Question

limits?

Answer 1

do you have any specific questions?

Answer 2

Maybe just a defintion first :S

Answer 3

I actually can make a "cheat sheet" and put whatever I want on the sheet

Answer 4

And my professor gave us a list of topics and the first is limits so I know I should put examples on there

Answer 5

but like idk where to start :S

Answer 6

:/

Answer 7

"the value that a function approaches as the domain approaches a certain value"?

Answer 8

if your prof. uses delta/epsilon definitions, include those as well

Answer 9

no he doesn't use those

Answer 10

its calc IA so

Answer 11

alright

what else do you want to know?

Answer 12

maybe some examples of problems with limits :S

Answer 13

Ugh I'm not sure

Answer 14

https://www.math.ucdavis.edu/~kouba/ProblemsList.html

try the first two links "limit of a function as x approaches a constant" and "limit of a function as x approaches infinity"

Answer 15

but the general approach: first you try plugging in x, if you get something undefined from that, try other techniques like l'hopital's or re-writing the function

Answer 16

SO for this i got the 0 over o part but Idk what is going on after could you just explain it to me

Answer 17

since 0/0 is undefined they factored the numerator and denominator, cancelled out

then they cancelled out the (x-2) from the numerator and denominator

after that, you can now plug in x = 2 and get something defined

Answer 18

special note: 0/0 is called indeterminate form

Answer 19

oh so basically just take the dactored form and subsitute in that

Answer 20

yes

Answer 21

Gonna just do a few more practice problems and then I'll be back in like 5 mins

Answer 22

Don't get how this wouldn't be zero

Answer 23

infinity's not really a number so "infinity minus infinity" isn't really 0

in this case the higher exponent, at very very high values, makes the lower exponent kind of irrelevant, so the function will just keep increasing at x ---> infinity

Answer 24

Next topic product rule =)

Answer 25

for derivatives?

Answer 26

Yes

Answer 27

if you have a specific example we can work through it together i gues

Answer 28

would def. recommend memorizing this or writing it down on your cheat sheet

Answer 29

Yeh, I just wrote it

Answer 30

OKay I have handouts he gave so I can just show that and we can work through that and I will just write them as "practice problems" on the cheat sheet?

Answer 31

Do you think that woul dbe good

Answer 32

or should I do something else to practice product rule

Answer 33

idk if you should put them on the cheat sheet (that takes up valuable space) but working through them would be good

Answer 34

I have the handouts all mixed up but this would be on where i would use product rule right

Answer 35

yes

Answer 36

a) is asking for the derivative of the sum, not the product

but for the rest, you should use the product rule

Answer 37

sorry trying to take notes as I go

Answer 38

the derivate of x^2 is 2x right

Answer 39

and the derivative of g(x) is um 3/2+1

Answer 40

g(x) = (1+x)^(3/2)

bring the exponent down --> (3/2)

reduce the original exponent by 1 ---> (1+x)^(1/2)

take the derivative of the base ---> derivative of 1 + x is just 1

so g'(x) = (3/2)(1+x)^(1/2) * 1

Answer 41

okay for the sum I just do( (3/2)(1+x)^(1/2) * 1) + (2x)

Answer 42

yes

Answer 43

do i actually solve or leave as is

Answer 44

there's nothing to solve for, it's asking for the derivative so that's it

Answer 45

okay b would be ((2x) * (1+x)^3/2) + (3/2(1+x)^1/2*1)*x^2

Answer 46

yeah

idk how picky they are with simplifying, but that's mathematically correct

Answer 47

im not sure how to do c

Answer 48

it's the same logic as c, but instead of g(x) you just use f(x) again

so derivative of f(x)*f(x) = f'(x)*f(x) + f(x)*f'(x)

Answer 49

remember that f and g are just arbitrary labels for the two component functions

Answer 50

2x*x^2 + x^2 * 2x

Answer 51

good

in this case, the algebraic simplification is very easy so they would probably expect you to simplify this down to 4x^3

Answer 52

okay so for D it woul just be (3/2(1+x)^1/2*1 * (1+x)^3/2) + (1+x)^3/2) * (3/2(1+x)^1/2*1

Answer 53

yes

since both terms are like terms you can add them to get

3(1+x)^1/2 * (1+x)^3/2) = 3(x+1)^2

Answer 54

if you're in a rush just try to get the derivative part right, and only simplify if you have time

Answer 55

Yeah thats what Im gonna do like at the end if I have time ill simplify

Answer 56

okay so for e : it woul djust be (f' * g )+(g'*f)(f' * g )+(g'*f)

Answer 57

?

Answer 58

good attempt but you need to treat "f(x) + g(x)" as one whole unit

derivative of [(f(x) + g(x)) * (f(x) + g(x))] --->

[ f(x) + g(x) ]' * [f(x) + g(x)] + [ f(x) + g(x) ] * [ f(x) + g(x) ]'

Answer 59

I'm not sure where i would use prime and where not :S

Answer 60

just fill in [ f(x) + g(x) ]' with your solution from part a) since you've already done that

Answer 61

oh okay

Answer 62

( (3/2)(1+x)^(1/2) * 1) + (2x) * ( (3/2)(1+x)^(1/2) * 1) + (2x)

Answer 63

gonna use different variables to make this more clear

d/dx of a(x) * b(x) --> a'(x)b(x) + a(x)b'(x)

replace a(x) with "f(x) + g(x)"
replace b(x) with "f(x) + g(x)"

d/dx of [f(x) + g(x)] * [f(x) + g(x)] --> [f(x) + g(x) ]'[f(x) + g(x)] + [ f(x) + g(x) ]'[f(x) + g(x)]

Answer 64

uh

Answer 65

it would just be the answer from part a being multiplies by each othr rigt

Answer 66

because part a is asking for f(x)+g(x)

Answer 67

no it's a little more complicated than that

Answer 68

so we know that

f(x) + g(x) = x^2 + (1+x)^(3/2)

and

[f(x) + g(x)]' = 2x + (3/2)(1+x)^(1/2)

so plugging both of these into the product rule:

[f(x) + g(x) ]'[f(x) + g(x)] + [ f(x) + g(x) ]'[f(x) + g(x)]

---> [x^2 + (1+x)^(3/2)]*[ 2x + (3/2)(1+x)^(1/2) ] + [ 2x + (3/2)(1+x)^(1/2) ] * [x^2 + (1+x)^(3/2)]

Answer 69

oh okay so we don't only take the answer from part one but combine them to get it

Answer 70

I see

Answer 71

do you think there will be problems with the product rule with actual numbers?

Answer 72

wait Water these are numbers

Answer 73

wth*

Answer 74

Sorry my brain is fried

Answer 75

okay i think I got the product rukle down

Answer 76

have you learned quotient rule yet? that's another good one to add

Answer 77

That was gonna be next =)

Answer 78

the lo d hi stuff is a mnemonic device, just write down the top part

Answer 79

which one is the qr one

Answer 80

the one where you're asked to take the derivative of a quotient (f(x)/g(x)) or something similar

Answer 81

just be aware you can't just take the derivative of the numerator/derivative of the denominator and call it a day, it follows a very specific format (see above)

Answer 82

ut would it be similar. I would start by just taking the individuals derivatives of f(x), g(x), and h(x) right

Answer 83

yes

Answer 84

the derivative of 1/x = 1/(x^2) :S ??

Answer 85

almost, -1/x^2

Answer 86

1/x = x^(-1)

bring the exponent down --> -1
reduce the exponent by 1 --> x^(-2)

so -1/x^2

Answer 87

and then 2x e ?

Answer 88

then we would do


for a


(2x/(-1/x^2))

Answer 89

again, you cannot just take

derivative of numerator/derivative of denominator

Answer 90

derivative of f(x) = 2x
derivative of g(x) = (-1/x^2)

so derivative of f(x)/g(x) ---> [ (1/x)*2x - x^2*(-1/x^2) ] / (1/x)^2

Answer 91

Oh I see

Answer 92

Basically just go by the formual and don't worry about whay they gvie

Answer 93

yes

Answer 94

Isn't this number supposed to be 2x because f' is 2x

Answer 95

in your specific situation, yes

Answer 96

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
derivative of f(x) = 2x
derivative of g(x) = (-1/x^2)

so derivative of f(x)/g(x) ---> [ (1/x)*2x - x^2*(-1/x^2) ] / (1/x)^2
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 97

wait a sec

Answer 98

oh wow that graphic actually has a typo

Answer 99

f(x)/g(x) ---> [ (1/x)*2x - 2x*(-1/x^2) ] / (1/x)^2

Answer 100

Lol what

Answer 101

the numerator should be g(x) * f'(x) - f(x) * g'(x) not g(x) * f'(x) - f'(x) * g(x)

Answer 102

so (1/x)*(2x) - (x^2)*(-1/x^2) / (1/x)^2

Answer 103

yes just be careful with parentheses

[(1/x)*(2x) - (x^2)*(-1/x^2)] / (1/x)^2

Answer 104

okay so for b same thing right

Answer 105

jusgt swtich the letters

Answer 106

the values I mean

Answer 107

yes

Answer 108

so would it be f(x)g'(x) - g(x)f'(x) / (f(x))^2

Answer 109

yes

Answer 110

okaay and what about if we were to use f(x)/h (x)

Answer 111

h would just take the place of g?

Answer 112

yes

Answer 113

Okay I think we can move on , do you think this is all I needed to know for quotient rule?

Answer 114

yes, just know what the rule is and how to use it

Answer 115

Okay should we up a new post ? or is this good

Answer 116

new post would be good