1a 1b

Table of Trig Functions

sin ( x ) cos ( x )

tan ( x )= sin ( x )cos ( x )

cot ( x )= cos ( x )sin (x )

sec ( x )= 1cos (x)

csc ( x )= 1sin (x )

Obtain co-functions in right column from left column by replacing sin ( x )→cos (x) and cos ( x )→sin (x).

Two Famous Trianglessketch 45°-45°-90° triangle, label angles and sides

Pythagorean theorem

( 1√2 )2

+( 1√2 )2

=1 or 12+12=1

sin( π4 )=cos( π4 )= 1√2

, tan( π4 )=1

sketch 30°-60°-90° triangle, label angles and sides

Pythagorean theorem

(√32 )2

+( 12 )2

=1 or 34 + 12=1

sin( π6 )=12 ,cos( π6 )=√32

, tan(π6 )= 1√3

sin( π3 )=√32

,cos ( π3 )=12 , tan(π3 )=√3

2a 2b

Derivatives of Trig Functions

ddxtan ( x )= d

dxsin ( x )cos ( x )

¿ (sin x )'¿¿

¿cos ( x )⋅cos (x )−sin ( x )⋅ (−sin ( x ) )

cos2 x

¿cos2 ( x )+sin2 ( x )

cos2 (x )

¿1

cos2 ( x ) ¿ sec2( x)

where sec ( x )=1 /cos (x ).

ddx

sec ( x )= ddx

1cos ( x )

¿ (1 )' ⋅cos x−1 ⋅ (cos x )'

cos2x

¿−¿¿

¿ 1cos x

⋅ sin xcos x

¿ sec x ⋅ tan x

similarly, obtain the following table (to memorize)

ddxsin ( x )=cos ( x ) d

dxcos ( x )=−sin ( x )

ddxtan ( x )=sec2 ( x ) d

dxcot ( x )=−csc2 ( x )

ddx

sec ( x )=sec (x ) ⋅ tan ( x ) ddx

csc ( x )=−csc ( x )⋅ cot(x)

to obtain the right hand column from the left hand column, replace functions by co-functions and multiply by −1.

?? ddx x sec(x)

?? ddθtan (θ)θ2+1

3a 3b

Example. Find an equation of the line tangent to the curve

y=x tan (x)

at the point ( π4 , π4 ).Solution.

Point-slope form of tangent line

y− y0=m(x−x0)

Need slope

dydx

=tan ( x )+x sec2(x)

dydx|π

4

=tan( π4 )+ π4sec 2( π4 )

where sec (π2 )= 1

cos( π4 )=√2

dydx|π

4

=1+ π4⋅2=1+ π

2

equation for tangent line

y− π4=(1+ π

2 )(x− π4 ) ■

§2.5 The Chain Rule

Composition

Example. Consider F ( x )=√x2+1

Let f (u )=√u, g ( x )=x2+1

F is the composition of f and g

Function machines:

x→ [g ]→x2+1=u→ [ f ]→√u=√ x2+1

■

Notation for composition

F ( x )=f (g (x ) )

4a 4b

f is the outer function,

g is the inner function

Derivatives of Compositions

Let y=F ( x )=f (g ( x ) )

or y=f (u), u=g (x).

The Chain Rule

If g is differentiable at x and f is differentiable at u, then

F ' ( x )= f ' (u ) g' (x)

The derivative of a composition of two functions is the product of their derivatives

Write another way

F ' ( x )= f ' (g ( x ) )g '(x )

Leibniz notation

Let y=f (u) and u=g (x) and

y=F ( x )=f (g ( x ) )

then

F ' ( x )= f ' (u ) g' (x)

dydx

=dydu

dudx

Easy to remember: looks as though a term “du” cancels on the right hand side!

Example. Let F ( x )=√x2+1. Find F '(x ).

Write as a composition

u=g (x )=x2+1

y=f (u )=√u where √u=u12

apply the chain rule

dydx

=dydu

dudx

5a 5b

¿ 12u

−12 ⋅2x

¿ (x2+1 )−12 x ¿

x√x2+1■

Example. Let y=cos (3 x ). Find dydx .

Write as a composition

y=cos (u) where u=3x

apply the chain rule

dydx

=dydu

dudx

¿−sin (u ) ⋅3

¿−3sin(3 x) ■

?? Class Practice

Let y=( x2+1 )10. Find dydx .

Let y=√sin ( x ). Find dydx .

Let y=tan (cos ( x ) )

6a 6b

Plausibility Argument for the Chain Rule

Recall the limit definition of derivative

dydx=f ' ( x )=lim

h→ 0

f ( x+h )−f (x)h

Change notation

Let Δ x=h

Δ y=f ( x+h )−f (x )

then

dydx

= limΔx→0

Δ yΔ x

Statement of Chain Rule

Let y=F ( x )=f (g ( x ) ) or y=f (u) with u=g (x)

If g is differentiable at x, and f is differentiable at u, then

F ' ( x )= f ' (u ) g' (x) or dydx=dydu

dudx

Plausibility Argument

Let Δu be the change in u corresponding to Δ x.

Δu=g ( x+Δ x )−g(x )

The corresponding change in y is

Δ y=f (u+Δu )−f (u)

Then

dydx

= limΔx→0

Δ yΔ x

¿ limΔx→0

Δ yΔu

ΔuΔ x as long as Δu≠0

¿ limΔx→0

Δ yΔu

⋅ limΔx→0

ΔuΔ x product rule for limits

but Δu→0 as Δ x→0, so

dydx

= limΔu→0

Δ yΔu

⋅ limΔx→0

Δ uΔ x

¿ dydu ⋅dudx

The only problem is that we may have Δu=0. ■

7a 7b

Generalized Power Rule

Combine the power rule and the chain rule

Consider

F ( x )=g ( x )n

write as a composition

u=g (x)

y=f (u )=un

apply the chain rule

dydx

=dydu

dudx

¿nun−1 dudx

this is frequently written

ddx

g ( x )n=ng (x )n−1g '( x)

Example. Consider

F ( x )=(x3+4 x )10

this has the form given above, with

g ( x )=x3+4 x

n=10

then

F ' ( x )=10 (x3+10x )9 ddx

(x3+4 x)

¿10 (x3+10x )9(3 x2+4 ) ■

Example.

F ( x )= 1√sin (x )

=(sin ( x ) )−12

F ' ( x )=−12

(sin ( x ) )−32 cos ( x) ■

WARNING It is a common mistake to forget the factor g' (x)!

8a 8b

Compositions of Three Functions

Consider

F (t )= f (g (h (t ) ) )

write as a composition

y=f (u ) ,u=g(x ) , x=h (t)

chain rule

dydt

=dydu

dudx

dxdt

it looks as though ‘du’ and ‘dx’ factors cancel

Example

F (t )=sin 2(4 t )

write as a composition

y=u2, u=sin ( x ) , x=4 t

chain rule

dydt

=dydu

dudx

dxdt

¿2u⋅ cos ( x )⋅ 4

¿8ucos (x)

¿8sin ( x )cos (x )

¿8 sin (4 t ) cos (4 t ) ■

9a 9b

Example

Find the line tangent to the graph of y=sin2 (4 t ) at

t= π16 .

Solution. Point slope form of a tangent line

y− y0=m ( t−t0 )

t 0=π /16 , y0=sin2( π4 )=( 1√2 )2

=12

slope? m= dydx|t= π

16

¿ 8sin (4 t ) cos (4 t )|t= π16

¿8sin( π4 )cos ( π4 )

¿8( 1√2 )( 1√2 ) ¿4

gives the tangent line

y−12=4(t− π

16 )

■

?? Class practice differentiating!

Find dydx

1. y=( x4−3 x2+6 )3

2. y=2x √x2+1

10a 10b

3. y=(x+ 1x2 )√7

4. y=t

1−t 2

5. y=sin ¿

6. y=cos (tan (√1+ t2 ))

11a 11b

§2.6 Implicit Differentiation

circle of radius 1…0…x-, …0… y-, circle

x2+ y2=1

defines y implicitly as a function of x

solve for y: y2=1−x2

two possibilities

y=(1−x2 )12=√1−x2 [1a]

y=−(1−x2 )12=−√1−x2 [1b]

derivatives

dydx

=12

(1−x2 )−12 (−2x )= −x

√1−x2[2a] [2a]

dydx

=−12

(1−x2 )−12 (−2x )= x

√1−x2[2b]

Using Implicit Differentiation is easier in many cases!

x2+ y2=1 [3][3]

regard y as a function of x

y=f (x ) or y2=f ( x )2

recall the generalized power rule

ddx

f ( x )2=2 f ( x ) f ' (x)

12a 12b

write this as

ddx

y2=2 y y '

must remember y '=dydx !

differentiate equation [3] implicitly with respect to x

2 x+2 y dydx

=0

solve for dy /dx

y dydx

=−x

dydx

=−xy

from [1], two possibilities

dydx

= −x√1−x2

dydx

= x√1−x2

these match [2]!

Example. Find dydx if

x−1+ y−1=1. [1]

Solution. Regard y as a function of x

y=f (x )

Use the generalized power rule

ddx

f ( x )−1=−f ( x )−2 f ' (x)

ddx

y−1=− y−2 dydx

differentiate [1] implicitly with respect to x

−x−2− y−2 dydx

=0

solve for dydx

13a 13b

− y−2 dydx

= 1x2

dydx

=− y2

x2 ■

Example. Find the equation of the line tangent to the curve

2cos ( x )sin ( y )=1 [1]

at (x0 , y0 )=( π4 , π4 ).

Solution. Recall sin( π4 )=cos( π4 )= 1√2 .

think of y as a function of x

ddxsin ( y )=cos ( y ) dy

dx

differentiate [1] implicitly

−2sin ( x )sin ( y )+2cos ( x )cos ( y ) dydx

=0

solve for dydx

2cos ( x )cos ( y ) dydx

=2 sin ( x ) sin( y)

dydx

=tan ( x ) tan( y )

slope of curve at x= y=π /4

dydx|x= y=π

4

= tan( π4 ) tan( π4 )=1 ⋅1=1 point-slope form of tangent line

y− y0=m(x−x0)

y− π4=x−π

4

y=x ■

14a 14b

?? Class practice with implicit differentiation

1. Let x3+ x2 y+4 y2=c, where c is a constant. Find dydx .

Solution dydx=−(3 x2+2xy )

x2+8 y

2. Let sin ( x )+cos ( y )=sin ( x ) cos( y) . Find dydx .

15a 15b

Solution dydx= cos (x)sin ( x )−1

cos ( y )−1sin ( y)

Find an equation of the line tangent to the following curve at the given point.

2 (x2+ y2 )2=25(x2− y2) at the point (3,1)

Solution. Differentiate implicitly with respect to x

4 (x2+ y2 ) (2 x+2 y y ' )=25(2x−2 y y ')

Solve for y '

4 (x2+ y2 )2 y y '+50 y y'=−8 (x2+ y2 ) x+50 x

y '¿

y '=−8 x3−8 x y2+50 x8 y (x2+ y2 )+50 y

¿ 2 x (−4 x2−4 y2+25)

2 y (4 x2+4 y2+25)

evaluate at x=3 and y=1

y '=6(−36−4+25)2(36+4+25)

=6(−15)2(65)

=−4565

=−913

point slope form of tangent line

y− y0=m ( x−x0 )

y−1=−913

( x−3 )

in slope intercept form

y=−913

x+ 2713

+1

¿ −913

x+ 4013

show Maple image of curve and tangent line ■

16a 16b

§2.7 Related Rates

Example. A descending balloonist lets helium escape at a rate of 10 cubic feet / minute. Assume the balloon is spherical. How fast is the radius decreasing when the radius is 10 feet?sketch balloon, basket, escaping gas

Given V=¿ volume of balloon

dVdt

=−10 (feet )3

minute

r=¿radius of balloon ¿10 feet

Find drdt when r=10

Equation relating these quantities

V= 43π r3 [1]

differentiate with respect to time using the chain rule

generalized power rule: ddt g (t )n=ng (t )n−1

consider radius r a function of time t

ddt

r3=3 r2 drdt

Differentiate [1] with respect to time to get

dVdt

=43π (3 r2 drdt )=4 π r2 drdt

Solve for dr /dt

17a 17b

drdt

= 14 π r2

dVdt

drdt |r=10= 1

4 π 1001

(feet )2(−10 ) (feet )3

min

¿ −140 π

feetmin

≈−0.008 feetmin

≈−0.1 inchesmin

■

Problem Solving Strategy

1. Read problem carefully

2. Draw picture if possible

3. Assign notation to given information and unknown rate

4. Write down an equation relating the given information and the function whose rate is unknown

5. Differentiate with respect to time

6. Solve for the unknown rate

Example. The length of a rectangle is increasing at a rate of 7 cm/s and its width is increasing at a rate of 7 cm/s. When the length is 6cm and the width is 4 cm, how fast is the area of the rectangle increasing?

2. picture

3. Given information

L(t)=6cm, W ( t)=4cm

L' ( t )=7 cm/s, W ' ( t )=7 cm/s

Unknown rate is A' (t), where A=LW .

18a 18b

4. A (t )=L ( t )W (t)

5. A' ( t )=L' ( t )W (t )+ L ( t )W '( t)

6. ¿7cms ⋅4 cm+6cm ⋅7cms

¿28cm2

s+42cm

2

s=70cm

2

s ■

19a 19b

Example. One end of a 13 foot ladder is on the ground and the other end rests on a vertical wall. If the bottom end is drawn from the wall at 3 feet/second, how fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?

…x…x-, … y… y-, ladder betw. x & y, l=¿ 13 ft

given information

l=13 ft , x=5 ft , dxdt =3 ft./sec.

unknown rate

dydt

=¿?

Pythagorean theorem

x2+ y2=l2

differentiate with respect to time

2 x dxdt

+2 y dydt

=0

solve for y

dydt

=−xy

dxdt

to evaluate dy /dt we need y

y2=l2−x2=169−25=144

y=√144=12

then

dydt |x=5 ft=−5

12⋅3 feetsec

=−54feetsec

■

Example. Pat walks 5 feet/second towards a street light whose lamp is 20 feet above the ground. If Pat

20a 20b

is 6 feet tall, find how rapidly Pat’s shadow changes in length.

ground, lamp, Pat, x, s, y, h

given information

y=20 feet, h=¿6 feet, dxdt =−5feetsec

unknown rate

dsdt

=?

by similar triangles

yx+s

=hs

y s=h (x+s)

( y−h ) s=hx

s= hy−h

x

differentiate wrt time

dsdt

= hy−h

dxdt

¿ 620−6 (−5feetsec )

¿−157feetsec ■

Example. The beacon of a lighthouse 1 mile from the shore makes 5 rotations per minute. Assuming that

21a 21b

the shoreline is straight, calculate the speed at which the spotlight sweeps along the shoreline as it lights up sand 2 miles from the lighthouse.

0…x…x-, 0… y… y-, lighthouse at (0 , y ), beam to (x ,0), length L, angle θ

given information

y=1 mile L=¿ 2 miles

dθdt

=5 revolutionsminute ¿5 revolutionsminute

2π radiansrevolution

¿10π radiansminute

unknown rate dxdt =?

from the definition of tangent

tan (θ )= xy

differentiate wrt time

sec2 (θ ) dθdt

= 1ydxdt using that y is a constant

dxdt

= y sec2 (θ ) dθdt

we need sec2 (θ )

cos (θ )= yL sec (θ )= L

y=2

then

dxdt

=(1mile) ⋅4 ⋅10π radiansminute

¿40 πmilesminute

¿40 πmilesminute60minuteshour

¿ (40 ) (60 )πmileshour

≈7539mileshour ■ STOP

22a 22b

§2.8 Linear Approximations and Differentials

Linear or Tangent Line Approximation

0…a…x-, 0… y-, y=f (x ), P (a , f (a ) )

Point-slope form of the tangent line

y= y0=m(x−x0)

Point (x0 , y0 )=(a , f (a ) )

Slope m=f ' (a )

Then

y−f (a )=f ' (a ) ( x−a ) or

y=L ( x )=f (a )+ f ' (a )(x−a)

L(x) approximates f (x) near x=a

Example. Find the linear approximation to f ( x )=√1+x near x=0.

f ( x )= (1+x )12

f ' ( x )=12(1+x )

−12

f ' (0 )=12

L ( x)=f (0 )+ f ' (0 ) ( x−0 )

¿1+ 12 x

this straight line approximates √1+x near x=0.

−1…0…1…x-, 0…1… y-, f (x), L(x)

■

23a 23b

Example (continued). Use linear approximation to estimate √1.1.

√1.1=√1+0.1=f (0.1)

L (0.1 )=1+ 12

(0.1 )=1.05

Exact value √1.1=10.488…■

Example (continued). For what values of x is the linear approximation

√1+x≈1+ 12 x

accurate to within 0.05? This means

|L ( x )−f ( x )|<0.05

or

L ( x)−0.05< f ( x )<L ( x )+0.05

show Maple output

From the plot, the linear approximation is accurate to within 0.05 for

−0.54<x<0.74■

Realistic Example. Estimate the amount of paint required to paint a spherical tank of radius 20 feet with a coat 0.01 inches thick.

Exact calculation: volume of sphere

V (r )=43π r3

volume between two concentric spheres of radii r and a

ΔV=V (r )−V (a )=43π r3−4

3π a3

a=20 feet=240 inches

r=240.01 inches

ΔV ≈7238.53cubic inches ≈31gallons

1 gallon ¿ 231 cubic inches

24a 24b

Approximate calculation:

Linear approximation, usual notation

f ( x )≈ f (a )+ f ' (a)(x−a)

Use V (r ) instead of f (x)

V (r )≈V (a )+V ' (a )(x−a)

ΔV=V (r )−V (a )≈V ' (a )(r−a)

V ' (r )= ddr43π r3 ¿ 43 π (3 r 2) ¿4 π r2

and

V ' (a )=4 π a2

then

ΔV=4 π a2 (r−a )

¿ surface area of sphere × thickness of paint

¿4 π (240 inches )2(0.01 inches)

≈7237.23 cubic inches

the error is only 0.3 cubic inches! ■

STOP Differentials

0…x-, 0… y-, f , P, L, x, x+Δ x

tangent line approximation

L ( x+ Δ x )=f ( x )+f ' ( x )Δ x

write

dx=Δ x

dy=L ( x+Δ x )− f (x )

¿ f ' ( x )dx

add dx=Δ x , Δ y ,dy=f ' (x )dx

Δ y=f ( x+Δ x )−f (x )

is approximated by

dy=f ' ( x)dx

Note that riserun=dydx

=f '(x ). This is clever notation.

25a 25b

Example. Estimate the amount of paint required to paint a spherical tank of radius 20 feet with a coat of paint 0.01 inches thick.

Volume of sphere

V (r )=43π r3

approximate amount of paint

dV=V ' (r )dr=4 π r2dr

if dr=0.01 inches, then

dV=4 π (240 inches )2(0.01 inches)

the same result as above. This is a shorthand for linear approximation. ■

Newton’s Method – An Application of Linear Approximation

0…r…x-, 0… y-, f

A root of f is a number r such that f (r )=0.

How to find roots geometrically?

guess x1add

let L1 be the line tangent to f at x1add

follow L1 to the x axis, get an intersection at x2add

repeat this process

if the initial guess was good, the sequence x1 , x2 ,… rapidly converges to r

How to find roots algebraically?

Make an initial guess x1.

Find the line tangent to f at x=x1.

L1 ( x )= f (x1 )+f '( x1)(x−x1) by the linear or tangent line approximation!

L1 intersects the x axis at x2. Find x2.

L1 (x2 )=0=f (x1 )+ f ' (x1)(x2−x1)

−f (x1 )=f '( x1)(x2−x1)

26a 26b

−f ( x1)f ' (x1)

=x2−x1

x2=x1−f (x1)f ' (x1)

Find the line tangent to f at x=x2.

L2 (x )= f (x2 )+f ' (x2)(x−x2)

L2 intersects the x axis at x3. Find x3.

L2 ( x3 )=0=f (x2 )+f ' (x2)(x3−x2)

−f (x2 )=f '( x2)(x3−x2)

−f ( x2)f ' ( x2)

=x3−x2

x3=x2−f (x2)f '( x2)

repeat this procedure to obtain a general formula for n=1 ,2 ,…

xn+1=xn−f (xn)f ' (xn)

If the initial guess x1 was good, the iterates x2 , x3 ,… will quickly approach r.

Example. f ( x )=x2−2

−2…−1…0…1…2…x-, −2…0…2…y-, f ,r=√2

f ' ( x )=2 x

from the boxed formula above

xn+1=xn−xn2−22 xn

guess x1=2

x2=2−4−24

=2−12=32

x3=32−

94−2

3=32−

143

=32− 112

=18−112

=1712

=1.4167…

Exact value of positive root

r=√2=1.4142…

Transparency – How Newton’s Method can go wrong!

27a 27b