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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
ChainRule, and Higher Derivatives
Chain Rule
We’ve got general procedures for differentiating expressions with addition, subtraction, and multiplication. What about composition?
Example 1. 𝑦 = 𝑓(𝑥) = 𝑠𝑖𝑛𝑥, 𝑥 = 𝑔(𝑡) = 𝑡 2 .
So, 𝑦 = 𝑓(𝑔(𝑡)) = sin⁡(𝑡 2 ). To find
𝑑𝑦
,
𝑑𝑡
write
𝑡0 = 𝑡0
𝑥0 = 𝑔(𝑡0 )
𝑥0 = 𝑓(𝑥0 )
𝑡 = 𝑡0 + ∆𝑡
𝑥 = 𝑥0 + ∆𝑥
𝑦 = 𝑦0 + ∆𝑦
Δ𝑦 Δ𝑦 Δ𝑥
=
∙
Δ𝑡 Δ𝑥 Δ𝑡
As ∆𝑡 → 0, ∆𝑥 → 0 too, because of continuity. So we get:
𝑑𝑦 𝑑𝑦 𝑑𝑥
=
⁡ ← 𝐓𝐡𝐞⁡𝐂𝐡𝐚𝐢𝐧⁡𝐑𝐮𝐥𝐞!
𝑑𝑡 𝑑𝑥 𝑑𝑡
𝑑𝑥
𝑑𝑦
In the example, 𝑑𝑡 = 2𝑡⁡and⁡ 𝑑𝑥 = cos 𝑥.
So,
𝑑
(sin(𝑡 2 ))
𝑑𝑡
𝑑𝑦
𝑑𝑥
𝑑𝑥
𝑑𝑡
= ( )( )
= (cos 𝑥)(2𝑡)
= (2𝑡)(cos(𝑡 2 ))
Another notation for the chain rule
𝑑
𝑓(𝑔(𝑡))
𝑑𝑡
Example 1. (continued)
= 𝑓 ′ (𝑔(𝑡))𝑔′(𝑡)
𝑑
( or 𝑑𝑥 𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥))𝑔′(𝑥) )
Composition of functions . 𝑦 = 𝑠𝑖𝑛𝑥⁡and⁡𝑔(𝑥) = 𝑥 2
(𝑓⁡⁡⃘𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = sin⁡(𝑥 2 )
(𝑔⁡⁡⃘𝑓 )(𝑥) = 𝑔(𝑓(𝑥)) = ⁡𝑠𝑖𝑛2 (𝑥)
Note:
𝑓⁡⁡⃘𝑔⁡ ≠ 𝑔⁡⁡⃘𝑓.⁡⁡⁡⁡⁡⁡⁡Not Commutative!
1
Lecture 4
Sept. 14, 2006
18.01 Fall 2006
Figure 1: Composition of functions: (f ⃘g)(x)=f(g(x))
Example 2.
𝒅
𝟏
𝐜𝐨𝐬 (𝒙)
𝒅𝒙
=?
1
Let 𝑢 = 𝑥
𝑑𝑦
𝑑𝑦 𝑑𝑢
⁡⁡ = ⁡⁡
𝑑𝑥
𝑑𝑢 𝑑𝑥
Example 3.
𝑑
⁡(𝑥 −𝑛 )
𝑑𝑥
𝑑𝑦
⁡=
𝑑𝑢
− sin ⁡(𝑢) ;
𝑑𝑦
𝑑𝑥
sin⁡(𝑢)
𝑥2
=
𝑑𝑦
𝑑𝑥
−1
= (− sin 𝑢)⁡( 𝑥 2 ) =
=?
1 𝑛
1
There are two ways to proceed. 𝑥 −𝑛 = (𝑥) , or⁡𝑥 −𝑛 = 𝑥 𝑛
1.
2.
𝑑
⁡(𝑥 −𝑛 )
𝑑𝑥
𝑑
⁡(𝑥 −𝑛 )
𝑑𝑥
𝑑
1 𝑛
1 𝑛−1 −1
( 𝑥 2 ) = −𝑛𝑥 −(𝑛−1) 𝑥 −2 = −𝑛𝑥 −𝑛−1
−1
𝑛𝑥 𝑛−1 ∙ (𝑥 2𝑛 ) = −𝑛𝑥 −𝑛−1 ⁡(Think⁡of⁡𝑥 𝑛 ⁡as⁡𝑢)
= 𝑑𝑥 (𝑥) = 𝑛 (𝑥)
𝑑
1
= 𝑑𝑥 (𝑥 𝑛 ) =
2
=−
1
𝑥
𝑥2
1
𝑥2
sin( )
Lecture 4
Sept. 14, 2006
18.01 Fall 2006
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 ′′ .
Notations
𝑓′(𝑥)
𝐷𝑓
𝑑𝑓
𝑑𝑥
𝑓′′(𝑥)
𝐷2𝑓
𝑑2 𝑓
𝑑𝑥 2
𝑓′′′(𝑥)
𝐷3𝑓
𝑓 (𝑛) (𝑥)
𝐷𝑛 𝑓
𝑑3 𝑓
𝑑𝑥 3
𝑑𝑛 𝑓
𝑑𝑥 𝑛
Higher derivatives are pretty straightforward ⇁ just keep taking the derivative!
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)
𝐷 𝑛 𝑥 𝑛 = 𝑛! ⁡ ← ⁡we⁡guess, based⁡on⁡the⁡pattern⁡we’re⁡seeing⁡here.
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 𝐷 𝑛 𝑥 𝑛 = 𝑛! (𝑛𝑡ℎ ⁡case).⁡Show it holds for the (𝑛 + 1)𝑠𝑡 case.
𝐷 𝑛+1 𝑥 𝑛+1 = 𝐷 𝑛 (𝐷𝑥 𝑛+1 ) = 𝐷 𝑛 ((𝑛 + 1)𝑥 𝑛 ) = (𝑛 + 1)𝐷 𝑛 𝑥 𝑛 = (𝑛 + 1)(𝑛!)
𝐷 𝑛+1 𝑥 𝑛+1 = (𝑛 + 1)!
Proved!
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