Calc 2.3b

Quotient rule applied to trig, finding second derivatives, third

derivatives, and so on!

Say what? Why would those be the derivatives? Let’s look at tan x:sin

tancos

xy x

x

2

cos ( [sin ]) sin ( [cos ])'

(cos )

d dx x x xdx dxyx

2 2

2 2

cos cos sin ( sin ) cos sin

cos cos

x x x x x x

x x

22

1sec

cosx

x

Now I want you students to figure out why the rest work. On the given sheet, show your group’s work. Step 1: Define as quotient. Step 2: DeriveStep 3: Simplify

Group 1: find y’ when y = csc x

Group 2: find y’ when y = sec xGroup 3: find y’ when y = cot x

2tan secd

x xdx

sin cosd

x xdx

sec sec tand

x x xdx

cos sind

x xdx

2cot cscd

x xdx

csc csc cotd

x x xdx

So how to memorize? Compare and contrast and come up with some patterns.

Ex 8 p.124 Differentiating Trigonometric Functions

Function Derivativea.y = x – tan x

b. y = x sec x

21 secdy

xdx

(sec tan ) sec (1)dy

x x x xdx

sec ( tan 1)x x x

Ex 9 p. 124 Different Forms of a Derivative

Differentiate both forms of

First form:

2nd Form:

Are these equivalent? Check it out!

1 coscsc cot

sin

xy x x

x

1 cos

sin

xy

x

2 2

2 2

sin (sin ) (1 cos )(cos ) sin cos cos'

sin sin

x x x x x x xy

x x

2

1 cos

sin

x

x

csc coty x x

2' csc cot cscy x x x

Much of the work in calculus comes AFTER taking the derivative. Characteristics of a simplified form?Absence of negative exponents

Combining of like terms

Factored forms

Higher-Order Derivatives

Just as velocity is the derivative of a position function, acceleration is the derivative of a velocity function.

s(t) Position Function. . . .v(t) = s’(t) Velocity Function. . . .

a(t) = v’(t) = s”(t) Acceleration Function.

a(t) is the second derivative of s(t) – which is the derivative of a derivative!

Notations for higher-order derivatives:

2 2

2 2

3 3

3

ddyFirst derivative: y' '( ) ( )dx dx

d2nd derivative: y'' ''( ) ( )

dx

d3rd derivative: y''' '''( )

d

f x f x

d yf x f x

dx

d yf x

dx

3

4 4(4) (4)

4 4

n n(n) ( )

n

( )x

d4th derivative: y ( ) ( )

dx

dnth derivative: y ( ) ( )

dxn

n

f x

d yf x f x

dx

d yf x f x

dx

Ex 10 p. 125 Finding Acceleration Due to GravityBecause the moon has no atmosphere, a falling object on the moon hits no air resistance. In 1971, astronaut David Scott showed that a hammer and a feather fell at the same rate on the moon. 2( ) 0.81 2s t t is the position

functionwhere s(t) is the height in meters and t is time in seconds. What is the ratio of the Earth’s gravitational force to the moon’s?To find acceleration due to gravity on moon, differentiate twice.2( ) 0.81 2 position

'( ) 1.62 velocity

''( ) 1.62 acceleration

s t t

s t t

s t

2

2

Earth's gravitational force 9.8 m/sec

Moon's gravitational force 1.62 m/sec

6.05

Assignment2.3b p. 126 #45, 61, 73-78, 83-87 odd, 93-101 odd