Chain Rule - Higher Derivatives

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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