Lesson 8: Basic Differentiation Rules (Section 21 handout)

Section 2.3Basic Differentiation Rules

V63.0121.021, Calculus I

New York University

September 29, 2010

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

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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.


Notes

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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.


Notation

Newtonian notation Leibnizian notation

f ′(x) y ′(x) y ′dy

dx

d

dxf (x)

df

dx


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


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2


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


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.


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


Notes

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3


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


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.


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)


Notes

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4


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.


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


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.


Notes

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5


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


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.


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


Notes

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6


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


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 .


The other Tower of Power


Notes

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7


Outline





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.


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

. . . . . .


Notes

Notes

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8


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.


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.


Calculus

Notes

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9


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. . . .


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) = 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.


Notes

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10


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) = 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)


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


Notes

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11


Outline





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


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


Notes

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12


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


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?


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


Notes

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13


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.


Notes

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Lesson 8: Basic Differentiation Rules (Section 21 handout)