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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
We’ve got general procedures for differentiating expressions with addition, subtraction, and multi­ plication. What about composition?
Example 1. y = f (x) = sin x, x = g(t) = t2 .
dy
, write
So, y = f (g(t)) = sin(t2 ). To find
dt
t0 = t0
x0 = g(t0 )
y0 = f (x0 )
t = t0 + Δt
x = x0 + Δx
y = y0 + Δy
Δy
Δy Δx
=
·
Δt
Δx Δt
As Δt → 0, Δx → 0 too, because of continuity. So we get:
dy
dy dx
← The Chain Rule!
=
dt
dx dt
In the example,
dx
dy
= 2t and
= cos x.
dt
dx
So,
�
d �
sin(t2 )
dt
dy dx
)( )
dx dt
(cos x)(2t)
�
�
(2t) cos(t2 )
=
(
=
=
Another notation for the chain rule
d
f (g(t)) = f � (g(t))g � (t)
dt
Example 1. (continued)
�
or
d
f (g(x)) = f � (g(x))g � (x)
dx
Composition of functions f (x) = sin x and g(x) = x2
(f ◦ g)(x)
=
f (g(x))
=
sin(x2 )
(g ◦ f )(x)
=
g(f (x))
=
sin2 (x)
f ◦g
�=
g ◦ f.
Not Commutative!
Note:
�
1
Lecture 4
Sept. 14, 2006
x
g(x)
g
18.01 Fall 2006
f
f(g(x))
Figure 1: Composition of functions: f ◦ g(x) = f (g(x))
Example 2.
Let u =
d
cos
dx
� �
1
=?
x
1
x
dy
dx
dy
du
dy
dx
Example 3.
=
dy du
du dx
=
− sin(u);
=
du
1
= − 2
dx
x
� �
1
�
�
sin
sin(u)
−1
x
= (− sin u)
=
x2
x2
x2
d � −n �
x
=?
dx
There are two ways to proceed. x−n =
� �n
1
1
, or x−n = n
x
x
� �n
� �n−1 �
�
1
1
−1
= n
= −nx−(n−1) x−2 = −nx−n−1
x
x
x2
� �
�
�
d � −n �
d
1
−1
n−1
2.
x
=
=
nx
= −nx−n−1 (Think of xn as u)
dx
dx xn
x2n
d � −n �
d
1.
x
=
dx
dx
2
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)
D2 f
d2 f
dx2
f ��� (x)
D3 f
d3 f
dx3
f (n) (x)
Dn f
dn f
dxn
Higher derivatives are pretty straightforward —- just keep taking the derivative!
Example. Dn xn = ?
Start small and look for a pattern.
Dx
=
1
2 2
D x
=
D(2x) = 2
D 3 x3
=
D2 (3x2 ) = D(6x) = 6
=
D (4x ) = D (12x ) = D(24x) = 24
n n
=
n! ← we guess, based on the pattern we’re seeing here.
D x
3
2
( = 1 · 2 · 3)
4 4
D x
3
( = 1 · 2)
2
( = 1 · 2 · 3 · 4)
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 xn = n! (nth case). Show it holds for the (n + 1)st case.
�
�
Dn+1 xn+1 = Dn Dxn+1 = Dn ((n + 1)xn ) = (n + 1)Dn xn = (n + 1)(n!)
Dn+1 xn+1
=
(n + 1)!
Proved!
3