help-17

[vast shale]

To f(y)?

[brittle minnow]

Yeah, yeah my bad

[vast shale]

Okay, now plugging in <a1, b1> ≠ <a2, b2> into h. How do I show that h(<a1, b1>) ≠ h(<a2, b2>). It seems to be pretty basic algebra...

[brittle minnow]

Notice that a_i goes to powers of 2 which is always even while the other part is always odd

[vast shale]

I figured it out. Thanks!

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[vast shale]

Logarithm help:

[vast shale]

How can I make h the subject of this ?

Well I need it to be in this format:

[hallow plover]

No logarithms involved in this

Just divide both sides by m^0.05

[vast shale]

I need it to be in this format though

So I can't divide

[hallow plover]

Why not?

It is in that format once you divide

[vast shale]

Because you would get 10^1.92 / m^0.05 which is not the multiplied format given

[hallow plover]

Note

1/(x^b) = x^(-b)

move the m up

change the exponnent sign

das it

[vast shale]

Oh I seeeeeeeeeeeeeeee

Thanks alot I get it now

I forgot about that law

Thanks again, I guess i'll close this now.

.close

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[verbal shuttle]

On the less triangular piece of paper i am unsure on how to find the perimeters in order to solve part b

I have solved part a as 19.2cm^2 i think

Not sure if its correct

[lone linden]

How did you get this?

[verbal shuttle]

Its shown on the triangular piece of paper

Sorry if its not clear

[lone linden]

It says the height is 21 cm tho

[lone linden]

This only has a height of 6

[verbal shuttle]

Yes

Above that i substituted the height for 6 in the formula for area

[lone linden]

…. But you need the height to be 21

Not 6

[verbal shuttle]

Huh why 21

[verbal shuttle]

Oh crap yeah

Ah my bad i thought the diagram wouldve been like accurate

Just didnt read properly

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[rugged idol]

So, this is a problem I was assigned, a while ago, My professor, gave very limited feedback, and said that I was wrong, but didnt really explain much futher, could anyone give me insight on what I could have improved on?

[subtle helm]

I would start by looking at the first few terms of a_k , what you have written is incorrect

[rugged idol]

trying to understand where went wrong, sorry not ignoring just looking over my work

[subtle helm]

As for the proof, it looks like you have the right ideas, not sure why your prof Said it's wrong unless he's nitpicking over something

[rugged idol]

I think my professor, invalidates a proof if even the beginning is incorrect, as in his words it has no base

a1=1x6=6, a2=26=12, a3=312=36, a4= 4*36=144?

$a_1=1*6=6$

[twin meteor]

grouchy

[rugged idol]

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[tulip parcel]

pick the 3 easiest tasks for me

[vocal sleet]

@tulip parcel Has your question been resolved?

[thin vale]

4,5,6

[tulip parcel]

hmm dunno everything on there looks kinda complicated

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[real niche]

Can someone please check my solution for this question. I need to prove by definition (epsilon-delta definition) the limit at the top.

Sorry for hand writing :>

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[robust summit]

Is there anything notably wrong with this?

I have:
1 gram of NaOH dissolved in 100 cm3 of distilled H2O.
Initial and Final Temperatures
Known Values

[robust summit]

I can provide more information if needed

[minor light]

First I think this is a maths discord server lol

But I think I can check whats going on

Is this gibbs

[robust summit]

Yeah, I asked in #discussion and they said to post it

[minor light]

k

my chem is kinda rusty

lol

only took ap chem

[robust summit]

thats alr

[minor light]

and it was like ages ago

[robust summit]

I just feel like I am doing something wrong

[minor light]

whats Q?

[robust summit]

The amount of heat gained or lost by a sample

q = mcΔT

[minor light]

ah

right

[robust summit]

c is like specific heat capacity

[minor light]

okok

[vast shale]

is 4.18 specific heat capacity?

oh nvm

[minor light]

ye

of water

[robust summit]

4.18 yeah

[minor light]

looks right tho

[robust summit]

and what im doing is dissolving naoh in water

i just worry theres like a unit mismatch

[minor light]

well do some unit cancellation

i usually do that

[robust summit]

at this point its nearing 40 hours no sleep i'll just hope its right and speak confidently

[vast shale]

\calc 1004.183.2

bruh im sleepy af

[robust summit]

you might need to multiply

[vast shale]

,calc 1004.183.2

[twin meteor]

Result:

1337.6

[robust summit]

100g ⋅ 4.18 J/g∘C ⋅ 3.2 K = 1337.6 J

[vast shale]

yea it seems like u messed up with the unit

[robust summit]

oo where

[vast shale]

u have to convert 3.2 into celcius right?

bruh i dont know how to spell it

[robust summit]

it's in kelvin

but it's delta t so either way its interchangable

Ti = 19.2 + 273.15 = 292.35 K
Tf = 22.4 + 273.15 = 295.55 K

the 19.2 and 22.4 are in celcius

295.55 - 292.35 = 3.2 K

[vast shale]

no i mean like heat capacity is given in J/g C or am i just missing something

[robust summit]

uhhhh

well
"The unit for T in the equation ΔG = ΔH - T ⋅ ΔS is kelvin (K)."

however you could be onto something

[vast shale]

yea maybe its been a year since i last studied thermo

mb

[robust summit]

thats alr it was a good idea

there is that rare chance i could be right

[vast shale]

i wish i could help in a better way

[robust summit]

that's alr!

[vast shale]

@robust summit btw u can close this channel by typing .close

if u think u r done

[robust summit]

.close

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[vast shale]

nice

[river kettle]

Yahat

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[vital wave]

I think the first one is yes but my teach just didn’t actually teach us this today so idk how to determine for b or c…

[minor light]

Correct

For the first diagram

actually its contradiction for b

[vital wave]

thanks but i dont know how to determine for b or c

[minor light]

(c) is also contradiction. Can you think of why?

[vital wave]

no i literally just said that

[minor light]

Oh.

What does it mean for a limit to exist?

@vital wave

[vital wave]

idk

i cant explain it

[minor light]

In simple terms, a limit exists when their side limits exist and their side limits converge to the same number.

[vital wave]

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[quasi hull]

how ?

[mild flower]

go step by step

common denominator is $(e^x-1)^2$

[twin meteor]

hayley table

[quasi hull]

huh

they don't have a common denominator tho

one is e^x-1 and the other is (e^x-1)^2

[mild flower]

remember how when you were adding fractions of numbers you'd multiply one of them by like 5/5 to get a common denominator?

[quasi hull]

so I multiply the fraction on the left by e^x-1 ?

[mild flower]

yes

well, (e^x-1)/(e^x-1)

[quasi hull]

yes

$\frac{e^x\cdot(e^x-1)}{(e^x-1)^2}$

[twin meteor]

Joeller

[mild flower]

great, now combine the fractions together

[quasi hull]

$\frac{e^x-e^x\cdot(e^x+1)}{(e^x-1)^2}$

[twin meteor]

Joeller

[mild flower]

you dropped a piece

[quasi hull]

oh

shoot

$\frac{e^x\cdot(e^x-1)-e^x\cdot(e^x+1)}{(e^x-1)^2}$

[twin meteor]

Joeller

[mild flower]

great, now simplify the numerator

[quasi hull]

wait

[mild flower]

if you want, you can stop writing the denominator at this point, to make it easier

[quasi hull]

$\frac{(e^{2x}-e^x)-(e^{2x}-e^x)}{(e^x-1)^2}$

[twin meteor]

Joeller

[quasi hull]

nah its fine

wait

nvm that

smth missing

$\frac{(e^{2x}-e^x)-(-e^{2x}-e^x)}{(e^x-1)^2}$

[twin meteor]

Joeller

[quasi hull]

we can lose the brackets

$\frac{e^{2x}-e^x+e^{2x}+e^x}{(e^x-1)^2}$

[twin meteor]

Joeller

[quasi hull]

$\frac{e^x+e^x}{(e^x-1)^2}$

[mild flower]

wasn't it e^2x

no it doesn't

[quasi hull]

oh wait nvm

okay

so -e^x will go with e^x

and we're left with e^2x+e^2x which is 2e^2x

$\frac{2e^{2x}}{(e^x-1)^2}$

[twin meteor]

Joeller

[mild flower]

hmm that doesn't seem to be what we wanted. check through your work

[quasi hull]

ye

wait

a sefc

wait

[quasi hull]

this is right

[mild flower]

nope

[quasi hull]

are u sure

[mild flower]

yes

[quasi hull]

I distributed the e^x

[mild flower]

signs

[quasi hull]

$\frac{e^{2x}-e^x-e^{2x}+e^x}{(e^x-1)^2}$

[twin meteor]

Joeller

[mild flower]

let's just write the numerators

[quasi hull]

ight

[mild flower]

$e^x(e^x-1) - e^x(e^x+1)$

[twin meteor]

hayley table

[quasi hull]

yes

[mild flower]

now we distribute the e^x

and what do we get?

[quasi hull]

$e^{2x}-e^x-e^{2x}-e^x$

[twin meteor]

Joeller

[mild flower]

ok

that seems right to me

[quasi hull]

so now $e^{2x}-e^{2x}$ is 0 and we're left with $-e^x-e^x$ which is $-2e^x$

[mild flower]

are you... using o for 0?

[twin meteor]

Joeller

[quasi hull]

mistype

[mild flower]

mmk

[quasi hull]

yeah

[mild flower]

anyway this seems to match now

[quasi hull]

$\frac{-2e^{x}}{(e^x-1)^2}$

[quasi hull]

it does

thank you

[twin meteor]

Joeller

[quasi hull]

Thank you sm

have a nice day

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[worn stone]

You wish to spray paint 25 spherical balloons for a party. Each balloon has a 20cm diameter. How many complete cans of spray paint must be used if one can covers 251.3 sq. cm?

[storm terrace]

! status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[worn stone]

2

[vast shale]

so

[storm terrace]

Where did you get stuck?

[vast shale]

yea

where did you get stuck

you typing a paragraph?

[worn stone]

I’m explaining myself

You know what

I’ll snap a photo

[storm terrace]

Probably faster

[vast shale]

yes

do thta

[storm terrace]

So far so good

[worn stone]

I’m trying to find the amount of spray cans used now from 31,415.9265 cm^2 of coverage

[storm terrace]

Is it the complete cans part that confuses you?

[worn stone]

Yea, I just need to know what the next steps are. Do I divide that big number per amount of coverage for one can? (251.3 cm^2)

[vast shale]

yes

i think

[worn stone]

You think?

[vast shale]

like

don't put $10m on that

[worn stone]

$10m?

How do you mean?

[vast shale]

10 million dollars

[worn stone]

Ok lol

I wish I had $10m

[vast shale]

me too

[worn stone]

So what are the next steps then?

[vast shale]

which would make it even worse if you put $10m on it

[worn stone]

Lmao

[vast shale]

divide it

[worn stone]

Okay

125.0136352403418

[twin meteor]

The following error occured while calculating:
Error: Unexpected operator , (char 3)

[vast shale]

,calc 31415.9265/251.3

[twin meteor]

Result:

125.01363509749

[vast shale]

yea

you got it

[worn stone]

Okay, so that means that over one hundred cans were used then?

[vast shale]

yea

125

it like

the oranges

[worn stone]

It’s just hard to visualize, that’s all

Which isn’t the important part.

[vast shale]

well

[worn stone]

I’ll just enter that as my answer

I’ll trust the process lol

Thank you for confirming

[vast shale]

yw

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@worn stone Has your question been resolved?

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[livid helm]

what this mean

[livid helm]

how do i even draw it ?

would it be like this?

[pale perch]

something like that, sure

do you know how to find the magnitude of a vector

[livid helm]

adding them /

[pale perch]

what do you mean by that

sounds like youre thinking of resultant vectors from multiple vectors

[livid helm]

yeah thats what i thinking

[pale perch]

thats not what you need to do here

you get the magnitude squared of a vector by dot producting it with itself

[livid helm]

im sorry i dont nkow what dot product means ?

[pale perch]

ie the square root of the sum of the square of each component
eg the magnitude of (a b c) is sqrt(a^2+b^2+c^2)

the dot product of two vectors (a b c) and (x y z) is ax+by+cz, a scalar

you can extend that to as many dimensions as you please

[livid helm]

oh ok

so

25, 100, 100 ?

[pale perch]

sqrt(25+100+100) yeah

[livid helm]

so sqrt 225

[pale perch]

yeah

[livid helm]

😭 sorry

cant add

ohh ok i see why its 15 now

[pale perch]

its an extension of the pythagorean theorem

[livid helm]

can you explain this more ? im still not sure what a dot product is or what its useful for

[pale perch]

one main use is to find the angle between two vectors, you can use this to extend to planes and whatnot

the dot product of a and b is the sum of the products of matching components but its also |a||b|cos(x) where x is the angle between them

it has the quirk that if theyre perpendicular ie x=pi/2 then the dot product is 0, which is a useful check when defining planes

since you need normal vectors

[livid helm]

im stupid

what are matching components

[pale perch]

hm like in (a b c) and (x y z) x and a are the first entries

they match

so do b and y

c and z

etc

[livid helm]

ohh ok

.close

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[strange crater]

What is the converse of the statement

And are you saying the answer to part (a) is supposed to be true?

<@&268886789983436800>

[mystic night]

i dont think they condone cheating bro...

they bouta ban u

tell @strange crater

[strange crater]

Ask your teacher to explain the answer then

[mystic night]

ur cooked

whar time is it for u

where is bro at

its 9:42 PM for me

[twin meteor]

The current time for nutgun. is 09:43 AM (AWST) on Tue, 14/05/2024.

[mystic night]

,time

[twin meteor]

You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.

[strange crater]

You're asking why it's true?

[mystic night]

ok imma stiop texting since this is for tutroing

[strange crater]

A square is a parallelogram

If you think I'm a liar, good luck then

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@finite hatch Has your question been resolved?

[regal parrot]

.close

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[waxen shell]

Could someone help me see where I messed up here? Am I misunderstanding?

[waxen shell]

here's my work so far:

I would actually rather do calc 2 than this, so many places to mess up arithmetic

[minor light]

Are you trying to do RREF?

Firstly, the final step is wrong

R1 would be in R3

R3 would be in R2

R2 would be in R1

Usually we would do it such that we have 1,0,0 on the first column, and then move on to the second one

[waxen shell]

I think rref is next class period, this is literally just solving the linear system haha

Thank you!

Yeah the prof said next class would be discussing how to do it more algorithmically

[minor light]

Well there is not much algorithm

lol

Do more practice

The complex RREF is harder lol

The one with complex numbers

[waxen shell]

That's both scary and fascinating haha

I fear arithmetic may be my downfall in advanced math

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[edgy peak]

Didn’t even go over this in class

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@edgy peak Has your question been resolved?

[edgy peak]

<@&286206848099549185>

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@edgy peak Has your question been resolved?

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[ivory sonnet]

How would I solve this?

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vast shale]

i'd assume 1

but

[ivory sonnet]

yes

[vast shale]

ok

so

do you know what the first derivative test is?

[ivory sonnet]

nope

[vast shale]

ok

that's just great

The first-derivative test examines a function's monotonic properties (where the function is increasing or decreasing), focusing on a particular point in its domain. If the function "switches" from increasing to decreasing at the point, then the function will achieve a highest value at that point. Similarly, if the function "switches" from decreasing to increasing at the point, then it will achieve a least value at that point. If the function fails to "switch" and remains increasing or remains decreasing, then no highest or least value is achieved.
is from wikipedia

[ivory sonnet]

ok

[ivory sonnet]

So using this how would I solve?

.close

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[shadow berry]

Help

[minor light]

?

[autumn plaza]

! Status

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What step are you on?
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4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[shadow berry]

I need help

[minor light]

First this is more like a chemistry question

[shadow berry]

Is there a chemistry chat here?

[minor light]

This is a maths server

[autumn plaza]

Check #old-network

[shadow berry]

Ok

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[rugged orchid]

Can every vector space be turned into an inner product space? In particular the K-vector fields where K is has a finite characteristic, meaning there isn’t really an ordering on K

[rugged orchid]

Somehow ℂ-vector spaces work but it seems to use the fact that it contains ℝ

A vector field over some other field like ℤ5 doesn’t seem to contain ℝ

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[cobalt crypt]

i mean half of the conditions stop making sense

[rugged orchid]

Well that’s why I’m asking

When I google this it seems to say every vector space can be turned into an inner product space but a lot of these answers considers only ℝ or ℂ vector spaces

[cobalt crypt]

you need to significantly modify the definition of an inner product

[rugged orchid]

So then no?

[cobalt crypt]

well like always in maths the answer will probably become yes after you change what the question is

but like say you have some random K-vector space where K is a subfield of C

call it V

just randomly giving V an inner product seems kinda rude surely

what if V comes with a topology

like C^\infty(\R) functions come with a topology

[rugged orchid]

So the fact that we want said inner product to work nicely with other structures like norms, topology, make it so even if you could define an inner product on a vector space over unordered fields it just won’t be useful

[cobalt crypt]

there are plenty of unordered subfields of C

but like the vector spaces where you care about having an inner product probably already have some kind of natural topology

randomly giving an inner product to a vector space doesnt seem like itll give you a topology that you'd want to work with

[rugged orchid]

Are these just field extensions of algebraic numbers

[cobalt crypt]

field extensions of Q even

[rugged orchid]

Ah

So vector spaces over Zp fields are not really inner product-able?

[cobalt crypt]

by the usual definition i wouldnt think so

[rugged orchid]

Since I guess these “unordered” sub fields of ℂ still has some concept of the real line and positivity

[cobalt crypt]

well you have the involution right

complex conjugation

and under that involution there is a fixed field

which will be some subfield of R

and thatll be ordered

[rugged orchid]

Well if you’re a sub field of C you’ll inherit that involution I suppose

[cobalt crypt]

you can probably mimic the construction

Let $K$ be a field with involution $\sigma : K \to K$ such that the fixed field
[ K^\sigma = \set {, x \in K \where \sigma(x) = x ,} ]
is ordered. Then you can define a sesquilinear form $\inner{}{} : V \by V \to K$ by
\begin{enumerate}
\ii $\inner xx \in K^{\sigma}_{\ge 0}$ and $\inner xx = 0$ iff $x = 0$.
\ii $\inner {ax + by} z = a\inner xz + b \inner yz$.
\ii $\inner xy = \sigma(\inner yx)$.
\end{enumerate}

[cobalt crypt]

or something like that

[rugged orchid]

Hmmm

So if you pretty much need to require some sort of ordering and an involution to make an inner product space from any vector space

[cobalt crypt]

for it to be what you're familiar with

bilinear forms exist though

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[idle tartan]

Hello

[idle tartan]

I need help

This question please

If you know the answer I’ll gift you 200 vbucks emote

[river kettle]

200 VBUX??

[idle tartan]

Yes

@river kettle

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[idle tartan]

Yes

[sullen shoal]

can u write it more clearly

i cant understand

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[minor folio]

why can you not say that the set of natural numbers is twice the size of the set of even numbers? Can you not do a Cantour mapping and say something like for every even number there are two natural numbers?

[chrome raptor]

No

The function f: N → 2N, given by f(n)=2n is a bijection

For every even number there are two natural numbers is okay but that's not an argument.

In fact for every natural numbers there are also two even numbers

1 → 2 and 4
2 → 6 and 8
3 → 10 and 12
...

That does not work for what you are asserting (|N|>|2N|) which is false

[minor folio]

hmmm

[minor folio]

Is there no use in calculating the density?
On every interval of 1000, e.g. [0,1000]. There are 500 even numbers and 1000 natural numbers and then do a proof by induction?

[chrome raptor]

No

There's bijection

That means you can ennumerate the even numbers with natural numbers

That is enough to prove that there is the same quantity of even numbers than that of natural numbers

You have the first even, the second even, the third even... So you can have a list of all even numbers indexed by natural numbers

1 ←→ 2
2 ←→ 4
3 ←→ 6
4 ←→ 8
...
n ←→ 2n

And you can go from one set to the other

That's why the cardinal or N is equal to the cardinal of 2N

[mild flower]

to do Cantor diagonalization, you need to show that no matter how you number them you'll always miss one

[chrome raptor]

(I like 0 being natural btw, don't bully me haha)

[mild flower]

this is a useful thing to do for some applications. but cardinality is not one of them.

[minor folio]

hmm, alright, because I am actually wondering about a physics question, namely how we can calculate the amount of stars of type A compared to the amount of stars of type B on the assumption that the universe is infinitely sized and homogenous and isotropic.

would calculating the density be useful then?

[hard atlas]

without knowing much about physics, density seems like the much better thing to compute here compared to cardinality

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[sand nova]

how do i pull out a constant while differentiating?

[sand nova]

in regards to this question specifically i can do differentiation where its a square root or a number but i have no clue how to do this

i tried to do -2x^-1 and go from there but thats wrong and do 2x^2-x^-3 but that doesnt make sense either and i dont know how to get rid of the constant

i also tride doing 6x^-2 and working from there but got a wrong answer too because i was trying to work off of this example

[bronze osprey]

3a is 3/2 * (derivative of 1/x^2)

You can pull constants out of the derivative; if you'd like to know why let me know

[sand nova]

yeah i dont know how or why

id love to know if you have the time to help

[bronze osprey]

Right, so you know that a derivative is a slope, (change in y)/(change in x) right?

[sand nova]

yeah its a gradient of an individual point

is what my teacher told me

[bronze osprey]

Yeah, so if f(x) is (change in y)/(change in x)

Then k * f(x) has slope (k * change in y)/(change in x)

[sand nova]

so the 3 on top is also a derivative?

[bronze osprey]

no, and I don't really know what you mean by that

[sand nova]

3 is the change in y

i mean

and what does K stand for?

[bronze osprey]

Oh k is just a constant

For question 3a this would be 3/2

[sand nova]

so 3 is K

[bronze osprey]

Nope, I never said that

[sand nova]

oh

[mild flower]

you might be better off proving that $\dv{x}(c\cdot f(x)) = c\cdot\dv{x}(f(x))$ using the definition of a derivative (the one with limits)

[twin meteor]

hayley table

[bronze osprey]

Like if we have f(x) = 1/x^2, then (change in y)/(change in x) would be approximately f'(x) = -2x^(-3)
where the changes in x and y are really, really small

So (change in y)/(change in x) is a ratio

[sand nova]

yeah

[bronze osprey]

Btw Zaya do you know the definition of the derivative?

$\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = f'(x)$?

[sand nova]

gradient of individual point

[twin meteor]

south

[bronze osprey]

this definition

[sand nova]

f'x is drivative

[sand nova]

ive never seen this ever

[bronze osprey]

Okay thanks for letting me know

it all comes down to (change in y)/(change in x) actually

a quick YouTube search didn't give any results with this image, but I'm sure there are videos on this online

So change in y = f(x + dx) - y = f(x + dx) - f(x)

y = f(x) is the definition of a function

And here we're using dx instead of h, to mean dx = a tiny change in x

[sand nova]

so the derivative is a small change of ratio

?

[bronze osprey]

Yes exactly

The slope dy/dx is a ratio between dy and dx

Which mean (small change in y) and (small change in x)

[sand nova]

Ok

so d/dx is small change in constant/small change in x

[bronze osprey]

No, d/dx is never on its own

[sand nova]

oh

[bronze osprey]

When we have d/dx, we always have d/dx (y)

or d/dx (f(x))

d/dx (y) is the same thing as dy/dx

d/dx actually doesn't mean anything on its own, it's just an operator saying: take the derivative of this function

[sand nova]

d/dx (3/2x^2)= 3/2x^2?

then

[bronze osprey]

No, that's not right

[sand nova]

is 2 in 2x^2 the small change in x?

[bronze osprey]

The correct statement is d/dx (3/2 * 1/x^2) = 3/2 * d/dx (1/x^2)

[bronze osprey]

No

Let me give you a simpler example

Say we have the line y = 2x

The slope is obviously 2

[sand nova]

Yeah

[bronze osprey]

We could have x = 1 and y =2

But we could also have x = 3 and y = 6

Or x = 11 and y = 22

And so on

[sand nova]

Yeah

[bronze osprey]

So just because we know dy/dx = 1/x^2 for example

Doesn't mean we know dy and dx individually

All we know is that the ratio is 1/x^2

Well, the ratio isn't a number cause it depends on what x we have

That's an additional complication

[sand nova]

so dy/dx is 2 in 2x

[bronze osprey]

Yes

[bronze osprey]

This means the slope of the function y = 2x is always 2

No matter which x-value you have

It's 2 everywhere

[sand nova]

but in other functions the slope isnt always 2 so thats why you dont know dy & dx individually?

[bronze osprey]

Yes, so calculus is when you have the idea about the slope not being a constant

The slope depends on the point you have

Which is the same as the slope varying based on the x-coordinate

[sand nova]

Ok

[bronze osprey]

Well, that's not the reason why you don't know dy and dx individually though

[sand nova]

so that is what it means when its said to find derivative in respect to the f(x)

[sand nova]

to find the slopes variation in respect to the x coordinate

[bronze osprey]

Even in the straight line example

We could let dx = 1 and dy = 2

Or dx = 4 and dy = 8

[bronze osprey]

Exactly!

So that once we are given the x-coordinate of the point

[sand nova]

i wish teachers told me this before throwing this stuff at me 🤦‍♂️

[bronze osprey]

We can just substitute x in and get the gradient at that point

So we get a number eventually don't worry

[bronze osprey]

that's sad

[sand nova]

yeah im far behind the rest of the class rn because i had a surgery and am trying to catch up but i often need to actually understand whats happening to do the math instead of just doing it

[bronze osprey]

Exactly and that's the best way to learn maths

[sand nova]

you actually enjoy it then asw lol

[bronze osprey]

Cause the formulas aren't actually that important, like they're online if you need to do your homework

[bronze osprey]

yeah I'm a maths major in fact

[sand nova]

i thought you were a teacher

because of the way you explain it

[bronze osprey]

But if you don't know how to apply your knowledge that means you're going to have a hard time

[bronze osprey]

Well many people I know tutor

I personally am not a tutor but it's pretty common if you can explain well enough

[sand nova]

yeah you could probably be a good tutor

[bronze osprey]

Also from the textbook I can tell you're doing IB maths

[sand nova]

yeah i am

[bronze osprey]

I actually did IB AAHL

[sand nova]

AASL

[bronze osprey]

ahhhh okay yeah that makes a lot of sense

[sand nova]

good to know youve suffered with the crappy textbooks aswell hahah

[bronze osprey]

yeah I hate the Oxford one

Schools should stop using it

Like why are the answers full of typos

😭

[sand nova]

all my teachers hate the books too but 'theres no better options' apparentally

[sand nova]

or are straight up wrong

without any reasonable doubt of a typo

[bronze osprey]

There are, they just aren't used to any other textbook

Haese is good but they have 2 books for dp1 and dp2

I actually had the Pearson textbook for extra practice and for AAHL it was really well explained

[sand nova]

my old school used haese when i did myp i miss those books

[bronze osprey]

ikr

[sand nova]

tbh i should probably just get a haese book

[bronze osprey]

Like I've heard from other students and they say anything is better than Oxford

Like literally any one

There's also Hodder

[sand nova]

oh yeah hodder does alot of subjects

their humanities books r pretty good

[bronze osprey]

Yeah unsurprisingly but no one seems to use Hodder for maths that much

[sand nova]

yeah i think haese and oxford are the most popular

[bronze osprey]

yeah

[sand nova]

but to apply wehat we did before

K is the specific point in the function

right

[bronze osprey]

k is the vertical stretch actually

k f(x) means you stretch f(x) in the y-direction by k

[sand nova]

oh yeah

so its a stretch of the function f(x) in the y direction by k and because its a vertical stretch

you can pull out the constants

or not X

at least

so its a vertical stretch by 3/2

then you have d/dx(3/2)

d/dx 3/2 (1/x^2

and you can turn the 1/x^2 into exponent form

x^-2

so 3/2(x^-2)

then u do power rule

so a*x^a-1

so

-2*x^-2-1

sp

-2x^-3

am i onto something

[bronze osprey]

Hold on

[sand nova]

and then

3/2(2x^-3)

[bronze osprey]

Yeah so the really important bit is that the change in y is now 3/2 * (original change in y)

But the change in x is just the original change in x

So (change in y)/(change in x) increases by 3/2 as well, even if you have a function with varying slope

So that explains why you can pull the 3/2 out of the slope or derivative

[sand nova]

and why you multiply 2x^-3 by 3/2

[bronze osprey]

What you're saying is all correct but I can't tell if you're just applying rules without understanding why or you're trying to convince yourself

[sand nova]

i have no idea

i probably dont fully understand it

but im using the knowledge from the other questions i can do by applying rules

[bronze osprey]

Yeah I guess you need to let it sink in

But you do seem to know how to apply the rule at least

Maybe give it some time

[sand nova]

because like in 5 * 5 root of x^3

which is a question i did before

i know i can turn the root into an exponent

[bronze osprey]

Yes so you get $5x^{3/5}$

[twin meteor]

south

[sand nova]

turning it into 5x^3/5

yeah

and then with the power rule

3/5*5x^3/5-1

[bronze osprey]

Yeah

[sand nova]

$3/5*5x^3/5-1$

[twin meteor]

Zaya

[bronze osprey]

So what is happening here is that $\frac{d}{dx} (5x^{3/5}) = 5 \frac{d}{dx} x^{3/5}$

[sand nova]

ok i missed it up but

[twin meteor]

south

[bronze osprey]

So yes $5 \cdot \frac{3}{5}x^{3/5 - 1}$

[twin meteor]

south

[bronze osprey]

Sometimes it helps to do things in small steps, like taking the constant out of the derivative

Cause you will have cases later on like the chain rule where you need to apply multiple steps from different differentiation rules at the same time

[bronze osprey]

It sounds scary but it's just a longer version of this with more steps

So it helps to write out each step by doing small steps each time

Otherwise you will be super confused

[sand nova]

so the derivative in respect to X (in respect to the x axis etc.) is $3x^2/5$

[twin meteor]

Zaya

[sand nova]

yeah

[sand nova]

yeah i was so lost doing this stuff and trying to learn it off the slides it took forever to get anywhere

[bronze osprey]

Nearly

3/5 - 1 = -2/5 so $\frac{3}{5}x^{-2/5}$

[twin meteor]

south

[sand nova]

oh

[bronze osprey]

types out maths expressions neatly

[bronze osprey]

Oh wait times 5

There's a 5 in front

[sand nova]

with that yeah

so 3x^-2/5

[bronze osprey]

mhm

Btw you can check your answer by integration

Don't worry about how to do it, but it's basically the reverse of differentiation, so it undoes differentiation and you should get your original answer back

,w integrate 3x^(-2/5)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sooouth.

[bronze osprey]

cool

[sand nova]

and you can turn this into 5 root x^-2 right

[bronze osprey]

Yes sorry

[sand nova]

alright

[sand nova]

oh

WWW

[bronze osprey]

You should say 5th root of x^(-2)

Cause 5 root x^(-2) can mean 5 * square root of x^(-2)

[sand nova]

ohhh yeah

mb

so i have 3/2x^2

[bronze osprey]

Careful, it's 3/2 * 1/x^2

So $\frac{3}{2}x^{-2}$

[twin meteor]

south

[bronze osprey]

always write it in this form to avoid confusion

[sand nova]

yeah im just writing out how to do it so i make sure i actually understand it

because i just did it in my book but i gotta make sure

[bronze osprey]

okay okay

[sand nova]

im finding derivative in respect to x

so i can remove the constants

or well

pull them out

[bronze osprey]

yes

[sand nova]

so i i got

3/2 derivative with respect to x 1/x^2

change 1/x^2 to exponent form to be able to apply power rule

x^-2

apply power rule to make it -2*x^-2-1

so -2x^-3

[bronze osprey]

yes

[sand nova]

3/2(2x^-3)

but i have to simplify this right

[bronze osprey]

yes, well (3/2) (-2 x^(-3))

but yeah you know how to do it

[sand nova]

i dont but im figuring it out rn

got it

ok ill do some other questions to practice thanks sm for the help south

ive asked a few people but nobody has explained it like you did ty

#1 math major

.close

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[edgy chasm]

In need of urgent help

[edgy chasm]

So uh in this vid

They transfer it right

Theey just changed the operation and its 96

BUT

In another video

They made it negativ

NOW

This is basic as hll

But its so confusing

Cause

Whats happening

js please some1 explain

[hallow plover]

Didn't they do the same thing both times?

[edgy chasm]

the other one

js changeed

the opration

Th eothr one

Made it ngative

I calculated both

Its different if they js changed the operation

if that makes sense

[hallow plover]

I don't understand

In the first one they had 84 + x = 180

they subtracted 84 from both sides

[edgy chasm]

its 96 hre

[hallow plover]

so 84 + x - 84 = 180 - 84

which gives x = 96

In the second example they had A + 117 = 180

they subtracted 117 from both sides

so now you have

A + 117 - 117 = 180 - 117

Which gives

A = 63

[edgy chasm]

but generally

when we move it

IT becomes negative right

man im alzhiemers

Ty for your help

.close

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[sand nova]

south

[sand nova]

i am trying to learn the chain rule

and kind of understand that its to differentiate functions of functions like f(g(x)

and that the formula is dy/dx=dy/du * du/dx

[bronze osprey]

Well it's easier to do f'(g(x)) * g'(x) tho

For two functions ofc

[sand nova]

or d/dx(y)=dy/du * du/dx

[bronze osprey]

https://www.youtube.com/watch?v=HaHsqDjWMLU

Try watching this video

[sand nova]

but the question is

3/(2x)^2

i can turn that into 3 d/dx 1/(2x)^2

what is the second fucntion in this

[bronze osprey]

you don't need chain rule lol

$\frac{3}{4x^2}$

[twin meteor]

south

[sand nova]

symbolab said so

[bronze osprey]

you don't need the chain rule anyways

you can use the chain rule but like why

[sand nova]

i did it and checked my answer

and symbolab gave this

[mild flower]

it's good to do things in multiple different ways! that way you can check your work

[sand nova]

or is it the thing tweaking out

oh wait i am really stupid

nevermind thank you

or am i

im confused

-3/2x^3 isnt the same as 3/4x^2 right?

[mild flower]

correct those are different

[sand nova]

would both answers be right?

[mild flower]

isn't 3/4x² what you started with

[sand nova]

i started with

[mild flower]

okay, let u = 2x

or like. show your own work.

[sand nova]

well i tried using the power rule because ive been trying to figure out the chain rule and been getting lost

wait

ill take a better phot

[mild flower]

,rcw

[sand nova]

i tried to relate 3 d/dx to dy/dx and stuff and then i remebered

these are notations

what am i doing

and then

yeah

[mild flower]

okay let's go through it like this

let u = 2x

[sand nova]

mhm

[mild flower]

so our original function is now 3/u²

aka 3u^{-2}. now we can differentiate with respect to u

using the power rule

[sand nova]

-2*3u^-3

then

and then from there on

[mild flower]

yep so now we need to multiply by du/dx

the way to see this is like $\dv{y}{x} = \dv{y}{u} \cdot \dv{u}{x}$ because the du's cancel out

[twin meteor]

hayley table

[mild flower]

so what's du/dx?

[sand nova]

(dy/dx) / dy/du?

[mild flower]

well, yes it is; but that's not what I mean

look at what we said u was

u = 2x

[sand nova]

d2x/dx)

dx

?

[mild flower]

hmm

[sand nova]

or no

wait

3u^-2

[mild flower]

du/dx just means the derivative of u with respect to x

so if u = 2x

then we differentiate both sides wrt x

du/dx on the left, what's on the right?

[sand nova]

dy/du?

[mild flower]

what's the derivative of 2x with respect to x

like if u(x) = 2x, then what's u ' (x)

[sand nova]

2?

[mild flower]

yes

so du/dx = 2

[mild flower]

so from here, we know that df/dx = this thing * du/dx

[sand nova]

what does u stand for

[mild flower]

u can be anything

[sand nova]

but when we made u 2x

[mild flower]

you can make it whatever you want

[sand nova]

why did we make 2x u

[mild flower]

because that was a subexpression

containing x

[sand nova]

what is a subexpression

[mild flower]

uhhh like

a part of the expression

like if I had 3(x-4)³ + 7

then (x-4)³ would be a subexpression

[sand nova]

so 2x=u is a part of f(x) =3/(2x)^2?

[mild flower]

see how there's a 2x in there

[sand nova]

like a part of the equation basically

[mild flower]

yeah

[sand nova]

ok

so i can make a part of the equation

the subexpression

preferably the X because du/dx

[mild flower]

yep

the idea behind the chain rule is that you differentiate the outer thing, treating that subexpression as a variable, and then you multiply by the derivative of that variable

so in this case we had like

y = 3/(2x)²
u = 2x; du/dx = 2
y = 3/u²

dy/dx = dy/du * du/dx
dy/dx = -6/u³ * 2

but you do still need to substitute u = 2x back in

so dy/dx = -12/(2x)³

[sand nova]

so

fx(x) or Y=3/(2x)^2

u=2x (subexpression of f(x)

give me a minute

sorry

y=3/u^2

i mean

3u^-2

then i can apply the power rule to that

to make it 6u^-3

nvm

-6u^-3

and that was dy/du

dy/du * du/dx

is -6u^-3

*2

-12u^-3

but i dont know where i am going wrong

because thats different than your answer

oh wait

why did u multiply -6u^3 by 2?

[sand nova]

and why isnt ^-3 in dy/dx negative?

[mild flower]

thats du/dx

[sand nova]

what

im confused

[mild flower]

umm

-6/u³ is dy/du

but we want dy/dx

so we need to multiply by du/dx

[sand nova]

oh

sorry i forgot

du/dx was 2

as u was 2x and the derivative in u in respect to x was therefore 2

so you did -6/u^3 (derivative of y in respect to u) * 2 (derivative of u in respect to x)

making dy/dx -12/u^3

and then to substitute x back in

(2x)^3

is that right?

[sand nova]

ty hayley!!!!!

.close

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[ember latch]

ANYONE FAMILIAR WITH MAPLE

[ember latch]

???

HOW DO I MAKE IT WRITE IT OUT WITH THE LIKE VARIABLES YK

noone? 😔

come on maan 😔

anyonee???

PLEAAASE

THIS WOULD MAKE MY WEEK

damn

😔

[cyan talon]

wdym, I don't understand your question

[ember latch]

uhmm

i want it to spit out this

im doing it manually rn but this feature can save me lots in the future

if it exists

if i just inputted it in a bad way

that could be an answer too

[cyan talon]

well r x r is just 0 for any vector r

that's normal

[ember latch]

ikik

ik the answer

i just want maple to do it for me