lesson 8: basic differentiation rules (section 21 handout)
TRANSCRIPT
Section 2.3Basic Differentiation Rules
V63.0121.021, Calculus I
New York University
September 29, 2010
Announcements
I Last chance for extra credit on Quiz 1: Do the get-to-know yousurvey and photo by October 1.
Announcements
I Last chance for extra crediton Quiz 1: Do theget-to-know you survey andphoto by October 1.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 2 / 42
Objectives
I Understand and use thesedifferentiation rules:
I the derivative of aconstant function (zero);
I the Constant MultipleRule;
I the Sum Rule;I the Difference Rule;I the derivatives of sine and
cosine.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 3 / 42
Notes
Notes
Notes
1
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Recall: the derivative
Definition
Let f be a function and a a point in the domain of f . If the limit
f ′(a) = limh→0
f (a + h)− f (a)
h= lim
x→a
f (x)− f (a)
x − a
exists, the function is said to be differentiable at a and f ′(a) is thederivative of f at a.The derivative . . .
I . . . measures the slope of the line through (a, f (a)) tangent to thecurve y = f (x);
I . . . represents the instantaneous rate of change of f at a
I . . . produces the best possible linear approximation to f near a.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 4 / 42
Notation
Newtonian notation Leibnizian notation
f ′(x) y ′(x) y ′dy
dx
d
dxf (x)
df
dx
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 5 / 42
Link between the notations
f ′(x) = lim∆x→0
f (x + ∆x)− f (x)
∆x= lim
∆x→0
∆y
∆x=
dy
dx
I Leibniz thought ofdy
dxas a quotient of “infinitesimals”
I We think ofdy
dxas representing a limit of (finite) difference quotients,
not as an actual fraction itself.
I The notation suggests things which are true even though they don’tfollow from the notation per se
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 6 / 42
Notes
Notes
Notes
2
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Outline
Derivatives so farDerivatives of power functions by handThe Power Rule
Derivatives of polynomialsThe Power Rule for whole number powersThe Power Rule for constantsThe Sum RuleThe Constant Multiple Rule
Derivatives of sine and cosine
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 7 / 42
Derivative of the squaring function
Example
Suppose f (x) = x2. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x
2
h= lim
h→0
2x�h + h�2
�h
= limh→0
(2x + h) = 2x .
So f ′(x) = 2x.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 42
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:
d2y
dx2
d2
dx2f (x)
d2f
dx2
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 9 / 42
Notes
Notes
Notes
3
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
The squaring function and its derivatives
x
y
f
f ′f ′′ I f increasing =⇒ f ′ ≥ 0
I f decreasing =⇒ f ′ ≤ 0
I horizontal tangent at 0=⇒ f ′(0) = 0
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42
Derivative of the cubing function
Example
Suppose f (x) = x3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x
3
h= lim
h→0
3x2�h + 3xh���
1
2 + h���2
3
�h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f ′(x) = 3x2.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 42
The cubing function and its derivatives
x
y
f
f ′f ′′I Notice that f is increasing,
and f ′ > 0 except f ′(0) = 0
I Notice also that the tangentline to the graph of f at(0, 0) crosses the graph(contrary to a popular“definition” of the tangentline)
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42
Notes
Notes
Notes
4
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Derivative of the square root function
Example
Suppose f (x) =√
x = x1/2. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
√x + h −
√x
h
= limh→0
√x + h −
√x
h·√
x + h +√
x√x + h +
√x
= limh→0
(�x + h)−�xh(√
x + h +√
x) = lim
h→0
�h
�h(√
x + h +√
x)
=1
2√
x
So f ′(x) =√
x = 12 x−1/2.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42
The square root function and its derivatives
x
y
f
f ′
I Here limx→0+
f ′(x) =∞ and f
is not differentiable at 0
I Notice also limx→∞
f ′(x) = 0
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42
Derivative of the cube root function
Example
Suppose f (x) = 3√
x = x1/3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h)−�xh((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)= lim
h→0
�h
�h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
) =1
3x2/3
So f ′(x) = 13 x−2/3.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42
Notes
Notes
Notes
5
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
The cube root function and its derivatives
x
y
f
f ′
I Here limx→0
f ′(x) =∞ and f is
not differentiable at 0
I Notice also limx→±∞
f ′(x) = 0
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42
One more
Example
Suppose f (x) = x2/3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)2/3 − x2/3
h
= limh→0
(x + h)1/3 − x1/3
h·(
(x + h)1/3 + x1/3)
= 13 x−2/3
(2x1/3
)= 2
3 x−1/3
So f ′(x) = 23 x−1/3.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 42
The function x 7→ x2/3 and its derivative
x
y
f
f ′
I f is not differentiable at 0and lim
x→0±f ′(x) = ±∞
I Notice also limx→±∞
f ′(x) = 0
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 42
Notes
Notes
Notes
6
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Recap: The Tower of Power
y y ′
x2 2x1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42
The Power Rule
There is mounting evidence for
Theorem (The Power Rule)
Let r be a real number and f (x) = x r . Then
f ′(x) = rx r−1
as long as the expression on the right-hand side is defined.
I Perhaps the most famous rule in calculus
I We will assume it as of today
I We will prove it many ways for many different r .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 20 / 42
The other Tower of Power
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 21 / 42
Notes
Notes
Notes
7
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Outline
Derivatives so farDerivatives of power functions by handThe Power Rule
Derivatives of polynomialsThe Power Rule for whole number powersThe Power Rule for constantsThe Sum RuleThe Constant Multiple Rule
Derivatives of sine and cosine
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 22 / 42
Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1h + (stuff with at least two hs in it)
Proof.
We have
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
The coefficient of xn is 1 because we have to choose x from eachbinomial, and there’s only one way to do that. The coefficient of xn−1h isthe number of ways we can choose x n− 1 times, which is the same as thenumber of different hs we can pick, which is n.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42
Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
(x + h)0 = 1
(x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
(x + h)3 = 1x3 + 3x2h + 3xh2 + 1h3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 42
Notes
Notes
Notes
8
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Proving the Power Rule
Theorem (The Power Rule)
Let n be a positive whole number. Then
d
dxxn = nxn−1
Proof.
As we showed above,
(x + h)n = xn + nxn−1h + (stuff with at least two hs in it)
So
(x + h)n − xn
h=
nxn−1h + (stuff with at least two hs in it)
h= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h→ 0.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 25 / 42
The Power Rule for constants
Theorem
Let c be a constant. Thend
dxc = 0
liked
dxx0 = 0x−1
(although x 7→ 0x−1 is not defined at zero.)
Proof.
Let f (x) = c . Then
f (x + h)− f (x)
h=
c − c
h= 0
So f ′(x) = limh→0
0 = 0.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42
Calculus
Notes
Notes
Notes
9
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Recall the Limit Laws
Fact
Suppose limx→a
f (x) = L and limx→a
g(x) = M and c is a constant. Then
1. limx→a
[f (x) + g(x)] = L + M
2. limx→a
[f (x)− g(x)] = L−M
3. limx→a
[cf (x)] = cL
4. . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 28 / 42
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 ′.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 29 / 42
Proof of the Sum Rule
Proof.
Follow your nose:
(f + g)′(x) = limh→0
(f + g)(x + h)− (f + g)(x)
h
= limh→0
f (x + h) + g(x + h)− [f (x) + g(x)]
h
= limh→0
f (x + h)− f (x)
h+ lim
h→0
g(x + h)− g(x)
h
= f ′(x) + g ′(x)
Note the use of the Sum Rule for limits. Since the limits of the differencequotients for for f and g exist, the limit of the sum is the sum of the limits.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 42
Notes
Notes
Notes
10
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
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 ′.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 31 / 42
Proof of the Constant Multiple Rule
Proof.
Again, follow your nose.
(cf )′(x) = limh→0
(cf )(x + h)− (cf )(x)
h
= limh→0
cf (x + h)− cf (x)
h
= c limh→0
f (x + h)− f (x)
h
= c · f ′(x)
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 42
Derivatives of polynomials
Example
Findd
dx
(2x3 + x4 − 17x12 + 37
)Solution
d
dx
(2x3 + x4 − 17x12 + 37
)=
d
dx
(2x3)
+d
dxx4 +
d
dx
(−17x12
)+
d
dx(37)
= 2d
dxx3 +
d
dxx4 − 17
d
dxx12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42
Notes
Notes
Notes
11
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Outline
Derivatives so farDerivatives of power functions by handThe Power Rule
Derivatives of polynomialsThe Power Rule for whole number powersThe Power Rule for constantsThe Sum RuleThe Constant Multiple Rule
Derivatives of sine and cosine
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 34 / 42
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
Proof.
From the definition:
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
( sin x cos h + cos x sin h)− sin x
h
= sin x · limh→0
cos h − 1
h+ cos x · lim
h→0
sin h
h
= sin x · 0 + cos x · 1 = cos x
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 35 / 42
Angle addition formulasSee Appendix A
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 36 / 42
Notes
Notes
Notes
12
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
Two important trigonometric limitsSee Section 1.4
θ
sin θ
1− cos θ
θ
−1 1
limθ→0
sin θ
θ= 1
limθ→0
cos θ − 1
θ= 0
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 38 / 42
Illustration of Sine and Cosine
x
y
π −π2
0 π2
π
sin xcos x
I f (x) = sin x has horizontal tangents where f ′ = cos(x) is zero.
I what happens at the horizontal tangents of cos?
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
d
dxcos x = − sin x
Proof.
We already did the first. The second is similar (mutatis mutandis):
d
dxcos x = lim
h→0
cos(x + h)− cos x
h
= limh→0
(cos x cos h − sin x sin h)− cos x
h
= cos x · limh→0
cos h − 1
h− sin x · lim
h→0
sin h
h
= cos x · 0− sin x · 1 = − sin x
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42
Notes
Notes
Notes
13
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010
What have we learned today?
I The Power Rule
I The derivative of a sum is the sum of the derivatives
I The derivative of a constant multiple of a function is that constantmultiple of the derivative
I The derivative of sine is cosine
I The derivative of cosine is the opposite of sine.
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 42 / 42
Notes
Notes
Notes
14
Section 2.3 : Basic Differentiation RulesV63.0121.021, Calculus I September 29, 2010