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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
ChainRule
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) = t2
So, y = f (g(t)) = sin(t2 ). To find dy
dt , write
t0 = t0
x0 = g(t0 )
y0 = f (x0 )
t = t0 + ∆t
x = x0 + ∆t
y = y0 + ∆y
∆y
∆y ∆x
=
·
∆t
∆x ∆t
As ∆t → 0, ∆x → 0 too, because of continuity. So we get:
dy
dy dx
=
← The Chain Rule!
dt
dx dt
In the example,
dx
dt
= 2t and
dy
dx
= cosx.
So,
dy dx
d
(sin(t2 )) = ( )( )
dt
dx dt
= (cosx)(2t)
= (2t)(cos(t2 ))
Another notation for the chain rule
d
f (g(t)) = f 0 (g(t))g 0 (t)
dt
!
d
0
0
or
f (f (x)) = f (g(x))g (x)
dx
Example 1. (continued) Composition of functions f (x) = sinx and g(x) = x2
(f ◦ g)(x) = f (g(x)) = sin(x2 )
(g ◦ f )(x) = g(f (x)) = sin2 (x)
Note: f ◦ g 6= g ◦ f. Not Commutative!
2
Lecture 4
Sept. 14, 2006
Figure 1: Composition of functions: f ◦ g(x) = f (g(x))
Example 2.
Let u =
d
1
cos( ) =?
dx
x
1
x
dy du
dy
=
dx
du dx
dy
= −sin(u);
du
du
−1
= 2
dx
x
1
sin( )
dy
sin(u)
−1
x
=
= (−sin(u))( 2 ) =
dx
x2
x
x2
Example 3.
d −n
(x ) =?
dx
1
1
There are two ways to proceed. x−n = ( )n , or x−n = n
x
x
d
d 1 n
1
−1
1. (x−n ) =
( ) = n( )n−1 ( 2 ) = −nx−(n−1) x−2 = −nx−n−1
dx
dx x
x
x
1.
d −n
d 1
−1
(x ) =
( n ) = nxn−1 ( 2n ) = −nx−n−1 (Think of xn as u)
dx
dx x
x
3
18.01 Fall 2006
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
df
Df
f 0 (x)
dx
f 00 (x)
D2 (f )
d2 f
dx2
f 000 (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 = (1.2)
D3 x3 = D2(3x2 ) = D(6x) = 6(= 1.2.3)
D4 x4 = D3(4x3 ) = D2(12x2 ) = D(24x) = 24(= 1.2.3.4)
Dn xn = 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 Dn xn = n!(nth case). Show it holds for the (n + 1)st case.
Dn+1 xn+1 = Dn (Dn+1 ) = Dn ((n + 1)xn ) = (n + 1)Dn xn = (n + 1)(n!)
Dn+1 xn+1 = (n + 1)!
Prove!
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