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18.01 Single Variable Calculus
Fall 2006
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Lecture 4
Sept. 14, 2006
18.01 Fall 2006
__________________________________________________________________________
Lecture 4
Chain Rule, and Higher Derivatives
Chain Rule
(𝑔 ° 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = sin2 (𝑥)
We’ve got general procedures for differentiating
expressions with addition, subtraction, and
multiplication. What about composition?
Note: 𝑓 ° 𝑔 ≠ 𝑔 ° 𝑓. Not Commutative!
Example 1. y = f(x) = sin x, x = g(t) = t
𝑑𝑦
So, y = f(g(t)) = sin(t2). To find, 𝑑𝑡 , write:
𝑡 = 𝑡0
𝑡 = 𝑡0 + ∆𝑡
𝑥 = 𝑥0
𝑥 = 𝑥0 + ∆𝑥
𝑦 = 𝑓(𝑥0 )
𝑦 = 𝑦0 + ∆𝑦
Figure 1: Composition of functions: 𝑓 ° g(x) =
f(g(x))
Δ𝑦
Δ𝑦 Δ𝑥
=
.
Δ𝑡
Δ𝑥 Δ𝑡
𝑑
1
As Δt → 0, Δx → 0 too, because of continuity.
So we get:
𝑑𝑦
𝑑𝑡
=
𝑑𝑦
𝑑𝑥
In the example,
So,
𝑑𝑥
. 𝑑𝑡
𝑑𝑥
𝑑𝑡
Let u = 𝑥
← The Chain Rule!
= 2𝑡, and
𝑑
(sin(𝑡 2 ))
𝑑𝑡
𝑑𝑦
𝑑𝑥
=
1
Example 2. 𝑑𝑥 cos (𝑥) = ?
= cos 𝑥.
𝑑𝑦
𝑑𝑥
𝑑𝑥
. 𝑑𝑡
𝑑𝑦
𝑑𝑥
=
𝑑𝑦 𝑑𝑢
𝑑𝑢 𝑑𝑥
𝑑𝑦
𝑑𝑢
= −𝑠𝑖𝑛(𝑢);
𝑑𝑦
𝑑𝑥
=
𝑠𝑖𝑛 (𝑢)
𝑥2
𝑑𝑢
𝑑𝑥
1
= − 𝑥2
−1
= (−𝑠𝑖𝑛 𝑢) ( 𝑥 2 ) =
1
𝑥
𝑠𝑖𝑛 ( )
𝑥2
= (cos 𝑥)(2𝑡)
= (2𝑡)(cos(𝑥 2 ))
Example 3.
𝑑
. (𝑥 −𝑛 )
𝑑𝑥
=?
Another notation for the chain rule
There are two ways to proceed.
𝑑
𝑑𝑡
𝑓(𝑔(𝑡)) = 𝑓 ′ (𝑔(𝑡))𝑔′(𝑡)
(or
𝑑𝑦
𝑑𝑥
1 𝑛
𝑥 −𝑛 = (𝑥) , or 𝑥 −𝑛 =
𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥))
1.
𝑑
𝑑 1 𝑛
1 𝑛−1 −1
(𝑥 −𝑛 ) = ( ) = 𝑛 ( )
( 𝑥2 )
𝑑𝑥
𝑑𝑥 𝑥
𝑥
−(𝑛−1) −2
−𝑛−1
−𝑛𝑥
Example 1. (continued)
Composition of
functions 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = 𝑥 2
(𝑓 ° 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = sin 𝑥
2.
1
𝑥
𝑑
(𝑥 −𝑛 )
𝑑𝑥
−(𝑛−1)
−𝑛𝑥
2
1
𝑥𝑛
= −𝑛𝑥
𝑑
1
−1
= 𝑑𝑥 (𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 (𝑥 2𝑛 ) =
(Think of 𝑥 𝑛 as u)
=
Lecture 4
Sept. 14, 2006
18.01 Fall 2006
__________________________________________________________________________
Start small and look for a pattern.
Higher Derivatives
𝐷𝑥 = 1
𝐷 𝑥 = 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)
derivatives are derivatives of derivatives. For
instance, if 𝑔 = 𝑓′, then ℎ = 𝑔′ is the second
derivative of f. We wrilte ℎ = (𝑓 ′ )′ = 𝑓′′.
2 2
Notations
𝑓 ′ (𝑥)
𝐷𝑓
𝑑𝑓
𝑑𝑥
The notation n! is called “n factorial” and
defined by n! = n(n − 1) · · · 2 · 1
𝑓′′(𝑥)
𝐷2𝑓
𝑑2 𝑓
𝑑𝑥 2
Proof by Induction: We’ve already checked the
𝑓′′′(𝑥)
𝐷3𝑓
𝑑3 𝑓
𝑑𝑥 3
base case (n = 1).
𝑓 𝑛 (𝑥)
𝐷𝑛𝑓
𝑑𝑛 𝑓
𝑑𝑥 𝑛
Induction step: Suppose we know 𝐷 𝑛 𝑥 𝑛 = 𝑛!
case. Show it holds for the (𝑛 + 1)𝑠𝑡 case.
𝐷 𝑛+1 𝑥 𝑛+1 = 𝐷 𝑛 (𝐷𝑥 𝑛+1 ) =
𝐷 ((𝑛 + 1)𝑥 𝑛 ) = (𝑛 + 1)𝐷 𝑛 𝑥 𝑛 = (𝑛 + 1)(𝑛!)
𝑛
Higher derivatives are pretty straightforward —-
𝐷 𝑛+1 𝑥 𝑛+1 = (𝑛 + 1)!
just keep taking the derivative!
Proved!
Example. 𝐷 𝑛 𝑥 𝑛 = ?
2