LECTURE 4
CHAIN RULE, AND HIGHER DERIVATIVES
CHAIN RULE
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MIT OpenCourseWare
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18.01 Single Variable Calculus
Fall 2006
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We’ve got general procedures for differentiating expressions with
addition, subtraction, and multiplication. What about composition?
Example 1. y = f(x) = sin x, x = g(t) = t
𝑑𝑦
So, y = f(g(t)) = sin(t2). To find,
, write:
𝑑𝑡
𝑡 = 𝑡0
𝑡 = 𝑡0 + ∆𝑡
𝑥 = 𝑥0
𝑥 = 𝑥0 + ∆𝑥
𝑦 = 𝑓(𝑥0 )
𝑦 = 𝑦0 + ∆𝑦
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Δ𝑦
Δ𝑦 Δ𝑥
=
.
Δ𝑡
Δ𝑥 Δ𝑡
As Δt → 0, Δx → 0 too, because of continuity.
So we get:
𝑑𝑦
𝑑𝑡
=
𝑑𝑦
𝑑𝑥
𝑑𝑥
.
𝑑𝑡
← The Chain Rule!
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
In the
𝑑𝑥
𝑑𝑡
𝑑𝑦
example,
= 2𝑡, and =
𝑑𝑥
𝑑
𝑑𝑦 𝑑𝑥
2
So, (sin(𝑡 )) =
.
𝑑𝑡
𝑑𝑥 𝑑𝑡
cos 𝑥.
= (cos 𝑥)(2𝑡)
= (2𝑡) cos(𝑥 2 )
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ANOTHER NOTATION FOR THE CHAIN RULE
𝑑
𝑓
𝑑𝑡
𝑔 𝑡
′
= 𝑓 𝑔 𝑡 𝑔′(𝑡)
(or
𝑑𝑦
𝑓
𝑑𝑥
𝑔 𝑥
= 𝑓 ′ 𝑔 𝑥 𝑔′ 𝑥 )
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EXAMPLE 1. (CONTINUED)
Composition of functions 𝑓(𝑥) = sin 𝑥 and
𝑔 𝑥 = 𝑥2
𝑓 ° 𝑔 𝑥 = 𝑓 𝑔 𝑥 = sin 𝑥 2
𝑔 ° 𝑓 𝑥 = 𝑔 𝑓 𝑥 = sin2 (𝑥)
Note: 𝑓 ° 𝑔 ≠ 𝑔 ° 𝑓. Not Commutative!
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COMPOSITION OF FUNCTIONS:
𝑓 ° G(X) = F(G(X))
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EXAMPLE 2.
Let u =
𝑑
1
COS
𝑑𝑥
𝑥
=?
1
𝑥
𝑑𝑦
𝑑𝑦 𝑑𝑢
=
𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑑𝑦
𝑑𝑢
1
= −𝑠𝑖𝑛 𝑢 ;
= − 2
𝑑𝑢
𝑑𝑥
𝑥
1
𝑠𝑖𝑛
𝑑𝑦
𝑠𝑖𝑛 (𝑢)
−1
𝑥
=
= (−𝑠𝑖𝑛 𝑢)
=
2
2
𝑑𝑥
𝑥
𝑥
𝑥2
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EXAMPLE 3.
𝑑
.
𝑑𝑥
𝑥 −𝑛 = ?
There are two ways to proceed. 𝑥 −𝑛 =
𝑥
−𝑛
=
1
𝑥𝑛
𝑑

𝑑𝑥
𝑥 −𝑛 =
−𝑛𝑥 − 𝑛−1 𝑥
𝑑

𝑑𝑥
−𝑛
𝑥
−𝑛𝑥 −
=
𝑛−1
𝑑 1 𝑛
𝑑𝑥 𝑥
−2
1 𝑛−1 −1
=𝑛
𝑥
𝑥2
−𝑛−1
1 𝑛
,
𝑥
or
=
= −𝑛𝑥
𝑑
1
𝑑𝑥 𝑥 𝑛
𝑛−1
= 𝑛𝑥
(Think of 𝑥 𝑛 as u)
−1
𝑥 2𝑛
=
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HIGHER DERIVATIVES
Higher derivatives are derivatives of
derivatives. For instance, if 𝑔 = 𝑓′, then ℎ =
𝑔′ is the second derivative of f. We write ℎ =
𝑓 ′ ′ = 𝑓′′.
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Notations
Higher derivatives are pretty straightforward —just
keep taking the derivative!
𝑓′ 𝑥
𝐷𝑓
𝑑𝑓
𝑑𝑥
𝑓′′(𝑥)
𝐷2 𝑓
𝑑2 𝑓
𝑑𝑥 2
𝑓′′′(𝑥)
𝐷3 𝑓
𝑑3 𝑓
𝑑𝑥 3
𝑓 𝑛 (𝑥)
𝐷𝑛 𝑓
𝑑𝑛 𝑓
𝑑𝑥 𝑛
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EXAMPLE. 𝐷 𝑛 𝑥 𝑛 = ?
Start small and look for a pattern.
𝐷𝑥 = 1
𝐷 2 𝑥 2 = 𝐷 2𝑥 = 2 ( = 1 ∙ 2)
𝐷 3 𝑥 3 = 𝐷 2 3𝑥 2 = 𝐷 6𝑥 = 6 ( = 1 ∙
2 ∙ 3)
𝐷 4 𝑥 4 = 𝐷 3 4𝑥 3 = 𝐷 2 12𝑥 2 = 𝐷 24𝑥 =
24 ( = 1 ∙ 2 ∙ 3 ∙ 4)
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The notation n! is called “n factorial” and defined
by n! = n(n − 1) · · · 2 · 1
Proof by Induction: We’ve already checked the
base case (n = 1).
Induction step: Suppose we know 𝐷𝑛 𝑥 𝑛 = 𝑛!
case. Show it holds for the (𝑛 + 1)𝑠𝑡 case.
𝐷𝑛+1 𝑥 𝑛+1 = 𝐷𝑛 (𝐷𝑥 𝑛+1 ) = 𝐷𝑛 (𝑛 +
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Thank You!
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