Jim Lambers
MAT 280
Spring Semester 2009-10
Lecture 6 Notes
These notes correspond to Section 11.5 in Stewart and Sections 2.5 and 3.5 in Marsden and Tromba.
The Chain Rule
Recall from single-variable calculus that if a function 𝑔(𝑥) is differentiable at 𝑥0 , and 𝑓 (𝑥) is
differentiable at 𝑔(𝑥0 ), then the derivative of the composition (𝑓 ∘ 𝑔)(𝑥) = 𝑓 (𝑔(𝑥)) is given by the
Chain Rule
(𝑓 ∘ 𝑔)′ (𝑥0 ) = 𝑓 ′ (𝑔(𝑥0 ))𝑔 ′ (𝑥0 ).
We now generalize the Chain Rule to functions of several variables. Let f : 𝐷 ⊆ ℝ𝑛 → ℝ𝑚 , and let
g : 𝑈 ⊆ ℝ𝑝 → 𝐷. That is, the range of g is the domain of f .
Assume that g is differentiable at a point p0 ∈ 𝑈 , and that f is differentiable at the point
q0 = g(p0 ). Then, f has a Jacobian matrix 𝐽f (q0 ), and g has a Jacobian matrix 𝐽g (p0 ). These
matrices contain the first partial derivatives of f and g evaluated at q0 and p0 , respectively.
Then, the Chain Rule states that the derivative of the composition (f ∘ g) : 𝑈 → ℝ𝑚 , defined
by (f ∘ g)(x) = f (g(x)), at p0 , is given by the Jacobian matrix
𝐽f ∘g (p0 ) = 𝐽f (g(p0 ))𝐽g (p0 ).
That is, the derivative of f ∘ g at p0 is the product, in the sense of matrix multiplication, of the
derivative of f at g(p0 ) and the derivative of g at p0 . This is entirely analogous to the Chain Rule
from single-variable calculus, in which the derivative of 𝑓 ∘ 𝑔 at 𝑥0 is the product of the derivative
of 𝑓 at 𝑔(𝑥0 ) and the derivative of 𝑔 at 𝑥0 .
It follows from the rules of matrix multiplication that the partial derivative of the 𝑖th component
function of f ∘ g with respect to the variable 𝑥𝑗 , an independent variable of g, is given by the dot
product of the gradient of the 𝑖th component function of f with the vector that contains the partial
derivatives of the component functions of g with respect to 𝑥𝑗 . We now illustrate the application
of this general Chain Rule with some examples.
Example Let 𝑓 : ℝ3 → ℝ be defined by
𝑓 (𝑥, 𝑦, 𝑧) = 𝑒𝑧 cos 2𝑥 sin 3𝑦,
and let g : ℝ → ℝ3 be a vector-valued function of one variable defined by
g(𝑡) = ⟨𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)⟩ = ⟨2𝑡, 𝑡2 , 𝑡3 ⟩.
1
Then, 𝑓 ∘ g is a scalar-valued function of 𝑡,
3
(𝑓 ∘ g)(𝑡) = 𝑒𝑧(𝑡) cos 2𝑥(𝑡) sin 3𝑦(𝑡) = 𝑒𝑡 cos 4𝑡 sin 3𝑡2 .
To compute its derivative with respect to 𝑡, we first compute
] [
]
[
∇𝑓 = 𝑓𝑥 𝑓𝑦 𝑓𝑧 = −2𝑒𝑧 sin 2𝑥 sin 3𝑦 3𝑒𝑧 cos 2𝑥 cos 3𝑦 𝑒𝑧 cos 2𝑥 sin 3𝑦 ,
and
g′ (𝑡) = ⟨𝑥′ (𝑡), 𝑦 ′ (𝑡), 𝑧 ′ (𝑡)⟩ = ⟨2, 2𝑡, 3𝑡2 ⟩,
and then apply the Chain Rule to obtain
𝑑𝑓
𝑑𝑡
= ∇𝑓 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) ⋅ g′ (𝑡)
⎡
=
[
𝑓𝑥 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) 𝑓𝑦 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) 𝑓𝑧 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡))
]
⎣
𝑑𝑥
𝑑𝑡
𝑑𝑦
𝑑𝑡
𝑑𝑧
𝑑𝑡
⎤
⎦
𝑑𝑦
𝑓𝑧
𝑑𝑥
+ 𝑓𝑦 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) + 𝑓𝑧 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡))
𝑑𝑡
𝑑𝑡
𝑑𝑡
= (−2𝑒𝑧(𝑡) sin 2𝑥(𝑡) sin 3𝑦(𝑡))(2) + (3𝑒𝑧(𝑡) cos 2𝑥(𝑡) cos 3𝑦(𝑡))(2𝑡) + (𝑒𝑧(𝑡) cos 2𝑥(𝑡) sin 3𝑦(𝑡))(3𝑡2 )
= 𝑓𝑥 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡))
3
3
3
= −4𝑒𝑡 sin 4𝑡 sin 3𝑡2 + 6𝑡𝑒𝑡 cos 4𝑡 cos 3𝑡2 + 3𝑡2 𝑒𝑡 cos 4𝑡 sin 3𝑡2 .
□
Example Let 𝑓 : ℝ2 → ℝ be defined by
𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 + 𝑥𝑦 2 ,
and let g : ℝ2 → ℝ2 be defined by
[
g(𝑠, 𝑡) =
𝑥(𝑠, 𝑡)
𝑦(𝑠, 𝑡)
]
[
=
2𝑠 + 𝑡
𝑠 − 2𝑡
]
.
Then, 𝑓 ∘ g is a scalar-valued function of 𝑠 and 𝑡,
(𝑓 ∘ g)(𝑠, 𝑡) = 𝑥(𝑠, 𝑡)2 𝑦(𝑠, 𝑡) + 𝑥(𝑠, 𝑡)𝑦(𝑠, 𝑡)2 = (2𝑠 + 𝑡)2 (𝑠 − 2𝑡) + (2𝑠 + 𝑡)(𝑠 − 2𝑡)2 .
To compute its gradient, which includes its partial derivatives with respect to 𝑠 and 𝑡, we first
compute
[
] [
]
∇𝑓 = 𝑓𝑥 𝑓𝑦 = 2𝑥𝑦 + 𝑦 2 𝑥2 + 2𝑥𝑦 ,
and
[
𝐽g (𝑠, 𝑡) =
𝑥𝑠 𝑥𝑡
𝑦𝑠 𝑦𝑡
2
]
[
=
2 1
1 −2
]
,
and then apply the Chain Rule to obtain
∇(𝑓 ∘ g)(𝑠, 𝑡) = ∇𝑓 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡))𝐽g (𝑠, 𝑡)
[
]
[
] 𝑥𝑠 𝑥𝑡
𝑓𝑥 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡)) 𝑓𝑦 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡))
=
𝑦𝑠 𝑦𝑡
[
𝑓𝑥 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡))𝑥𝑠 + 𝑓𝑦 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡))𝑦𝑠
=
]
𝑓𝑥 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡))𝑥𝑡 + 𝑓𝑦 (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡))𝑦𝑡
[
[2𝑥(𝑡)𝑦(𝑡) + 𝑦(𝑡)2 ](2) + [𝑥(𝑡)2 + 2𝑥(𝑡)𝑦(𝑡)](1)
=
]
[2𝑥(𝑡)𝑦(𝑡) + 𝑦(𝑡)2 ](1) + [𝑥(𝑡)2 + 2𝑥(𝑡)𝑦(𝑡)](−2)
[
4(2𝑠 + 𝑡)(𝑠 − 2𝑡) + 2(𝑠 − 2𝑡)2 + (2𝑠 + 𝑡)2 + 2(2𝑠 + 𝑡)(𝑠 − 2𝑡)
=
2(2𝑠 + 𝑡)(𝑠 − 2𝑡) + (𝑠 − 2𝑡)2 − 2(2𝑠 + 𝑡)2 − 4(2𝑠 + 𝑡)(𝑠 − 2𝑡)
]
.
□
Example Let 𝑓 : ℝ → ℝ be defined by
𝑓 (𝑥) = 𝑥3 + 2𝑥2 ,
and let 𝑔 : ℝ2 → ℝ be defined by
𝑔(𝑢, 𝑣) = sin 𝑢 cos 𝑣.
Then 𝑓 ∘ 𝑔 is a scalar-valued function of 𝑢 and 𝑣,
(𝑓 ∘ 𝑔)(𝑢, 𝑣) = (sin 𝑢 cos 𝑣)3 + 2(sin 𝑢 cos 𝑣)2 .
To compute its gradient, which includes partial derivatives with respect to 𝑢 and 𝑣, we first compute
𝑓 ′ (𝑥) = 3𝑥2 + 4𝑥,
and
∇𝑔 =
[
𝑔𝑢 𝑔𝑣
]
=
[
cos 𝑢 cos 𝑣 − sin 𝑢 sin 𝑣
]
,
and then use the Chain Rule to obtain
∇(𝑓 ∘ g)(𝑢, 𝑣) = 𝑓 ′ (𝑔(𝑢, 𝑣))∇𝑔(𝑢, 𝑣)
= [3(𝑔(𝑢, 𝑣))2 + 4𝑔(𝑢, 𝑣)]
[
]
cos 𝑢 cos 𝑣 − sin 𝑢 sin 𝑣
[
]
= [3 sin2 𝑢 cos2 𝑣 + 4 sin 𝑢 cos 𝑣] cos 𝑢 cos 𝑣 − sin 𝑢 sin 𝑣
[
]
(3 sin2 𝑢 cos2 𝑣 + 4 sin 𝑢 cos 𝑣) cos 𝑢 cos 𝑣 −(3 sin2 𝑢 cos2 𝑣 + 4 sin 𝑢 cos 𝑣) sin 𝑢 sin 𝑣 .
=
□
3
Example Let f : ℝ2 → ℝ2 be defined by
[
f (𝑥, 𝑦) =
𝑓1 (𝑥, 𝑦)
𝑓2 (𝑥, 𝑦)
]
[
=
𝑥2 𝑦
𝑥𝑦 2
]
,
and let g : ℝ → ℝ2 be defined by
g(𝑡) = ⟨𝑥(𝑡), 𝑦(𝑡)⟩ = ⟨cos 𝑡, sin 𝑡⟩.
Then f ∘ g is a vector-valued function of 𝑡,
f (𝑡) = ⟨cos2 𝑡 sin 𝑡, cos 𝑡 sin2 𝑡⟩.
To compute its derivative with respect to 𝑡, we first compute
[
]
] [
(𝑓1 )𝑥 (𝑓1 )𝑦
2𝑥𝑦 𝑥2
𝐽f (𝑥, 𝑦) =
,
=
(𝑓2 )𝑥 (𝑓2 )𝑦
𝑦 2 2𝑥𝑦
and g′ (𝑡) = ⟨− sin 𝑡, cos 𝑡⟩, and then use the Chain Rule to obtain
]
][ ′
[
𝑥 (𝑡)
(𝑓1 )𝑥 (𝑥(𝑡), 𝑦(𝑡)) (𝑓1 )𝑦 (𝑥(𝑡), 𝑦(𝑡))
′
′
(f ∘ g) (𝑡) = 𝐽f (𝑥(𝑡), 𝑦(𝑡))g (𝑡) =
𝑦 ′ (𝑡)
(𝑓2 )𝑥 (𝑥(𝑡), 𝑦(𝑡)) (𝑓2 )𝑦 (𝑥(𝑡), 𝑦(𝑡))
][
]
[
− sin 𝑡
2𝑥(𝑡)𝑦(𝑡)
𝑥(𝑡)2
=
𝑦(𝑡)2
2𝑥(𝑡)𝑦(𝑡)
cos 𝑡
= ⟨2 cos 𝑡 sin 𝑡(− sin 𝑡) + cos2 𝑡(cos 𝑡), sin2 𝑡(− sin 𝑡) + 2 cos 𝑡 sin 𝑡(cos 𝑡)⟩
= ⟨−2 cos 𝑡 sin2 𝑡 + cos3 𝑡, − sin3 𝑡 + 2 cos2 𝑡 sin 𝑡⟩.
□
The Implicit Function Theorem
The Chain Rule can also be used to compute partial derivatives of implicitly defined functions in
a more convenient way than is provided by implicit differentiation. Let the equation 𝐹 (𝑥, 𝑦) = 0
implicitly define 𝑦 as a differentiable function of 𝑥. That is, 𝑦 = 𝑓 (𝑥) where 𝐹 (𝑥, 𝑓 (𝑥)) = 0 for 𝑥 in
the domain of 𝑓 . If 𝐹 is differentiable, then, by the Chain Rule, we can differentiate the equation
𝐹 (𝑥, 𝑦(𝑥)) = 0 with respect to 𝑥 and obtain
𝐹𝑥 + 𝐹𝑦
𝑑𝑦
= 0,
𝑑𝑥
which yields
𝑑𝑦
𝐹𝑥
=− .
𝑑𝑥
𝐹𝑦
4
By the Implicit Function Theorem, the equation 𝐹 (𝑥, 𝑦) = 0 defines 𝑦 implicitly as a function of 𝑥
near (𝑥0 , 𝑦0 ), where 𝐹 (𝑥0 , 𝑦0 ) = 0, provided that 𝐹𝑦 (𝑥0 , 𝑦0 ) ∕= 0 and 𝐹𝑥 and 𝐹𝑦 are continuous near
(𝑥0 , 𝑦0 ). Under these conditions, we see that 𝑑𝑦/𝑑𝑥 is defined at (𝑥0 , 𝑦0 ) as well.
Example Let 𝐹 : ℝ2 → ℝ2 be defined by
𝐹 (𝑥, 𝑦) = 𝑥2 + 𝑦 2 − 4.
The equation 𝐹 (𝑥, 𝑦) = 0 defines 𝑦 implicitly as a function of 𝑥, provided that 𝐹 satisfies the
conditions of the Implicit Function Theorem.
We have
𝐹𝑥 = 2𝑥, 𝐹𝑦 = 2𝑦.
Since both of these partial derivatives are polynomials, and therefore are continuous on all of ℝ2 ,
it follows that if 𝐹𝑦 ∕= 0, then 𝑦 can be implicitly defined as a function of 𝑥 at a point where
𝐹 (𝑥, 𝑦, 𝑧) = 0, and
𝑑𝑦
𝐹𝑥
𝑥
=−
=− .
𝑑𝑥
𝐹𝑦
𝑦
For example, at the point (𝑥, 𝑦) = (0, 2), 𝐹 (𝑥, 𝑦) = 0, and 𝐹𝑦 = 4. Therefore, 𝑦 can be implicitly
defined as a function of 𝑥 near this point, and at 𝑥 = 0, we have 𝑑𝑦/𝑑𝑥 = 0. □
(0)
(0)
(0)
More generally, let 𝐹 : 𝐷 ⊆ ℝ𝑛+1 → ℝ, and let p0 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , 𝑦 (0) ) ∈ 𝐷 be such that
(0) (0)
(0)
𝐹 (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , 𝑦 (0) ) = 0. In this case, the Implicit Function Theorem states that if 𝐹𝑦 ∕= 0
near p0 , and all first partial derivatives of 𝐹 are continuous near p0 , then this equation defines 𝑦
as a function of 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , and
∂𝑦
𝐹𝑥
= − 𝑖,
∂𝑥𝑖
𝐹𝑦
(0)
𝑖 = 1, 2, . . . , 𝑛.
(0)
(0)
To see this, we differentiate the equation 𝐹 (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , 𝑦 (0) ) = 0 with respect to 𝑥𝑖 to obtain
the equation
∂𝑦
𝐹𝑥𝑖 + 𝐹𝑦
= 0,
∂𝑥𝑖
where all partial derivatives are evaluated at p0 , and solve for ∂𝑦/∂𝑥𝑖 at p0 .
Example Let 𝐹 : ℝ3 → ℝ be defined by
𝐹 (𝑥, 𝑦, 𝑧) = 𝑥2 𝑧 + 𝑧 2 𝑦 − 2𝑥𝑦𝑧 + 1.
The equation 𝐹 (𝑥, 𝑦, 𝑧) = 0 defines 𝑧 implicitly as a function of 𝑥 and 𝑦, provided that 𝐹 satisfies
the conditions of the Implicit Function Theorem.
We have
𝐹𝑥 = 2𝑥𝑧 − 2𝑦𝑧, 𝐹𝑦 = 2𝑦𝑧 − 2𝑥𝑧, 𝐹𝑧 = 𝑥2 + 2𝑦𝑧 − 2𝑥𝑦.
5
Since all of these partial derivatives are polynomials, and therefore are continuous on all of ℝ3 , it
follows that if 𝐹𝑧 ∕= 0, then 𝑧 can be implicitly defined as a function of 𝑥 and 𝑦 at a point where
𝐹 (𝑥, 𝑦, 𝑧) = 0, and
𝑧𝑥 = −
𝐹𝑥
2𝑦𝑧 − 2𝑥𝑧
= 2
,
𝐹𝑧
𝑥 + 2𝑦𝑧 − 2𝑥𝑦
𝑧𝑦 = −
𝐹𝑦
2𝑥𝑧 − 2𝑦𝑧
= 2
.
𝐹𝑧
𝑥 + 2𝑦𝑧 − 2𝑥𝑦
For example, at the point (𝑥, 𝑦, 𝑧) = (1, 0, −1), 𝐹 (𝑥, 𝑦, 𝑧) = 0, and 𝐹𝑧 = 1. Therefore, 𝑧 can be
implicitly defined as a function of 𝑥 and 𝑦 near this point, and at (𝑥, 𝑦) = (1, 0), we have 𝑧𝑥 = 2
and 𝑧𝑦 = −2. □
We now consider the most general case: let F : 𝐷 ⊆ ℝ𝑛+𝑚 → ℝ𝑚 , and let
(0)
(0)
(0)
(0)
(0)
p0 = (𝑥1 , 𝑥2 , . . . , 𝑥(0)
𝑛 , 𝑦1 , 𝑦2 , . . . , 𝑦𝑚 ) ∈ 𝐷
be such that
(0)
(0)
(0)
(0)
(0)
F(𝑥1 , 𝑥2 , . . . , 𝑥(0)
𝑛 , 𝑦1 , 𝑦2 , . . . , 𝑦𝑚 ) = 0.
If we differentiate this system of equations with respect to 𝑥𝑖 , we obtain the systems of linear
equations
∂𝑦𝑚
∂𝑦1
∂𝑦2
F 𝑥𝑖 + F 𝑦1
+ F 𝑦2
+ ⋅ ⋅ ⋅ + F 𝑦𝑚
= 0, 𝑖 = 1, 2, . . . , 𝑛,
∂𝑥𝑖
∂𝑥𝑖
∂𝑥𝑖
where all partial derivatives are evaluated at p0 .
(0) (0)
(0)
To examine the solvability of these systems of equations, we first define x0 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ),
and denote the component functions of the vector-valued function F by F = ⟨𝐹1 , 𝐹2 , . . . , 𝐹𝑚 ⟩. We
then define the Jacobian matrices
⎤
⎡
∂𝐹1
∂𝐹1
∂𝐹1
(p
)
(p
)
(p
)
0
0
0
∂𝑥1
∂𝑥2
∂𝑥𝑛
⎥
⎢ ∂𝐹
∂𝐹2
2
(p
)
⎢ ∂𝑥12 (p0 ) ∂𝐹
0
∂𝑥
∂𝑥𝑛 (p0 ) ⎥
2
𝐽x,F (p0 ) = ⎢
⎥,
⎦
⎣
∂𝐹𝑚
∂𝐹𝑚
∂𝐹𝑚
∂𝑥1 (p0 )
∂𝑥2 (p0 )
∂𝑥𝑛 (p0 )
⎡
⎢
⎢
𝐽y,F (p0 ) = ⎢
⎣
and
⎡
⎢
⎢
𝐽y (x0 ) = ⎢
⎣
∂𝑦1
∂𝑦1 (p0 )
∂𝑦2
∂𝑦1 (p0 )
∂𝐹1
∂𝑦2 (p0 )
∂𝐹2
∂𝑦2 (p0 )
∂𝐹1
∂𝑦𝑚 (p0 )
∂𝐹2
∂𝑦𝑚 (p0 )
∂𝑦𝑚
∂𝑦1 (p0 )
∂𝐹𝑚
∂𝑦2 (p0 )
∂𝐹𝑚
∂𝑦𝑚 (p0 )
∂𝑦1
∂𝑥1 (x0 )
∂𝑦2
∂𝑥1 (x0 )
∂𝑦1
∂𝑥2 (x0 )
∂𝑦2
∂𝑥2 (x0 )
∂𝑦1
∂𝑥𝑛 (x0 )
∂𝑦1
∂𝑥𝑛 (x0 )
∂𝑦𝑚
∂𝑥1 (x0 )
∂𝑦𝑚
∂𝑥2 (x0 )
∂𝑦1
∂𝑥𝑛 (x0 )
6
⎤
⎥
⎥
⎥,
⎦
⎤
⎥
⎥
⎥.
⎦
Then, from our previous differentiation with respect to 𝑥𝑖 , for each 𝑖 = 1, 2, . . . , 𝑛, we can concisely
express our systems of equations as a single system
𝐽x,F (p0 ) + 𝐽y,F (p0 )𝐽y (x0 ) = 0.
If the matrix 𝐽y,F (p0 ) is invertible (also nonsingular), which is the case if and only if its determinant is nonzero, and if all first partial derivatives of F are continuous near p0 , then the equation
F(p) = 0 implicitly defines 𝑦1 , 𝑦2 , . . . , 𝑦𝑚 as a function of 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , and
𝐽y (x0 ) = −[𝐽y,F (p0 )]−1 𝐽x,F (p0 ),
where [𝐽y,F (p0 )]−1 is the inverse of the matrix 𝐽y,F (p0 ).
Example Let F : ℝ4 → ℝ2 by defined by
[
] [
]
𝐹1 (𝑥, 𝑦, 𝑢, 𝑣)
𝑥𝑢 + 𝑦 2 𝑣
F(𝑥, 𝑦, 𝑠, 𝑡) =
=
.
𝐹2 (𝑥, 𝑦, 𝑢, 𝑣)
𝑥2 𝑣 + 𝑦𝑢 + 1
Then the vector equation F(𝑥, 𝑦, 𝑢, 𝑣) = 0 implicitly defines (𝑢, 𝑣) as a function of (𝑥, 𝑦), provided
that F satisifes the conditions of the Implicit Function Theorem. We will compute the partial
derivatives of 𝑢 and 𝑣 with respect to 𝑥 and 𝑦, at a point that satisfies this equation.
We have
[
] [
]
∂𝐹1
∂𝐹1
𝑢 2𝑦𝑣
∂𝑥
∂𝑦
𝐽(𝑥,𝑦),F (𝑥, 𝑦, 𝑢, 𝑣) = ∂𝐹2 ∂𝐹2 =
,
2𝑥𝑣 𝑢
∂𝑥
∂𝑦
[ ∂𝐹1 ∂𝐹1 ] [
]
𝑥 𝑦2
∂𝑢
∂𝑣
𝐽(𝑢,𝑣),F (𝑥, 𝑦, 𝑢, 𝑣) = ∂𝐹2 ∂𝐹2 =
.
𝑦 𝑥2
∂𝑢
∂𝑣
From the formula for the inverse of a 2 × 2 matrix,
]−1
[
]
[
1
𝑎 𝑏
𝑑 −𝑏
=
,
𝑐 𝑑
𝑎𝑑 − 𝑏𝑐 −𝑐 𝑎
we obtain
[
𝐽(𝑢,𝑣) (𝑥, 𝑦) =
𝑢𝑥 𝑢𝑦
𝑣𝑥 𝑣𝑦
]
= −[𝐽(𝑢,𝑣),F (𝑥, 𝑦, 𝑢, 𝑣)]−1 𝐽(𝑥,𝑦),F (𝑥, 𝑦, 𝑢, 𝑣)
[ 2
][
]
1
𝑥 −𝑦 2
𝑢 2𝑦𝑣
= − 3
𝑥
2𝑥𝑣 𝑢
𝑥 − 𝑦 3 −𝑦
[ 2
]
1
𝑥 𝑢 − 2𝑥𝑦 2 𝑣 2𝑥2 𝑦𝑣 − 𝑦 2 𝑢
=
.
2𝑥2 𝑣 − 𝑦𝑢
𝑥𝑢 − 2𝑦 2 𝑣
𝑦 3 − 𝑥3
These partial derivatives can then be evaluated at any point (𝑥, 𝑦, 𝑢, 𝑣) such that F(𝑥, 𝑦, 𝑢, 𝑣) = 0,
such as (𝑥, 𝑦, 𝑢, 𝑣) = (0, 1, 0, −1). Note that the matrix 𝐽(𝑢,𝑣),F (𝑥, 𝑦, 𝑢, 𝑣) is not invertible (that is,
singular) if its determinant 𝑥3 − 𝑦 3 = 0; that is, if 𝑥 = 𝑦. When this is the case, (𝑢, 𝑣) can not be
implicitly defined as a function of (𝑥, 𝑦). □
7
Practice Problems
1. For each of the following pairs of functions f and g, use the Chain Rule to compute all of the
first partial derivatives of f ∘ g with respect to the independent variables of g.
(a)
𝑓 (𝑥, 𝑦, 𝑧) =
√
𝑥2 + 𝑦 2 + 𝑧 2 ,
g(𝑡) = ⟨𝑒𝑡 cos 𝑡, 𝑒𝑡 sin 𝑡, 𝑡𝑒𝑡 ⟩
(b)
𝑓 (𝑥, 𝑦) = ln
√
𝑥2 + 𝑦 2 ,
g(𝑢, 𝑣) = ⟨𝑢 cos 𝑣, 𝑢 sin 𝑣⟩
(c)
f (𝑥, 𝑦) = ⟨𝑥3 + 3𝑦 2 , 𝑦 3 − 3𝑥2 ⟩,
g(𝑡) = ⟨cos 𝑡, sin 𝑡⟩
(d)
f (𝑢, 𝑣) =
〈(
1+
(
𝑣
𝑢)
𝑣
𝑢)
𝑉
𝑢
cos
cos 𝑢, 1 + cos
sin 𝑢, sin
2
2
2
2
2
2
〉
,
g(𝑥, 𝑦) = ⟨𝑥2 𝑦, 𝑥𝑦 2 ⟩
2. Compute the indicated derivatives or partial derivatives of the functions that are implicitly
defined by the following equations. Then obtain the equation of the tangent line (or tangent
plane) to the curve (surface) at the indicated point.
√
(a) 𝐹 (𝑥, 𝑦) = (𝑥2 + 𝑦 2 )2 − 2(𝑥2 − 𝑦 2 ) = 0, 𝑑𝑦/𝑑𝑥, (𝑥0 , 𝑦0 ) = ( 3/2, 1/2)
√
(b) 𝐹 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦 2 + 𝑧 2 − 8 𝑥2 + 𝑦 2 + 15 = 0, ∂𝑧/∂𝑥, ∂𝑧/∂𝑦, (𝑥0 , 𝑦0 , 𝑧0 ) = (4, 0, 1)
3. (Just for fun) Describe the graphs of the curves or surfaces defined by the equations in
Problem 2.
Additional Practice Problems
Additional practice problems from the recommended textbooks are:
∙ Stewart: Section 11.5, Exercises 1-11 odd, 17, 19 odd
∙ Marsden/Tromba: Section 2.5, Exercises 5, 9, 11; Section 3.5, Exercises 3, 5
8