Lesson 8: Basic Differentiation Rules (Section 41 slides)

Section 2.3
Basic Differentiation Rules

V63.0121.041, Calculus I

New York University

September 29, 2010

Announcements

Last chance for extra credit on Quiz 1: Do the get-to-know you
survey and photo by October 1.

. . . . . . .

Announcements

Last chance for extra credit
on Quiz 1: Do the
get-to-know you survey
and photo by October 1.

. . . . . .

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 2 / 42

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.

. . . . . .


Recall: the derivative

Definition
Let f be a function and a a point in the domain of f. If the limit

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

exists, the function is said to be differentiable at a and f′ (a) is the
derivative of f at a.
The derivative …
…measures the slope of the line through (a, f(a)) tangent to the
curve y = f(x);
…represents the instantaneous rate of change of f at a
…produces the best possible linear approximation to f near a.

. . . . . .


Notation

Newtonian notation Leibnizian notation
dy d df
f′ (x) y′ (x) y′ f(x)
dx dx dx

. . . . . .


Link between the notations

f(x + ∆x) − f(x) ∆y dy
f′ (x) = lim = lim =
∆x→0 ∆x ∆x→0 ∆x dx

dy
Leibniz thought of as a quotient of “infinitesimals”
dx
dy
We think of as representing a limit of (finite) difference
dx
quotients, not as an actual fraction itself.
The notation suggests things which are true even though they
don’t follow from the notation per se

. . . . . .


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.

. . . . . .


The second derivative

If f is a function, so is f′ , and we can seek its derivative.

f′′ = (f′ )′

It measures the rate of change of the rate of change!

. . . . . .


The second derivative

If f is a function, so is f′ , and we can seek its derivative.

f′′ = (f′ )′

It measures the rate of change of the rate of change! Leibnizian
notation:
d2 y d2 d2 f
f(x)
dx2 dx2 dx2

. . . . . .


The squaring function and its derivatives

y
.

. x
.

. . . . . .



y
.

f
.
. x
.

. . . . . .



y
.
.′
f
f increasing =⇒ f′ ≥ 0
f
. f decreasing =⇒ f′ ≤ 0
. x
. horizontal tangent at 0
=⇒ f′ (0) = 0

. . . . . .



y
.
.′
f
.′′
f f increasing =⇒ f′ ≥ 0
f
. f decreasing =⇒ f′ ≤ 0
. x
. horizontal tangent at 0
=⇒ f′ (0) = 0

. . . . . .


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
h→0 h h→0 h

. . . . . . .


.
Example
√

Solution

√ √
f(x + h) − f(x) x+h− x
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
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
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
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
. ... ...

. . . . . .


Proving the Power Rule
.

d n
x = nxn−1
dx

. . . . . . .


Proving the Power Rule
.

d n
x = nxn−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
= nxn−1 + (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

. . . . . .


Calculus

. . . . . .

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 of the Sum Rule

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 of the Constant Multiple Rule

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 A cos B + cos A sin B
.
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

.
. . . . . .


Lesson 8: Basic Differentiation Rules (Section 41 slides)

Lesson 8: Basic Differentiation Rules (Section 41 slides)

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