Lesson 8: Basic Differentiation Rules

Section 2.3
Basic Differentiation Rules

V63.0121.002.2010Su, Calculus I

New York University

May 24, 2010

Announcements

Homework 1 due Tuesday
Quiz 2 Thursday in class on Sections 1.5–2.5

. . . . . .

Announcements

Homework 1 due Tuesday
Quiz 2 Thursday in class
on Sections 1.5–2.5

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 2 / 36

Objectives

Understand and use these
differentiation rules:
the derivative of a
constant function (zero);
the Constant Multiple
Rule;
the Sum Rule;
the Difference Rule;
the derivatives of sine
and cosine.

. . . . . .


Outline

Derivatives so far
Derivatives of power functions by hand
The Power Rule

Derivatives of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Multiple Rule

Derivatives of sine and cosine

. . . . . .


Derivative of the squaring function

Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).

. . . . . .



Example

Solution

f(x + h) − f(x)
f′ (x) = lim
h→0 h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)2 − x2
f′ (x) = lim = lim
h→0 h h→0 h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)2 − x2
h→0 h h→0 h
2
+ 2xh + h −
x2 x2
= lim
h→0 h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)2 − x2
h→0 h h→0 h
2
+ 2xh + h −
x2 x2 2x + h¡
h 2
= lim = lim
h→0 h h→0 h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)2 − x2
h→0 h h→0 h
2
+ 2xh + h −
x2 x2 2x + h¡
h 2
= lim = lim
h→0 h h→0 h
= lim (2x + h) = 2x.
h→0

So f′ (x) = 2x.

. . . . . .


Derivative of the cubing function

Example

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)3 − x3
h→0 h h→0 h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)3 − x3
h→0 h h→0 h

2 3
+ 3x h + 3xh + h −
x3 2 x3
= lim
h→0 h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)3 − x3
h→0 h h→0 h
1 2

x3 + 3x2 h 2
+ 3xh + h − 3
x3 3x2 h ¡
!
2 !
¡
3
+ 3xh + h
= lim = lim
h→0 h h→0
h

. . . . . .



Example

Solution

f(x + h) − f(x) (x + h)3 − x3
h→0 h h→0 h
1 2

+ 3xh + h −
x3 + 3x2 h 2 3
x3 3x2 h ¡
!
2 !
¡
3
+ 3xh + h
= lim = lim
h→0
(
h
)
h→0
h
2 2 2
= lim 3x + 3xh + h = 3x .
h→0

So f′ (x) = 3x2 .
. . . . . .


The cubing function and its derivatives

y
.

. x
.

. . . . . .



y
.

f
.
. x
.

. . . . . .



y
. Notice that f is increasing,
.′
f and f′ 0 except f′ (0) = 0

f
.
. x
.

. . . . . .



y
.′′ .′
f f and f′ 0 except f′ (0) = 0

f
.
. x
.

. . . . . .



y
.′′ .′
f f and f′ 0 except f′ (0) = 0
Notice also that the
f
. tangent line to the graph of
. x
. f at (0, 0) crosses the
graph (contrary to a
popular “definition” of the
tangent line)

. . . . . .


Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x).

. . . . . .


Example
√

Solution

√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h

. . . . . .


Example
√

Solution

√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x

. . . . . .


Example
√

Solution

√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡
= lim (√ √ )
h→0 h x+h+ x

. . . . . .


Example
√

Solution

√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡
h
= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h→0
h x+h+ x

. . . . . .


Example
√

Solution

√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡
h
= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h→0
h x+h+ x
1
= √
2 x
√
So f′ (x) = x = 1 x−1/2 .
2
. . . . . .


The square root function and its derivatives

y
.

. x
.

. . . . . .



y
.

f
.
. x
.

. . . . . .



y
.

f
. Here lim+ f′ (x) = ∞ and f
x→0
. .′
f is not differentiable at 0
x
.

. . . . . .



y
.

f
. Here lim+ f′ (x) = ∞ and f
x→0
. .′
x
.
Notice also lim f′ (x) = 0
x→∞

. . . . . .


Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find f′ (x).

. . . . . .


Example
√
Suppose f(x) = 3

Solution

f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h

. . . . . .


Example
√
Suppose f(x) = 3

Solution

f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3

. . . . . .


Example
√
Suppose f(x) = 3

Solution

f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(x + h) − x
¡ ¡
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)

. . . . . .


Example
√
Suppose f(x) = 3

Solution

f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(x + h) − x
¡ ¡
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)

h
= lim ( )
h→0
h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
. . . . . .


Example
√
Suppose f(x) = 3

Solution

f(x + h) − f(x) (x + h)1/3 − x1/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(x + h) − x
¡ ¡
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)

h 1
= lim ( )=
h→0
h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3 3x2/3

So f′ (x) = 1 x−2/3 .
. . . . . .

3

The cube root function and its derivatives

y
.

. x
.

. . . . . .



y
.

f
.
. x
.

. . . . . .



y
.

Here lim f′ (x) = ∞ and f is
f
. x→0
not differentiable at 0
. .′
f
x
.

. . . . . .



y
.

Here lim f′ (x) = ∞ and f is
f
. x→0
not differentiable at 0
.′
f
. x
. Notice also lim f′ (x) = 0
x→±∞

. . . . . .


One more

Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).

. . . . . .


One more

Example

Solution

f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h

. . . . . .


One more

Example

Solution

f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 ( )
= lim · (x + h)1/3 + x1/3
h→0 h

. . . . . .


One more

Example

Solution

f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 ( )
= lim · (x + h)1/3 + x1/3
h→0 h
( )
1 −2/3 1/3
= 3x 2x

. . . . . .


One more

Example

Solution

f(x + h) − f(x) (x + h)2/3 − x2/3
h→0 h h→0 h
(x + h) 1/3 − x1/3 ( )
= lim · (x + h)1/3 + x1/3
h→0 h
( )
1 −2/3
= 3x 2x 1/3
= 2 x−1/3
3

So f′ (x) = 2 x−1/3 .
3

. . . . . .


The function x → x2/3 and its derivative

y
.

. x
.

. . . . . .



y
.

f
.
. x
.

. . . . . .



y
.

f
. and lim f′ (x) = ±∞
x→0±
. .′
f
x
.

. . . . . .



y
.

f
. and lim f′ (x) = ±∞
x→0±
. .′
f
x
. Notice also lim f′ (x) = 0
x→±∞

. . . . . .


Recap

y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

. . . . . .


Recap

y y′
x2 2x1
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

. . . . . .


Recap: The Tower of Power

y y′
x2 2x1 The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

. . . . . .



y y′
x2 2x The power goes down by
x 3
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x

. . . . . .



y y′
x2 2x The power goes down by
x 3
1/2 1 −1/2 The coefficient in the
x 2x derivative is the power of
1 −2/3
x1/3 3x
the original function
2 −1/3
x2/3 3x

. . . . . .


The Power Rule

There is mounting evidence for
Theorem (The Power Rule)
Let r be a real number and f(x) = xr . Then

f′ (x) = rxr−1

as long as the expression on the right-hand side is defined.

Perhaps the most famous rule in calculus
We will assume it as of today
We will prove it many ways for many different r.

. . . . . .


The other Tower of Power

. . . . . .


Outline

Derivatives so far
The Power Rule

The Sum Rule


. . . . . .


Remember your algebra
Fact
Let n be a positive whole number. Then

(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)

. . . . . .


Fact


Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0

. . . . . .


Fact


Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0

The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that.

. . . . . .


Fact


Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0

The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that. The coefficient of xn−1 h
is the number of ways we can choose x n − 1 times, which is the same
as the number of different hs we can pick, which is n. . . . . . .


Pascal's Triangle

..
1

1
. 1
.

1
. 2
. 1
.

1
. 3
. 3
. 1
.
(x + h)0 = 1
1
. 4
. 6
. 4
. 1
. (x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
1
. 5 1 1 5
. .0 .0 . 1
.
(x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3

1
. 6 1 2 1 6
. .5 .0 .5 . 1
. ... ...

. . . . . .


Let r be a positive whole number. Then

d r
x = rxr−1
dx

. . . . . .


Let r be a positive whole number. Then

d r
x = rxr−1
dx

Proof.
As we showed above,


So
(x + h)n − xn nxn−1 h + (stuff with at least two hs in it)
=
h h
n−1
= nx + (stuff with at least one h in it)

and this tends to nxn−1 as h → 0.
. . . . . .



Theorem
Let c be a constant. Then
d
c=0
dx

. . . . . .



Theorem d 0
l
.ike x = 0x−1
Let c be a constant. Then dx
d
c=0.
dx

(although x → 0x−1 is not defined at zero.)

. . . . . .



Theorem d 0
l
.ike x = 0x−1
Let c be a constant. Then dx
d
c=0.
dx

(although x → 0x−1 is not defined at zero.)
Proof.
Let f(x) = c. Then

f(x + h) − f(x) c−c
= =0
h h
So f′ (x) = lim 0 = 0.
h→0

. . . . . .


Recall the Limit Laws

Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
x→a x→a
1. lim [f(x) + g(x)] = L + M
x→a
2. lim [f(x) − g(x)] = L − M
x→a
3. lim [cf(x)] = cL
x→a
4. . . .

. . . . . .


Adding functions

Theorem (The Sum Rule)
Let f and g be functions and define

(f + g)(x) = f(x) + g(x)

Then if f and g are differentiable at x, then so is f + g and

(f + g)′ (x) = f′ (x) + g′ (x).

Succinctly, (f + g)′ = f′ + g′ .

. . . . . .


Proof.
Follow your nose:

(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h

. . . . . .


Proof.
Follow your nose:

(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h

. . . . . .


Proof.
Follow your nose:

(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h

. . . . . .


Proof.
Follow your nose:

(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
′ ′
= f (x) + g (x)

Note the use of the Sum Rule for limits. Since the limits of the
difference quotients for for f and g exist, the limit of the sum is the sum
of the limits.

. . . . . .


Scaling functions

Theorem (The Constant Multiple Rule)
Let f be a function and c a constant. Define

(cf)(x) = cf(x)

Then if f is differentiable at x, so is cf and

(cf)′ (x) = c · f′ (x)

Succinctly, (cf)′ = cf′ .

. . . . . .


Proof.
Again, follow your nose.

(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h

. . . . . .


Proof.

(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
cf(x + h) − cf(x)
= lim
h→0 h

. . . . . .


Proof.

(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
cf(x + h) − cf(x)
= lim
h→0 h
f(x + h) − f(x)
= c lim
h→0 h

. . . . . .


Proof.

(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
cf(x + h) − cf(x)
= lim
h→0 h
f(x + h) − f(x)
= c lim
h→0 h
′
= c · f (x)

. . . . . .


Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx

. . . . . .


Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx

Solution

d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx

. . . . . .


Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx

Solution

d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
d 3 d 4 d 12
= 2 x + x − 17 x + 0
dx dx dx

. . . . . .


Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx

Solution

d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
d 3 d 4 d 12
= 2 x + x − 17 x + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3

. . . . . .


Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx

Solution

d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
d 3 d 4 d 12
= 2 x + x − 17 x + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3

= 6x2 + 4x3 − 204x11
. . . . . .


Outline

Derivatives so far
The Power Rule

The Sum Rule


. . . . . .


Derivatives of Sine and Cosine
Fact

d
sin x = ???
dx

Proof.
From the definition:

. . . . . .


Fact

d
sin x = ???
dx

Proof.
d sin(x + h) − sin x
sin x = lim
dx h→0 h

. . . . . .


Angle addition formulas
See Appendix A

.
sin(A + B) = sin . cos B + cos A sin B
A
cos(A + B) = cos A cos B − sin A sin B

. . . . . .


Fact

d
sin x = ???
dx

Proof.
sin x = lim
dx h→0 h
( sin x cos h + cos x sin h) − sin x
= lim
h→0 h

. . . . . .


Fact

d
sin x = ???
dx

Proof.
sin x = lim
dx h→0 h
= lim
h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h

. . . . . .


Two important trigonometric limits
See Section 1.4

.
sin θ
lim
. =1
θ→0 θ

. in θ .
s θ cos θ − 1
lim =0
.
θ θ→0 θ
.
. − cos θ
1 1
.

. . . . . .


Fact

d
sin x = ???
dx

Proof.
sin x = lim
dx h→0 h
= lim
h→0 h
cos h − 1 sin h
h→0 h h→0 h
= sin x · 0 + cos x · 1
. . . . . .


Fact

d
sin x = cos x
dx

Proof.
sin x = lim
dx h→0 h
= lim
h→0 h
cos h − 1 sin h
h→0 h h→0 h
= sin x · 0 + cos x · 1 = cos x
. . . . . .


Illustration of Sine and Cosine

y
.

. x
.
.
π −2
. π 0
. .π .
π
2
s
. in x

. . . . . .


y
.

. x
.
.
π −2
. π 0
. .π .
π
2 c
. os x
s
. in x

f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.

. . . . . .



y
.

. x
.
.
π −2
. π 0
. .π .
π
2 c
. os x
s
. in x

f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
what happens at the horizontal tangents of cos?

. . . . . .


Fact

d d
sin x = cos x cos x = − sin x
dx dx

. . . . . .


Fact

d d
dx dx

Proof.
We already did the first. The second is similar (mutatis mutandis):

d cos(x + h) − cos x
cos x = lim
dx h→0 h

. . . . . .


Fact

d d
dx dx

Proof.

cos x = lim
dx h→0 h
(cos x cos h − sin x sin h) − cos x
= lim
h→0 h

. . . . . .


Fact

d d
dx dx

Proof.

cos x = lim
dx h→0 h
= lim
h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h

. . . . . .


Fact

d d
dx dx

Proof.

cos x = lim
dx h→0 h
= lim
h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
= cos x · 0 − sin x · 1 = − sin x
. . . . . .


Summary

d r
Power Rule: x = rxr−1
dx
Sum, Difference, Constant, and Constant Multiple Rules
Derivative of sin is cos. Derivative of cos is − sin.

. . . . . .


Lesson 8: Basic Differentiation Rules

Recommended

Recommended

More Related Content

What's hot

What's hot (17)

Viewers also liked

Viewers also liked (20)

Similar to Lesson 8: Basic Differentiation Rules

Similar to Lesson 8: Basic Differentiation Rules (20)

More from Matthew Leingang

More from Matthew Leingang (18)

Recently uploaded

Recently uploaded (20)

Lesson 8: Basic Differentiation Rules