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Section 2.3Basic Differentiation Rules

V63.0121.041, 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 surveyand photo by October 1.

V63.0121.041, 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.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 3 / 42

. . . . . .

Recall: the derivative

DefinitionLet 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

= limx→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 aI …produces the best possible linear approximation to f near a.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 4 / 42

. . . . . .

Notation

Newtonian notation Leibnizian notation

f′(x) y′(x) y′dydx

ddx

f(x)dfdx

V63.0121.041, 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

=dydx

I Leibniz thought ofdydx

as a quotient of “infinitesimals”

I We think ofdydx

as representing a limit of (finite) differencequotients, not as an actual fraction itself.

I The notation suggests things which are true even though theydon’t follow from the notation per se

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 6 / 42

. . . . . .

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

V63.0121.041, 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

= limh→0

(x+ h)2 − x2

h

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

h→0

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 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

= limh→0

(x+ h)2 − x2

h

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

h→0

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 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

= limh→0

(x+ h)2 − x2

h

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

h→0

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 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

= limh→0

(x+ h)2 − x2

h

= limh→0

��x2 + 2xh+ h2 −��x2

h

= limh→0

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 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

= limh→0

(x+ h)2 − x2

h

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

h→0

2x�h+ h�2

�h

= limh→0

(2x+ h) = 2x.

So f′(x) = 2x.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 8 / 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

= limh→0

(x+ h)2 − x2

h

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

h→0

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

V63.0121.041, 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!

Leibniziannotation:

d2ydx2

d2

dx2f(x)

d2fdx2

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 9 / 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! Leibniziannotation:

d2ydx2

d2

dx2f(x)

d2fdx2

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 9 / 42

. . . . . .

The squaring function and its derivatives

. .x

.y

.f

.f′.f′′ I f increasing =⇒ f′ ≥ 0

I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0

=⇒ f′(0) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42

. . . . . .

The squaring function and its derivatives

. .x

.y

.f

.f′.f′′ I f increasing =⇒ f′ ≥ 0

I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0

=⇒ f′(0) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42

. . . . . .

The squaring function and its derivatives

. .x

.y

.f

.f′

.f′′

I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0

=⇒ f′(0) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 10 / 42

. . . . . .

The squaring function and its derivatives

. .x

.y

.f

.f′.f′′ I f increasing =⇒ f′ ≥ 0

I f decreasing =⇒ f′ ≤ 0I horizontal tangent at 0

=⇒ f′(0) = 0

V63.0121.041, 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

= limh→0

(x+ h)3 − x3

h

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

h→0

3x2�h+ 3xh���1

2 + h���2

3

�h= lim

h→0

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 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

= limh→0

(x+ h)3 − x3

h

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

h→0

3x2�h+ 3xh���1

2 + h���2

3

�h= lim

h→0

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 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

= limh→0

(x+ h)3 − x3

h

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h

= limh→0

3x2�h+ 3xh���1

2 + h���2

3

�h= lim

h→0

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 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

= limh→0

(x+ h)3 − x3

h

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

h→0

3x2�h+ 3xh���1

2 + h���2

3

�h

= limh→0

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 11 / 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

= limh→0

(x+ h)3 − x3

h

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

h→0

3x2�h+ 3xh���1

2 + h���2

3

�h= lim

h→0

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

V63.0121.041, 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) = 0I Notice also that the

tangent line to the graph off at (0,0) crosses thegraph (contrary to apopular “definition” of thetangent line)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42

. . . . . .

The cubing function and its derivatives

. .x

.y

.f

.f′.f′′I Notice that f is increasing,

and f′ > 0 except f′(0) = 0I Notice also that the

tangent line to the graph off at (0,0) crosses thegraph (contrary to apopular “definition” of thetangent line)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 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 thetangent line to the graph off at (0,0) crosses thegraph (contrary to apopular “definition” of thetangent line)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 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 thetangent line to the graph off at (0,0) crosses thegraph (contrary to apopular “definition” of thetangent line)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42

. . . . . .

The cubing function and its derivatives

. .x

.y

.f

.f′.f′′I Notice that f is increasing,

and f′ > 0 except f′(0) = 0I Notice also that the

tangent line to the graph off at (0,0) crosses thegraph (contrary to apopular “definition” of thetangent line)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 12 / 42

. . . . . .

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

= limh→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 = 1

2x−1/2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42

. . . . . .

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

= limh→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 = 1

2x−1/2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42

. . . . . .

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

= limh→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 = 1

2x−1/2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42

. . . . . .

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

= limh→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)

= limh→0

�h�h(√

x+ h+√x)

=1

2√x

So f′(x) =√x = 1

2x−1/2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42

. . . . . .

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

= limh→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 = 1

2x−1/2.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 13 / 42

. . . . . .

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

= limh→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 = 1

2x−1/2.

V63.0121.041, 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 0I Notice also lim

x→∞f′(x) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42

. . . . . .

The square root function and its derivatives

. .x

.y

.f

.f′I Here lim

x→0+f′(x) = ∞ and f

is not differentiable at 0I Notice also lim

x→∞f′(x) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 14 / 42

. . . . . .

The square root function and its derivatives

. .x

.y

.f.f′

I Here limx→0+

f′(x) = ∞ and f

is not differentiable at 0I Notice also lim

x→∞f′(x) = 0

V63.0121.041, 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

= limh→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) = 13x

−2/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 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

= limh→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) = 13x

−2/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 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

= limh→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) = 13x

−2/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 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

= limh→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

)

= limh→0

�h�h((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3

) =1

3x2/3

So f′(x) = 13x

−2/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 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

= limh→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) = 13x

−2/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 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

= limh→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) = 13x

−2/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 15 / 42

. . . . . .

The cube root function and its derivatives

. .x

.y

.f.f′

I Here limx→0

f′(x) = ∞ and f isnot differentiable at 0

I Notice also limx→±∞

f′(x) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42

. . . . . .

The cube root function and its derivatives

. .x

.y

.f

.f′

I Here limx→0

f′(x) = ∞ and f isnot differentiable at 0

I Notice also limx→±∞

f′(x) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42

. . . . . .

The cube root function and its derivatives

. .x

.y

.f.f′

I Here limx→0

f′(x) = ∞ and f isnot differentiable at 0

I Notice also limx→±∞

f′(x) = 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 16 / 42

. . . . . .

The cube root function and its derivatives

. .x

.y

.f.f′

I Here limx→0

f′(x) = ∞ and f isnot differentiable at 0

I Notice also limx→±∞

f′(x) = 0

V63.0121.041, 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

= limh→0

(x+ h)2/3 − x2/3

h

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

)= 1

3x−2/3

(2x1/3

)= 2

3x−1/3

So f′(x) = 23x

−1/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 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

= limh→0

(x+ h)2/3 − x2/3

h

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

)= 1

3x−2/3

(2x1/3

)= 2

3x−1/3

So f′(x) = 23x

−1/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 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

= limh→0

(x+ h)2/3 − x2/3

h

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

)

= 13x

−2/3(2x1/3

)= 2

3x−1/3

So f′(x) = 23x

−1/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 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

= limh→0

(x+ h)2/3 − x2/3

h

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

)= 1

3x−2/3

(2x1/3

)

= 23x

−1/3

So f′(x) = 23x

−1/3.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 17 / 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

= limh→0

(x+ h)2/3 − x2/3

h

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

)= 1

3x−2/3

(2x1/3

)= 2

3x−1/3

So f′(x) = 23x

−1/3.

V63.0121.041, 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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 18 / 42

. . . . . .

Recap

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap

y y′

x2 2x1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 19 / 42

. . . . . .

Recap: The Tower of Power

y y′

x2 2x

1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

I The power goes down byone in each derivative

I The coefficient in thederivative is the power ofthe original function

V63.0121.041, 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) = xr. Then

f′(x) = rxr−1

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

I Perhaps the most famous rule in calculusI We will assume it as of todayI We will prove it many ways for many different r.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 20 / 42

. . . . . .

The other Tower of Power

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 21 / 42

. . . . . .

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 22 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 23 / 42

. . . . . .

Remember your algebra

FactLet 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−1his the number of ways we can choose x n− 1 times, which is the sameas the number of different hs we can pick, which is n.V63.0121.041, 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

. . . . . .

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 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

. . . . . .

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 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

. . . . . .

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 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

. . . . . .

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 24 / 42

. . . . . .

Proving the Power Rule.

.

Theorem (The Power Rule)

Let n be a positive whole number. Then

ddx

xn = 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.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 25 / 42

. . . . . .

Proving the Power Rule.

.

Theorem (The Power Rule)

Let n be a positive whole number. Then

ddx

xn = 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.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 25 / 42

. . . . . .

The Power Rule for constants

TheoremLet c be a constant. Then

ddx

c = 0

.

.likeddx

x0 = 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− ch

= 0

So f′(x) = limh→0

0 = 0.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42

. . . . . .

The Power Rule for constants

TheoremLet c be a constant. Then

ddx

c = 0 .

.likeddx

x0 = 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− ch

= 0

So f′(x) = limh→0

0 = 0.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42

. . . . . .

The Power Rule for constants

TheoremLet c be a constant. Then

ddx

c = 0 .

.likeddx

x0 = 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− ch

= 0

So f′(x) = limh→0

0 = 0.

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 26 / 42

Calculus

. . . . . .

. . . . . .

Recall the Limit Laws

FactSuppose lim

x→af(x) = L and lim

x→ag(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. . . .

V63.0121.041, 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′.

V63.0121.041, 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

+ limh→0

g(x+ h)− g(x)h

= f′(x) + g′(x)

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 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

+ limh→0

g(x+ h)− g(x)h

= f′(x) + g′(x)

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 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

+ limh→0

g(x+ h)− g(x)h

= f′(x) + g′(x)

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 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

+ limh→0

g(x+ h)− g(x)h

= f′(x) + g′(x)

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 30 / 42

. . . . . .

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′.

V63.0121.041, 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)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 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)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 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)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 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)

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 32 / 42

. . . . . .

Derivatives of polynomials

Example

Findddx

(2x3 + x4 − 17x12 + 37

)

Solution

ddx

(2x3 + x4 − 17x12 + 37

)=

ddx

(2x3

)+

ddx

x4 +ddx

(−17x12

)+

ddx

(37)

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42

. . . . . .

Derivatives of polynomials

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

ddx

(2x3 + x4 − 17x12 + 37

)=

ddx

(2x3

)+

ddx

x4 +ddx

(−17x12

)+

ddx

(37)

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42

. . . . . .

Derivatives of polynomials

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

ddx

(2x3 + x4 − 17x12 + 37

)=

ddx

(2x3

)+

ddx

x4 +ddx

(−17x12

)+

ddx

(37)

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42

. . . . . .

Derivatives of polynomials

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

ddx

(2x3 + x4 − 17x12 + 37

)=

ddx

(2x3

)+

ddx

x4 +ddx

(−17x12

)+

ddx

(37)

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42

. . . . . .

Derivatives of polynomials

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

ddx

(2x3 + x4 − 17x12 + 37

)=

ddx

(2x3

)+

ddx

x4 +ddx

(−17x12

)+

ddx

(37)

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 33 / 42

. . . . . .

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 34 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = ???

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cosh+ cos x sinh)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sinhh

= sin x · 0+ cos x · 1 = cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 35 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = ???

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cosh+ cos x sinh)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sinhh

= sin x · 0+ cos x · 1 = cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 35 / 42

. . . . . .

Angle addition formulasSee Appendix A

.

.

sin(A+ B) = sinA cosB+ cosA sinBcos(A+ B) = cosA cosB− sinA sinB

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 36 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = ???

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cosh+ cos x sinh)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sinhh

= sin x · 0+ cos x · 1 = cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 37 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = ???

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cosh+ cos x sinh)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sinhh

= sin x · 0+ cos x · 1 = cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 37 / 42

. . . . . .

Two important trigonometric limitsSee Section 1.4

..θ

.sin θ

.1− cos θ

.θ

.−1 .1

.

.

limθ→0

sin θθ

= 1

limθ→0

cos θ − 1θ

= 0

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 38 / 42

. . . . . .

Derivatives of Sine and Cosine

Fact

ddx

sin x = ???

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cos h+ cos x sin h)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sin hh

= sin x · 0+ cos x · 1

= cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 39 / 42

. . . . . .

Derivatives of Sine and Cosine

Fact

ddx

sin x = ???

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cos h+ cos x sin h)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sin hh

= sin x · 0+ cos x · 1

= cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 39 / 42

. . . . . .

Derivatives of Sine and Cosine

Fact

ddx

sin x = cos x

Proof.From the definition:

ddx

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

( sin x cos h+ cos x sin h)− sin xh

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sin hh

= sin x · 0+ cos x · 1 = cos x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 39 / 42

. . . . . .

Illustration of Sine and Cosine

. .x

.y

.π .−π2 .0 .π2 .π

.sin x

.cos x

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42

. . . . . .

Illustration of Sine and Cosine

. .x

.y

.π .−π2 .0 .π2 .π

.sin x

.cos x

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

I what happens at the horizontal tangents of cos?

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42

. . . . . .

Illustration of Sine and Cosine

. .x

.y

.π .−π2 .0 .π2 .π

.sin x

.cos x

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

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 40 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = cos xddx

cos x = − sin x

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

ddx

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

(cos x cos h− sin x sin h)− cos xh

= cos x · limh→0

cos h− 1h

− sin x · limh→0

sinhh

= cos x · 0− sin x · 1 = − sin x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = cos xddx

cos x = − sin x

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

ddx

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

(cos x cos h− sin x sin h)− cos xh

= cos x · limh→0

cos h− 1h

− sin x · limh→0

sinhh

= cos x · 0− sin x · 1 = − sin x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = cos xddx

cos x = − sin x

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

ddx

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

(cos x cos h− sin x sin h)− cos xh

= cos x · limh→0

cos h− 1h

− sin x · limh→0

sinhh

= cos x · 0− sin x · 1 = − sin x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = cos xddx

cos x = − sin x

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

ddx

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

(cos x cos h− sin x sin h)− cos xh

= cos x · limh→0

cos h− 1h

− sin x · limh→0

sinhh

= cos x · 0− sin x · 1 = − sin x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42

. . . . . .

Derivatives of Sine and Cosine.

.

Fact

ddx

sin x = cos xddx

cos x = − sin x

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

ddx

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

(cos x cos h− sin x sin h)− cos xh

= cos x · limh→0

cos h− 1h

− sin x · limh→0

sinhh

= cos x · 0− sin x · 1 = − sin x

V63.0121.041, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 29, 2010 41 / 42

. . . . . .

What have we learned today?

I The Power Rule

I The derivative of a sum is the sum of the derivativesI The derivative of a constant multiple of a function is that constant

multiple of the derivativeI The derivative of sine is cosineI The derivative of cosine is the opposite of sine.

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

. . . . . .

What have we learned today?

I The Power RuleI The derivative of a sum is the sum of the derivativesI The derivative of a constant multiple of a function is that constant

multiple of the derivative

I The derivative of sine is cosineI The derivative of cosine is the opposite of sine.

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

. . . . . .

What have we learned today?

I The Power RuleI The derivative of a sum is the sum of the derivativesI The derivative of a constant multiple of a function is that constant

multiple of the derivativeI The derivative of sine is cosineI The derivative of cosine is the opposite of sine.

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