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ALTERNATE FORMS OF THE DERIVATIVE:

The numeric value of the derivative at a point x = a is also found by:

f'( )1. f(a+h)-f(a) a = mtan = 1m

h-->O h or by f' (a) = lim f(x) - f(a)

x-->a x-a

provided the limit exists. For all a for which the limit exists, f' ( x) is a function of a.

EX #3: Use the alternate form of the derivative to find f' (a) for f (x) = .../3x + 1 at a = 1.

f(1)= J3Ct)+ I f(\)= cO.

f/ra'= \i(Y) [J3x+I -.al/✓_3x+I +.;2]

\. l .J X➔ I X- I JLV3x+J f-cl-

+/(CAJ = \im X➔ 1

+/(a)� llM X➔l

f/(\)::: *

3X+I- 4 (x-0( f3x+I -t o1. ')

3 (� .=;;, -;::::::3=;=-�

�(✓3x+I +.:2) v'3(1)+1 + l

Observations for using the derivative formulas:

su bstifu-frlh' X=j.

When finding the derivative at a point, there are two approaches. You can either find the general formula first and substitute your x-value, or your can substitute the x-value from the start.

Question Types and Equivalent Terminology:

If given one point , ( a, f (a)), you are finding:

□ Instantaneous rate of change at x = a

□ Slope of the tangent line

□ Derivative at x = c

D f'(c) = lim f(a+h)-f(a)

h-->O h

© Jean Adams Flamingo Math, LLC

If given two points, (a,f(a)) and (b,f(b)),you are finding:

□ Average rate of change on the interval [ a, b]

□ Slope of the secant line

□ f(b)-f(a)

b-am =

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Understanding the Derivative S'/mbolicall'I

Oftentimes you will be asked to extend your knowledge base for operations in calculus where you need to recognize functions, procedures, processes, theorems, and properties. You might also need to "decompose or deconstruct" mathematical operations symbolically. Here is a sample of such instances.

EX #6: Each of the following expressions represents the derivative of some function y = f (x) forsome value, x = c. Identify the function y = f(x) and the value for x = c.

A. ...fftTT-2lim h

C.

h->O

+c)(1= rx

C, = �

tan2(2 + h) - tan(4)lim

h h->O

-t()() � w �)C

C,:::: �

E . (x -1)2 - 4· hm-----

x->3 X - 3

{(x) = ( :x-l);i

C= 3

© Jean Adams Flamingo Math, LLC

B. ecos(rc+h) _ e-1

D.

F.

limh->O h

lim-ln --

1 (

6 + h)

h->Oh 6

+()()::: Ln X

C, ::: lo

Vx-6 - v-slim ------

x->-2 X + 2

-{:()() = ?ix-6 C.== -:i

__

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Derivatives from a Table

One last thing before we go! We've seen that it's possible to estimate the numerical value of a derivative at a single point by finding the slope of the secant line passing through two points on the graph. Let's explore how this can be done using a table of values for a continuous function.

EX#9: Use the table of values for y = f(x) to find the derivative for eachx-value. Describe the behavior of the function and explain your reasoning.

x = a

-2

-1

0

2

3

Derivative Estimate

(show work)

/

-Ff-J) = -6,-'t - -J_Q

-lj+(+J)- -3

+(-1) = J+(+I) = -2.. -�-o -�

fco)= 4-'+ = 0-,-�

I f' 'l ,, _ - .f -: cl -t 3_rt" LrJJ - _ _.;;..... __ + 0-3 3

+ '(3):: Lt- 0�-8

= --�

Important Facts and Findings:

Function Behavior

atx = a

CJJns+ant

. . l nCfeaJIAj

decrea.s 1/\9

Justification

At (-d,�) f1-J) >0

At (01 - 1) --F.fd)=o rel . m 111 b�c -fC� OOP< -from nee to, nc:.

1. If f' (x) < 0, then the graph of the function is d_ e C(e_as 1'{) 3 atx = a.

2. If f'(x) > 0, then the graph of the function is J n�reas l/\g atx = a.

3. A relative maximum occurs on a function if f'(x) = 0 and the derivative changes from

pos,:bve to n�a:h ve4. A relative minimum occurs on a function if f' (x) = 0 and the derivative changes from

ne9qiJve to pos t:hVL

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EX #3: Given the graph of f(x) = -C 1

)2 x+2

A. Describe the derivative off as x approaches negativetwo from the left and the right.

as x ➔ -J+) -P tx)➔ -co

as x ➔-;i-J -Ptx)➔ QJ

. �--� ·;· .; .. .i.. . 1'···10 .,·.;. )--+-+··i .. .):::; �:./.J:: . -I /� ::J:::;:::C) :·;:-:... : .. : .. � .. : .) .. I .; .. . ;. J ) .•• ; ••• : •... • i .i .•. , ... , .! ... : I ; 6 ! .. i .•. i.. ,. �-· •. ; ··! .,. -: .. � . . , ·I· :·· .. , ... , ... ) .. ..:. i · ··,··-··:·:·.J·.�+·\4 ., '·······i··:···• � •• ..i •• � •• i i ... J .. l. : .. ·-�---l---� --�·-·'····-�---�- -J--�--� ·· ··-l··-· 2

·'.-·-1 • � +· , ...···'.···� --; •-:--J. �--1 .:.

· j �--·;·-·i --�-x

.-8 ;.1.;.-� .: .. ~t -i ··i---f--····� .] ..

B. Given the derivative f'(x) = cx:�)3 ,use theformula ···j·� .:. i··: ,-·,1,· !·2 ·;··1·' -r-j--

to evaluate the derivative when x = -2 f t-�-) -:. - � -:::: +oo

-P /(-�)= DN( (-J.r+-�)3

-f (-J.+

) � ( rttf P :-:- -.::i:iDif f erentiabilit-y Implies Continuit-y:

If/ is differentiable at x = c, then/ is continuous atx = c.

1. If a function is differentiable at x = c , then it is continuous at x = c .So, differentiability implies continuity.

2. It is possible for a function to be continuous at x = c and not be differentiable at x = c .Continuity does not imply differentiability.

SHARP CORNER, CUSP, VERTICAL TANGENT LINE or DISCONTINUITY

Write the contrapositive to the statement, "If a function is differentiable at x=c, then the function

is continuous at x=c.

1-F f(x) ,·s NOT C,6ntJnuous a;t, X=C

EX #4: Describe the x-values at which the function g(x) = lx2 - SI is differentiable.

Justify your answer.

x;2- 5-==0

x == :±fs

E:-eo,- J5) u (-J51 Js) U ( f5/XJ)

Sif\ce.. 3Cx) h� (A- sharpCJ>rner a:-1 x� -J5 o.NlX-= v5

© Jean Adams Flamingo Math, LLC

' a g(.v) • 7

• 6

X

-? -4 -� ·?- ·J -1 ? � 4 :i

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Higher Order Derivatives= To12ic 2.2: Defining the Derivativ� of g Function and !.!�ing Derivative NQtatiQn

First derivative: y' f'(x)

Second derivative: y" f"(x)

Third derivative: y"' f"' (x)

Fourth derivative: yC4) fC4)(x)

dy -dx

dzy -

dx2

d3ydx3

d4ydx4

d dx [t(x)]

dz

dx2 [f(x)]

d3

dx3 [t(x)]

d4 dx4 [f(x)]

EX #6: For each of the following, find the stated derivative.

A. Find/"(x) given f(x) = f x3 + Sx2

- 6x + 1

-F /= i x J + IDx-� +//(x)= qx+IO-

d32

d 3�➔ 8x1-C.�[smx]dx

-f/ ::: COJX 1 f /1 == -Sil' X 1

r pa.ttefl\-f /tt: -CDS)( rJ.. C3J.) _ ,

f(-lf-)= s if\x .. d)\(:3,1) -sin�

B. Find f<5)(x) if f(x) = -3x4 + 7x3 -10x

f,✓()(): - J � X 3+JI X .;2_ I 0

f//(x)= -31.ax�+�x -f/tt(x) =�7Jx + 4-Q..

-04->�x) "'-7;)..-fC5 Cx) = 0

D. Find y<867) (x) given y = cos x

� 1= -St'()X

<4 11= -eosx

� 1/I= 5ill Xt(if)-= ca s x

8�7 _ 6),/(p + 34 .... •- .

��1{:x)o: S1()X'_

EX #7: A potato launcher fires a potato downward, from the top of a 112-foot tall building. The potato hits the ground in two seconds. Find the initial velocity of the potato. The position function is given as s(t) = -16t2

+ v0 t + 112, where distance s(t) is in feet and time tis in seconds.

${�):o :)Vo = - '-f 3 - l� (J) � + Vo(J)+ 11;;2 ==-o Yo= - �411

� � -t ';JVo =o sec...

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