When gear A makes x turns, gear B makes u turns and gear C makes y turns.,

3.6 Chain rule

1

2

dy

du

3du

dx

y turns ½ as fast as u

u turns 3 times as fast as xSo y turns 3/2 as fast as x

dy dy du

dx du dx

Rates are multiplied

The Chain Rule for composite functions

If y = f(u) and u = g(x) then y = f(g(x)) and

dy dy du

dx du dx multiply rates

( ( ) ( ( )) ( )dy d

f g x f g x g xdx dx

multiply rates

Find the derivative (solutions to follow)

2 71) ( ) (3 5 )f x x x

2 232) ( ) ( 1)f x x

2

73) ( )

(2 3)f t

t

4) ( ) sin(2 )f x x

25) ( ) tan( 1)f x x

Solutions

2 71) ( ) (3 5 )f x x x

2 232) ( ) ( 1)f x x

dy dy du

dx du dx

2

7 6

3 5 3 10

7

u x x x

y

du

dy

du

x

u u

d

6 3 10 )7 (dy

ud

xx

2 67(3 2 ) (3 10 )dy

x x xdx

2

2 1

3 3

1 2

2

3

u x xd

dy u u

y

du

u

dx

1

32

3(2 )

du

y

dx

x

12 32

( 1)3

(2 )dy

dxx x

12 3

4

3( 1)

dy x

dxx

Solutions dy dy du

dx du dx

2 3

2 3 2

7 14

u t

y u

d

d

d

u

d

u

x

uy

314 2udy

dx

33

2814(2 3) (2)

(2 3)

dyt

dx t

2 2

sin( ) cos( )

du

d

u x

yy

d

u udu

x

2co 2 (2 )s( )dy

ud

cos xx

22

73) ( ) 7(2 3)

(2 3)f t t

t

4) ( ) sin(2 )f x x

25) ( ) tan( 1)f x x 2

2

1 2

tan( ) sec ( )

u x x

y udy

du

du

dx

u

2 2 22 secsec ( )(2 )) ( 1dy

u x xx

xd

Outside/Inside method of chain rule

( ( ) ( ( )) ( )dy d

f g x f g x g xdx dx

insideoutside derivative of outside wrt inside

derivative of inside

think of g(x) = u

Outside/Inside method of chain rule example

1

2 33 1 ( ( )) ( )d

x x f g x g xdx

inside

outside

derivative of outside wrt inside

derivative of inside

1 2

2 23 313 1 3 1 6 1

3

dx x x x x

dx

2

2 3

6 1

3 3 1

x

x x

Outside/Inside method of chain rule

33sin sin ( ( )) ( )d d

f g x g xdx dx

inside

outside

derivative of outside wrt inside

derivative of inside

3 2sin 3 sin sin

d d

dx dx

23sin cos

Outside/Inside method of chain rule

2csc( 3) ( ( )) ( )d

f g x g xdx

insideoutsidederivative of outside wrt inside

derivative of inside

2 2 2csc( 3) csc( 3)cot( 3)2d

dx

2 22 csc( 3)cot( 3)

1

2 2 2( ) 1f x x x

More derivatives with the chain rule2 2( ) 1f x x x

1 1

2 2 2 22 2( ) 1 1d d

f x x x x xdx dx

1 1

2 2 22 21( ) 1 ( 2 ) 1 2

2f x x x x x x

132 2

12 2

( ) 1 2

1

xf x x x

x

1 12 22 23 3 2 3 3

1 1 1 12 2 2 22 2 2 2

1 2 1 (1 )2 2 2( )

1 1 1 1

x x xx x x x x x xf x

x x x x

3

12 2

3 2( )

1

x xf x

x

product

Simplify terms

Combine with common denominator

More derivatives with the chain rule3

2

3 1( )

3

xf x

x

2

2 2

3 1 3 1( ) 3

3 3

x d xf x

dxx x

2 2

2 22

3 1 ( 3)3 (3 1)(2 )3

3 3

x x x x

x x

2 2 2

2 22

3 1 (3 9) (6 2 )3

3 3

x x x x

x x

2 2 2

2 22

2 2

42

3(3 1) ( 3 2 9)3 1 3 9 6 23

3 3 3

x x x x

x x

x x x

x

Quotient rule

The formulas for derivatives assume x is in radian measure.sin (x°) oscillates only /180 times as often as sin (x) oscillates. Its maximum slope is /180.

Radians Versus Degrees 1180

radians

d/dx[sin (x)] = cos (x)

d/dx [sin (x°) ] = /180 cos (x°)

3.7 Implicit Differentiation

Although we can not solve explicitly for y, we can assume that y is some function of x and use implicit differentiation to find the slope of the curve at a given point

y=f (x)

If y is a function of x then its derivative is dy

dx

y2 is a function of y, which in turn is a function of x.

2 2d dy

y ydx dx

sind

ydx

2 3dx y

dx

1

2dy

dx

1

21

2y

dy

dx

cos ydy

dx

2 2 33 2dy

dxx y y x

using the chain rule:

Find the following derivatives wrt x

Use product rule

Implicit Differentiation

4. Solve for dy/dx

2 2 sin( )y x xy

2 2 cos( ) (1)dy dy

y x xy x ydx dx

2 cos( )

2 cos( )

dy x y xy

dx y x xy

2 2 cos( )( ) cos( )dy dy

y x xy x xy ydx dx

2 cos( )( ) 2 cos( )dy dy

y xy x x xy ydx dx

(2 cos( )) 2 cos( )dy

y x xy x y xydx

1. Differentiate both sides of the equation with respect to x, treating y as a function of x. This requires the chain rule.

2. Collect terms with dy/dx on one side of the equation.

3. Factor dy/dx

Find equations for the tangent and normal to the curve at (2, 4).

Use Implicit Differentiation

find the slope of the tangent at (2,4)find the slope of the normal at (2,4)

1. Differentiate both sides of the equation with respect to x, treating y as a function of x. This requires the chain rule.

2. Collect terms with dy/dx on one side of the equation.

3. Factor dy/dx

4. Solve for dy/dx

3 3 9 0x y xy Solution

2 23 3 (9 9) 0dy dy

x y x ydx dx

2 23 9 9 3dy dy

y x y xdx dx

2 23 3 9 9 ) 0dy dy

x y x ydx dx

2

2

9 3

3 9 )

dy y x

dx y x

2 2(3 9 ) 9 3dy

y x y xdx

2

tan 2

9(4) 3(2) 24 4

30 53(4) 9(2)m

5

4normalm

Find dy/dx2 2 2( )x y x y

1. Write the equation of the tangent line at (0,1)

2. Write the equation of the normal line at (0,1)

1. Differentiate both sides of the equation with respect to x, treating y as a function of x. This requires the chain rule.

2. Collect terms with dy/dx on one side of the equation.

3. Factor dy/dx

4. Solve for dy/dx

Solution 2 2 2( )x y x y

2 21 2( )(2 2 )dy dy

x y x ydx dx

3 2 2 31 4 4 4 4dy dy dy

x x y xy ydx dx dx

2 3 3 24 4 1 4 4dy dy dy

x y y x xydx dx dx

2 3 3 2(1 4 4 ) 1 4 4dy

x y y x xydx

3 2

2 3

1 4 4

1 4 4

dy x xy

dx x y y

Find dy/dx2 2 2( )x y x y

1. Write the equation of the tangent line at (0,1)

2. Write the equation of the normal line at (0,1)

3 2

2 3

1 4 4

1 4 4

dy x xy

dx x y y

1 11 ( 0) 1

3 3y x or y x

1 3( 0) 3 1y x or y x

3.8 Higher Derivatives

The derivative of a function f(x) is a function itself f ´(x). It has a derivative, called the second derivative f ´´(x)

If the function f(t) is a position function, the first derivative f ´(t) is a velocity function and the second derivative f ´´(t) is acceleration.

2

2( )

d yf x

dx

The second derivative has a derivative (the third derivative) and the third derivative has a derivative etc.

3

3( )

d yf x

dx

4(4)

4( )

d yf x

dx ( ) ( )

nn

n

d yf x

dx

Find the second derivative for ( )1

xf x

x

2 2

( 1)(1) (1) 1( )

( 1) ( 1)

x xf x

x x

22

1( ) ( 1)

( 1)

d df x x

dx x dx

33

2( ) 2( 1) (1)

( 1)f x x

x

Find the third derivative for ( )1

xf x

x

In algebra we study relationships among variables

•The volume of a sphere is related to its radius •The sides of a right triangle are related by Pythagorean Theorem•The angles in a right triangle are related to the sides.

In calculus we study relationships between the rates of change of variables.

How is the rate of change of the radius of a sphere related to the rate of change of the volume of that sphere?

Examples of rates-assume all variables are implicit functions of t = time

Rate of change in radius of a sphere

Rate of change in volume of a sphere

Rate of change in length labeled x

Rate of change in area of a triangle

dr

dt

dV

dt

dx

dt

d

dt

Rate of change in angle,

dA

dt

3.9

Solving Related Rates equations

1. Read the problem at least three times.2. Identify all the given quantities and the quantities

to be found (these are usually rates.)3. Draw a sketch and label, using unknowns when

necessary.4. Write an equation (formula) that relates the

variables.5. ***Assume all variables are functions of time and

differentiate wrt time using the chain rule. The result is called the related rates equation.

6. Substitute the known values into the related rates equation and solve for the unknown rate.

Figure 2.43: The balloon in Example 3.Related RatesA hot-air balloon rising straight up from a level field is tracked by a range finder 500 ft from the liftoff point. The angle of elevation is increasing at the rate of 0.14 rad/min. How fast is the balloon rising when the angle of elevation is is /4?

Given: 0.14 / min

drad

dt

500x ft

500x

Find:

4

dywhen

dt

y

Figure 2.43: The balloon in Example 3.Related RatesA hot-air balloon rising straight up from a level field is tracked by a range finder 500 ft from the liftoff point. At the moment the range finder’s elevation angle is /4, the angle is increasing at the rate of 0.14 rad/min. How fast is the balloon rising at that moment?

x

tan500

y

y

tan500

d d y

dt dt

2 1sec

500

d dy

dt dt

2 1sec ( )(.14)

4 500

dy

dt

2

1 1(.14)

500cos ( )4

dy

dt 140 / sec

dyft

dx

Figure 2.44: Figure for Example 4.Related RatesA police cruiser, approaching a right angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. The cruiser is moving at 60 mph and the police determine with radar that the distance between them is increasing at 20 mph. When the cruiser is .6 mi. north of the intersection and the car is .8 mi to the east, what is the speed of the car?

Given: 20

60

dsmph

dtdy

mphdt

Find:.8, .6

dxwhen x y

dt

Figure 2.44: Figure for Example 4.

Given: 20

60

dsmph

dtdy

mphdt

Find:.8, .6

dxwhen x y

dt

2 2 2s x y

2 2 2ds dx dy

s x ydt dt dt

then s = 1

2(1)(20) 2(.8) 2(.6)( 60)dx

dt

70dx

mphdt

Figure 2.45: The conical tank in Example 5.Related RatesWater runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base of radius 5 ft. How fast is the water level rising when the water is 6 ft. deep?

Given: 39 / min

10 , 5

dVft

dtH ft R ft

Find:

6 .dy

when y ftdt

Figure 2.45: The conical tank in Example 5.Water runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base of radius 5 ft. How fast is the water level rising when the water is 6 ft. deep?

Given: 39 / min

10 , 5

dVft

dtH ft R ft

Find:

6 .dy

when y ftdt

21

3V x y

5

10

x

y 2y x

2 31 2(2 )

3 3V x x x

22dV dx

xdt dt

29 2 (3)dx

dt

x=3

1

2

dx

dt

1 12* 2* .32 / min

2

dy dxft

dt dt

3. .10 The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent.

Differentiability

Differentiability and Linearization

Approximating the change in the function f by the change in the tangent line of f.

Linearization

Write the equation of the straight line approximation co (0 1)s ,y x x at

1 sin 1 0 (0,1)1 ay x t

1 1( )y y m x x

( ) ( )( )y f a f a x a

Point-slope formula

y=f(x)

1 1( 0)

1

y x

y x