzdx ·
∂x∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 15 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂s∂t = ∂
∂s
[∂z∂t
]= ∂
∂s
[ dzdx ·
∂x∂t
]PR= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂s
[∂x∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 16 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂s∂t = ∂
∂s
[∂z∂t
]= ∂
∂s
[ dzdx ·
∂x∂t
]PR= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂s
[∂x∂t
]D2P= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂2x∂s∂t
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 17 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂s∂t = ∂
∂s
[∂z∂t
]= ∂
∂s
[ dzdx ·
∂x∂t
]PR= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂s
[∂x∂t
]D2P= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂2x∂s∂t
1−2=
( ddx
[ dzdx
]∂x∂s
)· ∂x∂t +
dzdx
∂2x∂s∂t
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 18 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
1st, realize that∂z∂t
,dzdx
still depend on the intermediate variable (x).
∂2z∂s∂t = ∂
∂s
[∂z∂t
]= ∂
∂s
[ dzdx ·
∂x∂t
]PR= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂∂s
[∂x∂t
]D2P= ∂
∂s
[ dzdx
]· ∂x∂t +
dzdx ·
∂2x∂s∂t
1−2=
( ddx
[ dzdx
]∂x∂s
)· ∂x∂t +
dzdx
∂2x∂s∂t
D2D= d2z
dx2∂x∂s
∂x∂t +
dzdx
∂2x∂s∂t
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 19 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
∴∂2z∂s∂t
=d2zdx2
∂x∂s
∂x∂t
+dzdx
∂2x∂s∂t
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 20 / 44
1-2 Chain Rule
Proposition(1-2 Chain Rule)
Let z = f (x) ∈ C2 where x = g(s, t) ∈ C(2,2). Then:
∂z∂s
=dzdx
∂x∂s
∂z∂t
=dzdx
∂x∂t
EXERCISE: Starting from scratch, show that
∂2z∂s2 =
d2zdx2
(∂x∂s
)2
+dzdx
∂2x∂s2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 21 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
”2-1” means 2 intermediate var’s (x, y) and 1 independent variable (t).
How to compute the 2nd-order derivative:d2zdt2 ??
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 22 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
1st, realize thatdzdt
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 23 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
1st, realize thatdzdt
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 24 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
1st, realize thatdzdt
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]PR=
( ddt
[∂z∂x
]· dx
dt +∂z∂x ·
ddt
[ dxdt
])+(
ddt
[∂z∂y
]· dy
dt +∂z∂y ·
ddt
[ dydt
])
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 25 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
1st, realize thatdzdt
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]PR=
( ddt
[∂z∂x
]· dx
dt +∂z∂x ·
ddt
[ dxdt
])+(
ddt
[∂z∂y
]· dy
dt +∂z∂y ·
ddt
[ dydt
])D2D= d
dt
[∂z∂x
]· dx
dt +∂z∂x
d2xdt2 + d
dt
[∂z∂y
]· dy
dt +∂z∂y
d2ydt2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 26 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
1st, realize thatdzdt
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]PR=
( ddt
[∂z∂x
]· dx
dt +∂z∂x ·
ddt
[ dxdt
])+(
ddt
[∂z∂y
]· dy
dt +∂z∂y ·
ddt
[ dydt
])D2D= d
dt
[∂z∂x
]· dx
dt +∂z∂x
d2xdt2 + d
dt
[∂z∂y
]· dy
dt +∂z∂y
d2ydt2
2−1=
(∂∂x
[∂z∂x
] dxdt +
∂∂y
[∂z∂x
] dydt
)dxdt +
(∂∂x
[∂z∂y
] dxdt +
∂∂y
[∂z∂y
] dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 27 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]PR=
( ddt
[∂z∂x
]· dx
dt +∂z∂x ·
ddt
[ dxdt
])+(
ddt
[∂z∂y
]· dy
dt +∂z∂y ·
ddt
[ dydt
])D2D= d
dt
[∂z∂x
]· dx
dt +∂z∂x
d2xdt2 + d
dt
[∂z∂y
]· dy
dt +∂z∂y
d2ydt2
2−1=
(∂∂x
[∂z∂x
] dxdt +
∂∂y
[∂z∂x
] dydt
)dxdt +
(∂∂x
[∂z∂y
] dxdt +
∂∂y
[∂z∂y
] dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
D2P=
(∂2z∂x2
dxdt +
∂2z∂y∂x
dydt
)dxdt +
(∂2z∂x∂y
dxdt +
∂2z∂y2
dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 28 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]PR=
( ddt
[∂z∂x
]· dx
dt +∂z∂x ·
ddt
[ dxdt
])+(
ddt
[∂z∂y
]· dy
dt +∂z∂y ·
ddt
[ dydt
])D2D= d
dt
[∂z∂x
]· dx
dt +∂z∂x
d2xdt2 + d
dt
[∂z∂y
]· dy
dt +∂z∂y
d2ydt2
2−1=
(∂∂x
[∂z∂x
] dxdt +
∂∂y
[∂z∂x
] dydt
)dxdt +
(∂∂x
[∂z∂y
] dxdt +
∂∂y
[∂z∂y
] dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
D2P=
(∂2z∂x2
dxdt +
∂2z∂y∂x
dydt
)dxdt +
(∂2z∂x∂y
dxdt +
∂2z∂y2
dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
= ∂2z∂x2
( dxdt
)2+ ∂2z
∂y∂xdxdt
dydt +
∂2z∂x∂y
dxdt
dydt +
∂2z∂y2
( dydt
)2+ ∂z
∂xd2xdt2 + ∂z
∂yd2ydt2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 29 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
d2zdt2 = d
dt
[∂z∂x ·
dxdt
]+ d
dt
[∂z∂y ·
dydt
]PR=
( ddt
[∂z∂x
]· dx
dt +∂z∂x ·
ddt
[ dxdt
])+(
ddt
[∂z∂y
]· dy
dt +∂z∂y ·
ddt
[ dydt
])D2D= d
dt
[∂z∂x
]· dx
dt +∂z∂x
d2xdt2 + d
dt
[∂z∂y
]· dy
dt +∂z∂y
d2ydt2
2−1=
(∂∂x
[∂z∂x
] dxdt +
∂∂y
[∂z∂x
] dydt
)dxdt +
(∂∂x
[∂z∂y
] dxdt +
∂∂y
[∂z∂y
] dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
D2P=
(∂2z∂x2
dxdt +
∂2z∂y∂x
dydt
)dxdt +
(∂2z∂x∂y
dxdt +
∂2z∂y2
dydt
)dydt +
∂z∂x
d2xdt2 + ∂z
∂yd2ydt2
= ∂2z∂x2
( dxdt
)2+ ∂2z
∂y∂xdxdt
dydt +
∂2z∂x∂y
dxdt
dydt +
∂2z∂y2
( dydt
)2+ ∂z
∂xd2xdt2 + ∂z
∂yd2ydt2
EMP= ∂2z
∂x2
( dxdt
)2+ 2 ∂2z
∂x∂ydxdt
dydt +
∂2z∂y2
( dydt
)2+ ∂z
∂xd2xdt2 + ∂z
∂yd2ydt2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 30 / 44
2-1 Chain Rule
Proposition(2-1 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(t) ∈ C2 and y = h(t) ∈ C2. Then:
dzdt
=∂z∂x
dxdt
+∂z∂y
dydt
∴d2zdt2 =
∂2z∂x2
(dxdt
)2
+ 2∂2z∂x∂y
dxdt
dydt
+∂2z∂y2
(dydt
)2
+∂z∂x
d2xdt2 +
∂z∂y
d2ydt2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 31 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
”2-2” means 2 intermediate var’s (x, y) and 2 independent var’s (s, t).
How to compute the 2nd-order partial:∂2z∂t2 ??
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 32 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
1st, realize that∂z∂t
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 33 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
1st, realize that∂z∂t
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 34 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
1st, realize that∂z∂t
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]PR=
(∂∂t
[∂z∂x
]· ∂x∂t +
∂z∂x ·
∂∂t
[∂x∂t
])+(
∂∂t
[∂z∂y
]· ∂y∂t +
∂z∂y ·
∂∂t
[∂y∂t
])
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 35 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
1st, realize that∂z∂t
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]PR=
(∂∂t
[∂z∂x
]· ∂x∂t +
∂z∂x ·
∂∂t
[∂x∂t
])+(
∂∂t
[∂z∂y
]· ∂y∂t +
∂z∂y ·
∂∂t
[∂y∂t
])D2P= ∂
∂t
[∂z∂x
]· ∂x∂t +
∂z∂x
∂2x∂t2 + ∂
∂t
[∂z∂y
]· ∂y∂t +
∂z∂y
∂2y∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 36 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
1st, realize that∂z∂t
,∂z∂x
,∂z∂y
still depend on all intermediate variables (x, y).
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]PR=
(∂∂t
[∂z∂x
]· ∂x∂t +
∂z∂x ·
∂∂t
[∂x∂t
])+(
∂∂t
[∂z∂y
]· ∂y∂t +
∂z∂y ·
∂∂t
[∂y∂t
])D2P= ∂
∂t
[∂z∂x
]· ∂x∂t +
∂z∂x
∂2x∂t2 + ∂
∂t
[∂z∂y
]· ∂y∂t +
∂z∂y
∂2y∂t2
2−2=
(∂∂x
[∂z∂x
]∂x∂t +
∂∂y
[∂z∂x
]∂y∂t
)∂x∂t +
(∂∂x
[∂z∂y
]∂x∂t +
∂∂y
[∂z∂y
]∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 37 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]PR=
(∂∂t
[∂z∂x
]· ∂x∂t +
∂z∂x ·
∂∂t
[∂x∂t
])+(
∂∂t
[∂z∂y
]· ∂y∂t +
∂z∂y ·
∂∂t
[∂y∂t
])D2P= ∂
∂t
[∂z∂x
]· ∂x∂t +
∂z∂x
∂2x∂t2 + ∂
∂t
[∂z∂y
]· ∂y∂t +
∂z∂y
∂2y∂t2
2−2=
(∂∂x
[∂z∂x
]∂x∂t +
∂∂y
[∂z∂x
]∂y∂t
)∂x∂t +
(∂∂x
[∂z∂y
]∂x∂t +
∂∂y
[∂z∂y
]∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
D2P=
(∂2z∂x2
∂x∂t +
∂2z∂y∂x
∂y∂t
)∂x∂t +
(∂2z∂x∂y
∂x∂t +
∂2z∂y2
∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 38 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]PR=
(∂∂t
[∂z∂x
]· ∂x∂t +
∂z∂x ·
∂∂t
[∂x∂t
])+(
∂∂t
[∂z∂y
]· ∂y∂t +
∂z∂y ·
∂∂t
[∂y∂t
])D2P= ∂
∂t
[∂z∂x
]· ∂x∂t +
∂z∂x
∂2x∂t2 + ∂
∂t
[∂z∂y
]· ∂y∂t +
∂z∂y
∂2y∂t2
2−2=
(∂∂x
[∂z∂x
]∂x∂t +
∂∂y
[∂z∂x
]∂y∂t
)∂x∂t +
(∂∂x
[∂z∂y
]∂x∂t +
∂∂y
[∂z∂y
]∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
D2P=
(∂2z∂x2
∂x∂t +
∂2z∂y∂x
∂y∂t
)∂x∂t +
(∂2z∂x∂y
∂x∂t +
∂2z∂y2
∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
= ∂2z∂x2
(∂x∂t
)2+ ∂2z
∂y∂x∂x∂t
∂y∂t +
∂2z∂x∂y
∂x∂t
∂y∂t +
∂2z∂y2
(∂y∂t
)2+ ∂z
∂x∂2x∂t2 + ∂z
∂y∂2y∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 39 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
∂2z∂t2 = ∂
∂t
[∂z∂x ·
∂x∂t
]+ ∂
∂t
[∂z∂y ·
∂y∂t
]PR=
(∂∂t
[∂z∂x
]· ∂x∂t +
∂z∂x ·
∂∂t
[∂x∂t
])+(
∂∂t
[∂z∂y
]· ∂y∂t +
∂z∂y ·
∂∂t
[∂y∂t
])D2P= ∂
∂t
[∂z∂x
]· ∂x∂t +
∂z∂x
∂2x∂t2 + ∂
∂t
[∂z∂y
]· ∂y∂t +
∂z∂y
∂2y∂t2
2−2=
(∂∂x
[∂z∂x
]∂x∂t +
∂∂y
[∂z∂x
]∂y∂t
)∂x∂t +
(∂∂x
[∂z∂y
]∂x∂t +
∂∂y
[∂z∂y
]∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
D2P=
(∂2z∂x2
∂x∂t +
∂2z∂y∂x
∂y∂t
)∂x∂t +
(∂2z∂x∂y
∂x∂t +
∂2z∂y2
∂y∂t
)∂y∂t +
∂z∂x
∂2x∂t2 + ∂z
∂y∂2y∂t2
= ∂2z∂x2
(∂x∂t
)2+ ∂2z
∂y∂x∂x∂t
∂y∂t +
∂2z∂x∂y
∂x∂t
∂y∂t +
∂2z∂y2
(∂y∂t
)2+ ∂z
∂x∂2x∂t2 + ∂z
∂y∂2y∂t2
EMP= ∂2z
∂x2
(∂x∂t
)2+ 2 ∂2z
∂x∂y∂x∂t
∂y∂t +
∂2z∂y2
(∂y∂t
)2+ ∂z
∂x∂2x∂t2 + ∂z
∂y∂2y∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 40 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
∴∂2z∂t2 =
∂2z∂x2
(∂x∂t
)2
+ 2∂2z∂x∂y
∂x∂t
∂y∂t
+∂2z∂y2
(∂y∂t
)2
+∂z∂x
∂2x∂t2 +
∂z∂y
∂2y∂t2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 41 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
EXERCISE: Starting from scratch, show that
∂2z∂s2 =
∂2z∂x2
(∂x∂s
)2
+ 2∂2z∂x∂y
∂x∂s
∂y∂s
+∂2z∂y2
(∂y∂s
)2
+∂z∂x
∂2x∂s2 +
∂z∂y
∂2y∂s2
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 42 / 44
2-2 Chain Rule
Proposition(2-2 Chain Rule)
Let z = f (x, y) ∈ C(2,2) where x = g(s, t) ∈ C(2,2) and y = h(s, t) ∈ C(2,2). Then:
∂z∂s
=∂z∂x
∂x∂s
+∂z∂y
∂y∂s
∂z∂t
=∂z∂x
∂x∂t
+∂z∂y
∂y∂t
EXERCISE: Starting from scratch, show that
∂2z∂s∂t
=∂2z∂x2
∂x∂s
∂x∂t
+∂2z∂x∂y
(∂y∂s
∂x∂t
+∂x∂s
∂y∂t
)+
∂2z∂y2
∂y∂s
∂y∂t
+∂z∂x
∂2x∂s∂t
+∂z∂y
∂2y∂s∂t
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 43 / 44
Fin
Fin.
Josh Engwer (TTU) Multivariable Chain Rules: 2nd -order Derivatives 10 December 2014 44 / 44