Chapter 3 – Differentiation Rules 3.1 – Derivatives of Polynomials and Exponential Functions Aka – The Short Cuts! Yay! f(x) = c f’(x) = 0 g(x) = x g’(x) = 1 h(x) = xn h’(x) = n xn-1 (The Power Rule) k(x) = c f(x) k’(x) = c f’(x) m(x) = f(x) + g(x) m’(x) = f’(x) + g’(x) (The Sum Rule) n(x) = f(x) – g(x) n’(x) = f’(x) – g’(x) (The Difference Rule) p(x) = ex p’(x) = ex

Ex: Prove !!" #% & = #%′(&) Note: Proof of Power Rule on page 175 uses the Binomial Theorem.

:et g ( x ) = c f ( x )

guttingeaten thing CFAIHIHD= c I in fkthf.fi#

.

his 0

= c f '

Cx )

Ex: Differentiate a) y = &+ b) f( r) = ,- +/0 c) 1"2

=P's (w4?xa=fxyttxtt ( xgttrd

''

=sx÷ x3a=x

= # =3# or

35¥ )

s÷aFE=3E3 3X

now

f G) =e''

+ xeorfcrkertre

ftp.extexetorftrterteret

a= 4×-2

y'=4fz)×→ ' '

= - 8×-3=-8×3

BewMath0#abinFM square Bear

T.jp#=2yrxx

34M¥I 3 Run

,

1 survives

y

y3-y×

Escape from New York

¥1 ¥9¥

¥*r÷x¥= 2×7

y

?¥Fa÷Fix¥¥⇒in

= 213ft .

¥#÷FEIZIZZIZZ

= 34€n¥F¥t±€¥I÷

= 84¥Zxy 23

Definition of e It’s clear from the study of exponential functions that the slope of the tangent to f(x) = ax at x = 0 is 1 for some value of a between 2 and 3. Mathematicians wanted to find that value. The slope of the tangent line at x = 0 is given by the first derivative at 0: f’(0) = lim

6→89 8:6 ;9(8)

6 = lim6→8

<=;<>6 = lim

6→8<=;?6 = 1

So the definition of e is 2345→6

75;85 = 1

3.2 The Product and Quotient Rules

99: ; ∙ = : =?

Ex: Find m’(x) if m(x) = x3x4. Is the derivative of m(x) equal to the product of the derivatives of its factors?

mcx )=x3x4=×7. MK ) = 7×6

£x' = 3×2 ¥ 4=4×3

*xD(4×3)=12×5 # 7×6

Geometric Explanation of the Product Rule (via Gottfried Wilhelm Leibniz, one of the two fathers of calculus along with Sir Isaac Newton)

Start with f(x)g(x), which can be visualized as the white rectangle above. Let’s imagine that we change x by an amount equal to ∆:. Now we have a new product, f(x0 + ∆:) g(x0 + ∆:), which is the larger rectangle. We see from the picture that Whole Rectangle = White Rectangle + Side Blue + Top Blue + Grey f(x0 + ∆:)g(x0 + ∆:) = f(x0)g(x0) + f(x0)∆= + g(x0)∆;+ ∆;∆= So the change in the product is just the blue and grey rectangles: ∆ ;= =f(x0 + ∆:)g(x0 + ∆:) – f(x0)g(x0) = f(x0)∆= + g(x0)∆;+ ∆;∆= If we divide the whole thing by ∆:, we get ∆(LM)∆" = L(N>)∆O∆" + M(N>)∆9∆" + ∆; ∆O

∆" Then, if we let ∆: → 0,

99: ; : = : = ; : =R : + = : ;′(:)

(Note: ∆; = f(x0 + ∆:) − ;(:) à 0 as ∆: à 0)

The Product Rule

KKL M L N L = M L K

KLN L + N L KKLM L

See page 187 for a similar explanation of what happens with the quotient of two functions:

The Quotient Rule

KKL

M LN(L) =

N L KKLM L − M L K

KLN L[N L ]Q

Ex: Differentiate a) f(x) = (x2 – 2x)(1 + ex)

Lowd high highd low

low low

f'

( x ) .

- ( x '.2x)F×a+e×)+£×( xtzx ) . ( HE )←

=(xtz×)( Hey'

+ (xtzxj ( He )

fEx)=(xZ2x)( 0+0+(2×-2)(HED= e×× '

. 2e×x +2×+20×-2-22= x2e×¥e×+2×+2*2-2 ex

= ×2e×+2× . 2e× -2

f ( × ) . # = f ( × ) .[gC×D"

b) g(x) = "2;["?:0\

Gtexkx '. ZD

'- ( xtzx )( Hey

'

gkxk -

( ltexjz

=( HE )( 2×-2 ) - ( x ?zD(e×)

TIE=

2×-2+2×5 '- 2e× -

x2e×t2xe×

ftp.#IKeI2e*@=2Cxy+eexxzIxd

3.3 Derivatives of Trigonometric Functions

))* (sin *) = cos *

))* (cos *) = −sin *

Ex: Using the above, find the derivative of y = tan x = abc "dea "

!!" (tan *) = sec[ * !!" (cot *) = −csc[ *

!!" (csc *) = − csc * cot * !!" (sec *) = sec * tan *

y =

wstz

¥

Egypt.

IT -1-

1

II 0 0

y1=Gnxkcosx ) - Csiwxkasxj

'

#=

Co5× - sinxfsnx )

#

=6s2x+sN2×t.ws#=seix

3.4 The Chain Rule

The Chain Rule If g is differentiable at x and f is differentiable at g(x), then the composite function F = >?@ defined by F(x) = f(g(x)) is differentiable at x and F' is given by the product

F'(x) = f'(g(x)) ·g'(x) In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions, then !j

!" =!j!k

!k!"

Ex: Differentiate 1) f(x) = (x+1)2 2) g(x) = 3 − *[+ 3) y = sin (x2)

*deriugtlusiofe

f'

G) = 2 #' '

. (1+0)der iv #

of inside

= 2 ( × + D'

. I = 2 (x + D= 2×+2

= ( 3. xy 's

g' G) = 's G- xI÷f2x)

= EYE or js¥ay

= cos ( xD . ZX

= 2 × cos ×2

Using the Chain Rule to come up with the derivatives of exponential functions: !!" E

" = E", but what is !!" F"?

First: ax = (eln a)x

Then: !!" F

" = !!" (Eno<)" = (Eno<)" !

!" (KLF)* = (Eno<)"KLF = ax ln a Question: What happens when a = e? £ a

×= a× In a

⇐ e×=e× he

= ex . 1 = e×

3.5 Implicit Differentiation Ex: If y = x2 + 4, find !j!" Ex: If y – x2 = 4, find !j!" Ex: If y2 – x2 = 4, find !j!" (Hint: Use the Chain Rule)

d£, = Zx + O

II .

- 2 x

¥ G) - E. GY = ¥ th

Th - 2 x = 0

+ 2 × + 2 x

+

date = 2 ×

€44 - £,

ay = £,

a )

2 y. ¥ - 2 × = 0

Zy ¥ = Zx

¥ = ¥¥

,= ¥

Ex: a) Find y’ if x2 + y2 = 25 b) Find the slope of the tangent line to the curve at x = 3 c) Looking at the graph, does part b make sense?

⇐ kY+¥G4=£,tD2×+241.31=02×+2/1×1=0

Zy ¥x= - Zx

E÷=¥=¥

×2ty2=25 (3,4 ) m=¥32+42=25 =-39+42=254

y2=l6 ( 3,

. 4) m= ¥y=±4

=÷=¥

.Easiesm=÷

Inverse Trig Functions If x = sin y, then y = sin-1x for -r[ < y < r[ (prounounced “arcsine”) But what’s the derivative? Use x = sin y, and solve for !j!": so !j!" = ?

deaj = ??;sto2j = ?

?;"2 Derivatives of Inverse Trig Functions: !!" sin-1x = ?

?;"2 !!" csc-1x = − ?

" "2;? !!" cos-1x = − ?

?;"2 !!" sec-1x = ?

" "2;? !!" tan-1x = ?

?:"2 !!" cot-1x = − ?

?:"2

X = SINY

¥( x ) = ¥ ( sny )1 = cosy . £¥

cosy + s n2y= 1

This is Bob f ( x )Bob is a guy f ( x ) = y

I threw a ball to Bob fYx )I threw a ball to a guy y

'

I threw a red ball to aguy dayI threw a green ball to a guy day

I threw a red ball to d-d ×

f ( xth ) -f⇐f4x)=y' = }¥=hsnop

difference

M = K¥ .

.

fcb ) . f(a )quotient

b=a=fcxz ) - f(× , )

ISlopes give vs change

Specifically the average rate

of changeXz= X. th for some h

m= f( x. th ) - f( xD th) - flxX+h=×,=fk#, )

And then limits !

f(x,t#)f(×)← Average rate of

Ling fcxith ) - f(× ,)change

=← Instantaneous

velocity

tin tC×'th#K)← usecharts

h→ 0 find

And then . Limit

Laws←

Tort

's !

di→mo*h¥ = high= lim 1 =/

h→0

%m h(×'# = lim ( xt's )

h→o h→o

= Xt 5

or ×2h+5h+h2lim ==dijn ( x

'+ 5th )

.

h→O

= × 't 5

3.6 Derivatives of Logarithmic Functions What is !!" #$%<' ? Start with y = #$%<', then (j = x. **' (j = *

*' (') (j ln (*1*' = 1 *1*' =

1(j ln (

So, !j!" =

?<xyz{\ |c < = ?

" |c < If y = ln x, then !j!" = ?

" |c 0 = ?"

In general, **'(ln <) =

<′<

Ex: Find f’(x) if f(x) = ln(x3 + 1)

4=109a×

"u is special

"

÷aftnsise °

f '(x)= ×}¥Ex : g

( × ) .

- In

(46×-3×2)find g' ( × ) f4fsnx)

gtxk-4*6×-1

465×-3×2

Cool Tricks: Ex: Find y’ if y = ln "

+:~";B["Ä;1"

Yuck – this one looks horrible, until you remember that: ln "

+:~";B["Ä;1" = ln ('~ + 3' − 5) - ?[ ln (2'1 − 4')

Ex: Find !!" '

[" (Hint: Try this y = '[" ln y = ln '[" ln y = 2x ln x)

-

÷¥e¥¥¥¥g¥¥*I÷÷i*¥In (F) =ha . KB xts

- 45k¥,

In A' = Una

±×[mu]= I

lny= Zxlnx

Y'

S§×(×¥ 'T= #

'

1 n' + 2× and'

y, ¥= Zlnx + Zx ( ⇒

¥= Zlnx +2

y '= ( Zlnxtz )yZX

= ( Zlnxt D ×

= zx" ( Inx + 1 )

3.7 Rates of Change in the Natural and Social Sciences Whoo hooo! Word problems!!!!! Ex: (#8) If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s(t) = 80t – 16t2. a) What is the maximum height reached by the ball? b) What is the velocity of the ball when it is 96 feet above the ground on its way up? On its way down? Notes: Marginal Cost = C’(x) ≈ The change in cost when you produce one more item Velocity Gradient = Instantaneous rate of change of velocity with respect to r ≈ dv/dr

µvertex x= ten

Ff '

:#Instant setV ( t ) = 80 -32T

= 0 200 - 100

¥¥¥t=÷=¥⇒.

} Taft.

SA )=ax2+bx+c✓ (E) =2ax+b

v (f) =O : 2axtb=O

-bZa×=- b

X=-b=

3.8 Exponential Growth and Decay Radioactive Decay Fact: Radioactive substances decay at a rate proportional to the remaining mass. Simply speaking, a big pile decays faster than a small pile. So, for m = mass and t = time, !Ñ!Ö = %&, for some constant k. In other words, m(t) = m0 ekt, where m0 is the initial mass. Ex: (#8) Strontium-90 has a half-life of 28 days. a) A sample has a mass of 50 mg initially. Find a formula for the mass remaining after t days. b) Find the mass remaining after 40 days. c) How long does it take the sample to decay to a mass of 2 mg? d) Look at the graph of the function. Does it make sense?

3.9 Related Rates This section offers applications of Implicit Differentiation. We’ll use our understanding of one change to compute another change – possibly one that’s hard to measure directly. Ex: Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/s. How fast is the radius of the balloon increasing in general? How about when the diameter reaches 50 cm? Hint: Volume of a Sphere = 1~ )*

~ We know: !â!Ö = 100 cm3/s We want to find: !-!Ö

Ex: Car A is traveling west at 50 miles per hour and car B is traveling north at 60 miles per hour. Both are headed for the intersection of the two roads. At what rate are the cars approaching one another when car A is 0.3 miles and car B is 0.4 miles from the intersection? Ex: A water tank has the shape of an inverted circular cone with base radius 2m and height 4 m. If water is being pumped into the tank at a rate of 2 m3/minute, find the rate at which the water level is rising when the water is 3 m deep. Hint: Volume of a cone = ?~ )*

[ℎ, and use similar triangles to eliminate r.

3.10 Linear Approximations and Differentials Let f(x) be a differentiable function. What is the equation of line tangent to f(x) at x = a? 1) We want the point (a, f(a)) to lie on the tangent line. 2) We want the slope of the line to equal f’(a). y – y1 = m (x – x1) y – f(a) = f’(a) (x – a) y = f(a) + f’(a) (x – a) This is called the linearization of f at a. Ex: Find the linearization of f(x) = x3/4 at x = 16

Differentials If f(x) is differentiable, then f’(x) = !j!" By multiplying both sides by dx, we get dy = f’(x) dx We call dx and dy differentials. But what do dx and dy represent? Since f’(x) gives us the slope of the tangent line, dy and dx represent values related to the tangent line. On the other hand, ∆x and ∆y represent values related to the original curve, y = f(x).

Ex: Let f(x) = x3 + x2 – 2x + 1. Compare ∆y and dx when x changes from 2 to 2.05. f(2.05) = f(2) = So ∆y = f(2.05) – f(2) =

But dy = f’(x) dx f’(x) = f’(2) = dx = ∆x = so dy = Note that dy is slightly easier to calculate than ∆y. Ex: a) The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere? If the radius of the sphere is r, then its volume is V = 1~ )*

~ If the error in the measured value of r is denoted by dr = ∆r, how can we calculate ∆V? Let’s approximate it with dV: dV = b) What’s the relative error in this example? Relative error =

∆ãâ ≈ !â

â This can be expressed either as a decimal or a percentage.

3.11 Hyperbolic Functions In the same way that sin x, cos x and tan x are related to the unit circle, x2 + y2 = 1,

sinh x = 0\;0ç\

[ csch x = ?abcé "

cosh x =

0\:0ç\[ sech x = ?

deaé "

tanh x = abcé "deaé " coth x = deaé "abcé " sinh , cosh and tanh are related to the hyperbola, x2 - y2 = 1

In this case, though, the argument isn’t equal to the angle, but to twice the area of the shaded region. Ex: Find !!" sinh E