Advanced Calculus (II)
WEN-CHINGLIEN
Department of Mathematics National Cheng Kung University
2009
WEN-CHINGLIEN Advanced Calculus (II)
Ch11: Differentiability on R
11.4: Chain Rule
Theorem (11.28 Chain Rule)
Suppose thata ∈ Rn, that g is a vector function from n variables to m variables, and that f is a vector function from m variables to p variables. If g is differentiable ata and f is differentiable at g(a), then f ◦ g is differentiable at a and
(20) D(f ◦ g)(a) = Df (g(a))Dg(a).
(The product Df (g(a))Dg(a) is a matrix multiplication.)
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g.By the
uniqueness of the total derivative, we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the
uniqueness of the total derivative,we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g.By the
uniqueness of the total derivative, we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the
uniqueness of the total derivative,we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the
uniqueness of the total derivative, we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the
uniqueness of the total derivative, we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the
uniqueness of the total derivative, we must show that
(21) lim
h→0
f (g(a + h)) − f (g(a)) − T (h)
khk =0.
Letb = g(a). Set
(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and
(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn,and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a).Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm.Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a))=f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a).Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b)=Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a))=f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b)=Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k)=T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k)=T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h)=Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h)=Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and
δ(k)
kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply
f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)
=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have
f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)
=:T1(h) + T2(h).
It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.
WEN-CHINGLIEN Advanced Calculus (II)
Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication,it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm.we have
kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k
≤ kDg(a)k · khk + kε(h)k.
WEN-CHINGLIEN Advanced Calculus (II)
Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0.On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have
kkk := kg(a + h) − g(a)k= kDg(a)(h) + ε(h)k
≤ kDg(a)k · khk + kε(h)k.
WEN-CHINGLIEN Advanced Calculus (II)
Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm.we have
kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k
≤ kDg(a)k · khk + kε(h)k.
WEN-CHINGLIEN Advanced Calculus (II)
Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have
kkk := kg(a + h) − g(a)k= kDg(a)(h) + ε(h)k
≤ kDg(a)k · khk + kε(h)k.
WEN-CHINGLIEN Advanced Calculus (II)
Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have
kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k
≤ kDg(a)k · khk + kε(h)k.
WEN-CHINGLIEN Advanced Calculus (II)
Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have
kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k
≤ kDg(a)k · khk + kε(h)k.
WEN-CHINGLIEN Advanced Calculus (II)
Thus kkkkhk is bounded forh sufficiently small.Sincek → 0 inRm ash → 0 in Rn, it follows that
kT2(h)k
khk = kkk
khk· kδ(k)k kkk → 0
ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).
WEN-CHINGLIEN Advanced Calculus (II)
Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn,it follows that
kT2(h)k
khk = kkk
khk· kδ(k)k kkk → 0
ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).
WEN-CHINGLIEN Advanced Calculus (II)
Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn, it follows that
kT2(h)k
khk = kkk
khk· kδ(k)k kkk → 0
ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).
WEN-CHINGLIEN Advanced Calculus (II)
Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn, it follows that
kT2(h)k
khk = kkk
khk· kδ(k)k kkk → 0
ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).
WEN-CHINGLIEN Advanced Calculus (II)
Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn, it follows that
kT2(h)k
khk = kkk
khk· kδ(k)k kkk → 0
ash → 0. We conclude that f ◦ g is differentiable at a and the differentiable ata and the derivative is Df (g(a))Dg(a).
WEN-CHINGLIEN Advanced Calculus (II)
Let f :R2→ R be C2 onR2and set u(r , θ)
=f (r cos θ, r sin θ). If f satisfies the Laplace equation, i.e., if
∂2f
∂x2 + ∂2f
∂y2 =0 prove for each r 6= 0 that
1 r2
∂2u
∂θ2 +1 r
∂u
∂r + ∂2u
∂r2 =0.
WEN-CHINGLIEN Advanced Calculus (II)
Thank you.
WEN-CHINGLIEN Advanced Calculus (II)
