lesson 8: basic differentiation rules

Section 2.3Basic Differentiation Rules

V63.0121.002.2010Su, Calculus I

New York University

May 24, 2010

Announcements

I Homework 1 due TuesdayI Quiz 2 Thursday in class on Sections 1.5–2.5

. . . . . .

Announcements

I Homework 1 due TuesdayI Quiz 2 Thursday in class

on Sections 1.5–2.5

V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 2 / 36

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 10 / 36

. . . . . .

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

. . . . . .

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

. . . . . .

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

. . . . . .

Let r be a positive whole number. Then

xr = rxr−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.

. . . . . .

Let r be a positive whole number. Then

xr = rxr−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.V63.0121.002.2010Su, Calculus I (NYU) Section 2.3 Basic Differentiation Rules May 24, 2010 20 / 36

. . . . . .

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.

. . . . . .

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.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.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 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

. . . . . .

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 hh

. . . . . .

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

. . . . . .

Two important trigonometric limitsSee Section 1.4

.sin θ

.1− cos θ

.−1 .1

limθ→0

sin θθ

limθ→0

cos θ − 1θ

. . . . . .

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 = ???

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 cosh− sin x sinh)− cos xh

= cos x · limh→0

cosh− 1h

− sin x · limh→0

sin hh

= 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

cosh− 1h

sin hh

= 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

cosh− 1h

sin hh

= 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

cosh− 1h

sin hh

= 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

cosh− 1h

sin hh

= cos x · 0− sin x · 1 = − sin x

. . . . . .

Summary

I Power Rule:ddx

xr = rxr−1

I Sum, Difference, Constant, and Constant Multiple RulesI Derivative of sin is cos. Derivative of cos is − sin.

lesson 8: basic differentiation rules

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