lesson 8: basic differentiation rules (section 21 slides)

Section 2.3Basic Differentiation Rules

V63.0121.021, Calculus I

New York University

September 30, 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.

V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 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.

. . . . . .

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.

. . . . . .

Notation

Newtonian notation Leibnizian notation

f′(x) y′(x) y′dydx

f(x)dfdx

. . . . . .

Link between the notations

f′(x) = lim∆x→0

f(x+∆x)− f(x)∆x

= lim∆x→0

∆y∆x

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

. . . . . .

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

= limh→0

(x+ h)2 − x2

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2 − x2

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2 − x2

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2 − x2

= limh→0

��x2 + 2xh+ h2 −��x2

= limh→0

2x�h+ h�2

�h= lim

h→0(2x+ h) = 2x.

So f′(x) = 2x.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2 − x2

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

2x�h+ h�2

= limh→0

(2x+ h) = 2x.

So f′(x) = 2x.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2 − x2

= limh→0

��x2 + 2xh+ h2 −��x2

h= lim

2x�h+ h�2

�h= lim

h→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!

Leibniziannotation:

d2ydx2

dx2f(x)

d2fdx2

. . . . . .

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

dx2f(x)

d2fdx2

. . . . . .

The squaring function and its derivatives

.f′.f′′ I f increasing =⇒ f′ ≥ 0

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

=⇒ f′(0) = 0

. . . . . .

=⇒ f′(0) = 0

. . . . . .

.f′′

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

=⇒ f′(0) = 0

. . . . . .

=⇒ f′(0) = 0

. . . . . .

Derivative of the cubing function

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)3 − x3

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

3x2�h+ 3xh��1

2 + h��2

�h= lim

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)3 − x3

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

3x2�h+ 3xh��1

2 + h��2

�h= lim

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)3 − x3

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

= limh→0

3x2�h+ 3xh��1

2 + h��2

�h= lim

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)3 − x3

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

3x2�h+ 3xh��1

2 + h��2

= limh→0

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)3 − x3

= limh→0

��x3 + 3x2h+ 3xh2 + h3 −��x3

h= lim

3x2�h+ 3xh��1

2 + h��2

�h= lim

(3x2 + 3xh+ h2

)= 3x2.

So f′(x) = 3x2.

. . . . . .

The cubing function and its derivatives

.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)

. . . . . .

.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)

. . . . . .

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)

. . . . . .

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−

= limh→0

√x+ h−

h·√x+ h+

√x√

x+ h+√x

= limh→0

(�x+ h)− �xh(√

x+ h+√x) = lim

�h�h(√

x+ h+√x)

So f′(x) =√x = 1

2x−1/2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

√x+ h−

= limh→0

√x+ h−

h·√x+ h+

√x√

x+ h+√x

= limh→0

(�x+ h)− �xh(√

x+ h+√x) = lim

�h�h(√

x+ h+√x)

So f′(x) =√x = 1

2x−1/2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

√x+ h−

= limh→0

√x+ h−

h·√x+ h+

√x√

x+ h+√x

= limh→0

(�x+ h)− �xh(√

x+ h+√x) = lim

�h�h(√

x+ h+√x)

So f′(x) =√x = 1

2x−1/2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

√x+ h−

= limh→0

√x+ h−

h·√x+ h+

√x√

x+ h+√x

= limh→0

(�x+ h)− �xh(√

x+ h+√x)

= limh→0

�h�h(√

x+ h+√x)

So f′(x) =√x = 1

2x−1/2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

√x+ h−

= limh→0

√x+ h−

h·√x+ h+

√x√

x+ h+√x

= limh→0

(�x+ h)− �xh(√

x+ h+√x) = lim

�h�h(√

x+ h+√x)

So f′(x) =√x = 1

2x−1/2.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

√x+ h−

= limh→0

√x+ h−

h·√x+ h+

√x√

x+ h+√x

= limh→0

(�x+ h)− �xh(√

x+ h+√x) = lim

�h�h(√

x+ h+√x)

So f′(x) =√x = 1

2x−1/2.

. . . . . .

The square root function and its derivatives

.f.f′

I Here limx→0+

f′(x) = ∞ and f

is not differentiable at 0I Notice also lim

x→∞f′(x) = 0

. . . . . .

.f′I Here lim

x→0+f′(x) = ∞ and f

x→∞f′(x) = 0

. . . . . .

.f.f′

I Here limx→0+

f′(x) = ∞ and f

is not differentiable at 0

I Notice also limx→∞

f′(x) = 0

. . . . . .

.f.f′

I Here limx→0+

f′(x) = ∞ and f

x→∞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

= limh→0

(x+ h)1/3 − x1/3

= 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�h((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3

So f′(x) = 13x

−2/3.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)1/3 − x1/3

= 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�h((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3

So f′(x) = 13x

−2/3.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)1/3 − x1/3

= 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�h((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3

So f′(x) = 13x

−2/3.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)1/3 − x1/3

= 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

So f′(x) = 13x

−2/3.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)1/3 − x1/3

= 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�h((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3

So f′(x) = 13x

−2/3.

. . . . . .

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)1/3 − x1/3

= 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�h((x+ h)2/3 + (x+ h)1/3x1/3 + x2/3

So f′(x) = 13x

−2/3.

. . . . . .

The cube root function and its derivatives

.f.f′

I Here limx→0

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

I Notice also limx→±∞

f′(x) = 0

. . . . . .

I Here limx→0

f′(x) = 0

. . . . . .

.f.f′

I Here limx→0

f′(x) = 0

. . . . . .

.f.f′

I Here limx→0

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

= limh→0

(x+ h)2/3 − x2/3

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

3x−2/3

(2x1/3

3x−1/3

So f′(x) = 23x

−1/3.

. . . . . .

One more

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2/3 − x2/3

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

3x−2/3

(2x1/3

3x−1/3

So f′(x) = 23x

−1/3.

. . . . . .

One more

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2/3 − x2/3

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

−2/3(2x1/3

3x−1/3

So f′(x) = 23x

−1/3.

. . . . . .

One more

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2/3 − x2/3

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

3x−2/3

(2x1/3

−1/3

So f′(x) = 23x

−1/3.

. . . . . .

One more

Example

Solution

f′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

(x+ h)2/3 − x2/3

= limh→0

(x+ h)1/3 − x1/3

h·((x+ h)1/3 + x1/3

3x−2/3

(2x1/3

3x−1/3

So f′(x) = 23x

−1/3.

. . . . . .

The function x 7→ x2/3 and its derivative

.f.f′

I f is not differentiable at 0and lim

x→0±f′(x) = ±∞

f′(x) = 0

. . . . . .

x→0±f′(x) = ±∞

f′(x) = 0

. . . . . .

.f.f′

x→0±f′(x) = ±∞

f′(x) = 0

. . . . . .

.f.f′

x→0±f′(x) = ±∞

f′(x) = 0

. . . . . .

y y′

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

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x2 2x1

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

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

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

y y′

x3 3x2

x1/2 12x

−1/2

x1/3 13x

−2/3

x2/3 23x

−1/3

. . . . . .

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.

. . . . . .

The other Tower of Power

. . . . . .

Outline

. . . . . .

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

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.

. . . . . .

Proof.We have

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.

. . . . . .

Proof.We have

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.

. . . . . .

Proof.We have

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.V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42

. . . . . .

Proof.We have

ckxkhn−k

. . . . . .

Proof.We have

ckxkhn−k

. . . . . .

Proof.We have

ckxkhn−k

. . . . . .

Pascal's Triangle

.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

. . . . . .

Pascal's Triangle

.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

. . . . . .

Pascal's Triangle

.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

. . . . . .

Pascal's Triangle

.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

. . . . . .

Proving the Power Rule.

Let n be a positive whole number. Then

xn = nxn−1

Proof.As we showed above,

(x+ h)n − xn

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.

. . . . . .

Proving the Power Rule.

Let n be a positive whole number. Then

xn = nxn−1

Proof.As we showed above,

(x+ h)n − xn

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

TheoremLet c be a constant. Then

.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

So f′(x) = limh→0

0 = 0.

. . . . . .

c = 0 .

.likeddx

x0 = 0x−1

f(x+ h)− f(x)h

=c− ch

0 = 0.

. . . . . .

c = 0 .

.likeddx

x0 = 0x−1

f(x+ h)− f(x)h

=c− ch

0 = 0.

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. . . .

. . . . . .

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

+ 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.

. . . . . .

(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)

. . . . . .

(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)

. . . . . .

(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)

. . . . . .

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)

. . . . . .

= limh→0

cf(x+ h)− cf(x)h

= c limh→0

f(x+ h)− f(x)h

= c · f′(x)

. . . . . .

= limh→0

cf(x+ h)− cf(x)h

= c limh→0

f(x+ h)− f(x)h

= c · f′(x)

. . . . . .

= limh→0

cf(x+ h)− cf(x)h

= c limh→0

f(x+ h)− f(x)h

= c · f′(x)

. . . . . .

Derivatives of polynomials

Example

Findddx

(2x3 + x4 − 17x12 + 37

Solution

(2x3 + x4 − 17x12 + 37

x4 +ddx

(−17x12

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

. . . . . .

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

(2x3 + x4 − 17x12 + 37

x4 +ddx

(−17x12

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

. . . . . .

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

(2x3 + x4 − 17x12 + 37

x4 +ddx

(−17x12

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

. . . . . .

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

(2x3 + x4 − 17x12 + 37

x4 +ddx

(−17x12

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

. . . . . .

Example

Findddx

(2x3 + x4 − 17x12 + 37

)Solution

(2x3 + x4 − 17x12 + 37

x4 +ddx

(−17x12

= 2ddx

x3 +ddx

x4 − 17ddx

x12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

. . . . . .

Outline

. . . . . .

Derivatives of Sine and Cosine.

sin x = ???

Proof.From the definition:

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

= sin x · 0+ cos x · 1 = cos x

. . . . . .

sin x = ???

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

. . . . . .

Angle addition formulasSee Appendix A

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

. . . . . .

sin x = ???

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

. . . . . .

sin x = ???

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

. . . . . .

Two important trigonometric limitsSee Section 1.4

.sin θ

.1− cos θ

.−1 .1

limθ→0

sin θθ

limθ→0

cos θ − 1θ

. . . . . .

Derivatives of Sine and Cosine

sin x = ???

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

. . . . . .

sin x = ???

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sin hh

= sin x · 0+ cos x · 1

= cos x

. . . . . .

sin x = cos x

sin x = limh→0

sin(x+ h)− sin xh

= limh→0

= sin x · limh→0

cos h− 1h

+ cos x · limh→0

sin hh

. . . . . .

Illustration of Sine and Cosine

.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?

. . . . . .

.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?

. . . . . .

.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?

. . . . . .

sin x = cos xddx

cos x = − sin x

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

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

= cos x · 0− sin x · 1 = − sin x

. . . . . .

sin x = cos xddx

cos x = − sin x

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

= cos x · limh→0

cos h− 1h

= cos x · 0− sin x · 1 = − sin x

. . . . . .

sin x = cos xddx

cos x = − sin x

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

= cos x · limh→0

cos h− 1h

= cos x · 0− sin x · 1 = − sin x

. . . . . .

sin x = cos xddx

cos x = − sin x

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

= cos x · limh→0

cos h− 1h

= cos x · 0− sin x · 1 = − sin x

. . . . . .

sin x = cos xddx

cos x = − sin x

cos x = limh→0

cos(x+ h)− cos xh

= limh→0

= cos x · limh→0

cos h− 1h

= cos x · 0− sin x · 1 = − sin x

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.

lesson 8: basic differentiation rules (section 21 slides)

derivative of f

h fx x

h0so f x

f decreasing

derivativeif f

x fx y dyf x

h fxf x

lim h0hh0 h

Documents

differentiation rules -...

3.3 rules for differentiation colorado national monument

differentiation 3 basic rules of differentiation the product...

+ follow lesson 11: implicit differentiation (section 21...

differentiation rules

2-2: differentiation rules objectives: learn basic...

lesson 11: implicit differentiation (section 41 slides)

differentiation rules - university of alaska...

tensor differentiation slides

differentiation rules 3. before starting this section, you...

rules for differentiation

11x1 t09 03 rules for differentiation (2011)

lesson 8: basic differentiation rules

lesson 11: implicit differentiation (slides)

chapter 3 – differentiation rules

3 algebraic rules of differentiation · differentiation we...

3.3 –differentiation rules

3 differentiation rules. 3.2 the product and quotient rules...

differentiation presentation slides

differentiation rules -...