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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