help-39

[glass meadow]

No, 0 doesn't have to exist either

1/x is odd

[boreal crescent]

but if 0 doesn't exists it's nothing

[glass meadow]

f(x) = tan(pi(x+1/2)) is odd and 1-periodic, yet f(0) and f(1/2) don't exist

[boreal crescent]

okay I see yes

[glass meadow]

Sorry f(1/2) exists there but not f(1)

Anyway you can make it so by changing the constants

tan(2pi(x+1/4))

[boreal crescent]

it's 2 periodic in this case

[glass meadow]

1/2-periodic you mean? That's also 1-periodic

[boreal crescent]

but ok I see, so what's the necessary condition for like f(0) and f(1/2) exists ?

[glass meadow]

The necessary condition would be that they exist

In other words that the function is defined at these points

[boreal crescent]

?

[glass meadow]

tan(2pi(x+1/4)) repeats every 1/2 interval

[boreal crescent]

oh yes mb

[glass meadow]

1/cos(2pi(x+1/4)) is odd and 1-periodic (and not 1/2)

I think you get the point

[boreal crescent]

yes ty a lot

ty @ember cloak too

.close

[pearl pond]

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[ember cloak]

You're welcome

[pearl pond]

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[nimble osprey]

why does convexity implies continuity ?

[nimble osprey]

where $f:\mathbb R\to\mathbb R$ is a convex function

[jolly parrot]

everg

[pearl pond]

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[cobalt pilot]

What is the derivative of (x^3 + 7)^4 ?

[cobalt pilot]

And how do I work out using chain rule

[last summit]

write it as a composition of functions first

from the chain rule, do you know what the derivative of f(g(x)) is?

Also, could you delete your messages in #help-1 so they don't clog up my channel please

[cobalt pilot]

How do I do this

[last summit]

well

can you tell what the "inner" function is?

and the outer function?

[cobalt pilot]

4x^3 outer function

Right?

And x^3 + 7

Inner

[last summit]

I'm not asking for the derivative

let f(x)=x^4

and let g(x)=x^3+7

then what is f(g(x))?

[cobalt pilot]

Is it not just (x^3 + 7)^4?

[last summit]

exactly

and do you know the derivative of f(g(x)) using the chain rule?

[pearl pond]

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[pearl pond]

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

[formal tiger]

1

[warm current]

What are the dimensions of the plot, in terms of x?

[formal tiger]

-2x^2 + 200x

[formal tiger]

^

[warm current]

Equation for area, but yeah

Better to write it as A=-2x²+200x

Do you know about first derivative test?

[formal tiger]

Nope

[warm current]

you absolutely need that for this problem

[formal tiger]

O

[warm current]

https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-4/a/applying-the-first-derivative-test-to-find-extrema

[pearl pond]

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[pearl pond]

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[midnight haven]

Hi, I'm stuck... what am I supposed to do after finding the t-score I'll be working with?

[midnight haven]

The second image is the (incomplete) answer that I came up with by the way

[rough stream]

You've got your null hypothesis and alternative hypothesis mixed

If you have a t-score and your degrees of freedom, it's time to get your t-table out

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

Really?

I'm using this table

I get a one-tail value of 0.15

Rounding the my t-score

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

Does this make sense?

I have no idea what I'm doing

I instinctively looked at the one-tail value in the t-chart but I might be doing this whole thing wrong

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

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[wary silo]

welll, i need another help today with complex numbers, but i dont really understand them i think
i have two tasks to calculate
z^2 = 4 + 3i

z^3 = -64

[merry carbon]

Do you have a picture of the original question? Are you supposed to find z given those properties, or calculate z^2 for z = 4 + 3i (and similarly for the other)?

[wary silo]

well, what we got is just those two exercises

i assume we gotta calculate z so i guess in first should be 2 solutions and in other 3 solutions

[glass meadow]

Do you know how to use the polar form?

[pearl pond]

@wary silo Has your question been resolved?

[wary silo]

no, not really

[glass meadow]

Tough luck

I'm kidding but z^2 = 4 + 3i looks like a pretty evil one

z^3 = -64 has an obvious real solution

[wary silo]

okay yeah the z^3 i figured out it seems easy

but the first one is 💀

[pearl pond]

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[shut linden]

What property of solenoid are they using in cancelling the second and 5th term?

[shut linden]

.close

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[robust oar]

$\sqrt[3]{24z^5x^9}$

[jolly parrot]

Akira E-Girl

[robust oar]

I've began but stuck midway

ill show work

$\sqrt[3]{4} \sqrt{6} z^3 x^7$

[jolly parrot]

Akira E-Girl

[robust oar]

I did it like this

is that right

im stuck here

[shrewd basin]

How did it become a square root

And how did z^5 become z^3

how did x^9 become x^7

[robust oar]

it took by root because it has 2

[midnight haven]

what is the cube root of 24 first

[robust oar]

root 4 and root 6

[shrewd basin]

That’s not really helpful

Try factorising it

[robust oar]

idk

[shrewd basin]

Well okay 24=6*4

24 = 2•2•2•3

Does that help

[robust oar]

2^3 and root 3?

[rustic gate]

this is a bit hm

you're on the right track

[robust oar]

I mean like this

$\sqrt[3]{2}$

[jolly parrot]

Akira E-Girl

[robust oar]

idk

[rustic gate]

can you simplify $\cbrt{24} = \cbrt{2\times2\times2\times3}$?

[robust oar]

$2 \sqrt[3]{3}$

[jolly parrot]

Akira E-Girl

[rustic gate]

great

so what about [ \cbrt{24 z^5 x^9}?]

[cinder flower]

@rustic gate @shrewd basin

[jolly parrot]

layla

[shrewd basin]

Check out #help-31 @cinder flower

[robust oar]

$2 \sqrt[3]{3} z \sqrt[3]{z^2} x^3$

ded latex

[jolly parrot]

Akira E-Girl

[rustic gate]

ye pretty much

that's about it

[robust oar]

the answer sheet says $2zx^3 \sqrt[3]{3z^2}$

[rustic gate]

uh

[jolly parrot]

Akira E-Girl

[robust oar]

fixed

[rustic gate]

oh

[rustic gate]

you see the two cube roots

you can bring them together

to match the answer sheet

[robust oar]

oh I see

[rustic gate]

otherwise it's the same

[robust oar]

.close

[pearl pond]

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[true charm]

I've got the following question: "let f: X -> Y and g: Y -> Z, where g after f is injective. Must g be injective too?". Is it sufficient to provide a counter example, or do i need to go into more detail?

[timid spindle]

a counter example is fine

to show that it is false

[true charm]

ah ok, just to check, in general if I want to prove that a "for all..." is false, can I just provide a single counter example?

[timid spindle]

yep!

[true charm]

even if its got nested for alls, which i think is whats happening in the question

[timid spindle]

yeah

[true charm]

wow thats neat to know

cheers

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[covert ivy]

is d correct?

[wet osprey]

The task is to find the incorrect statement

What is the definition of sec(x)?

[covert ivy]

what

1/cos

[wet osprey]

So what is sec²(x)

[covert ivy]

1/cos^2

[wet osprey]

So is d a correct statement?

[covert ivy]

no cause the deriv of sec^2x is not that

https://d138zd1ktt9iqe.cloudfront.net/media/seo_landing_files/derivative-of-sec-square-x-1640098105.png

@wet osprey so?

[copper kestrel]

Well d claims the derivative of f(x) is 1 / cos^2x

Not the derivative of sec^2

[covert ivy]

but f(x) is equal to tanx

[copper kestrel]

Right

So d claims the derivative of tanx is 1 / cos^2

[covert ivy]

yea

thats wrong tho

cause derivative of tanx is sec^2x

[wet osprey]

But you just told me that sec² x is 1/cos² x

[wet osprey]

Right here you said that

[covert ivy]

yea thats what it is not the deriv of it

[wet osprey]

But they are the same thing

If d/dx tanx = sec² x

And sec² x = 1/cos² x

[covert ivy]

oh wait

im tripping

[wet osprey]

Then by the transitive property of equality, d/dx tanx = 1/cos² x

[covert ivy]

so then is b wrong?

[copper kestrel]

How so

[covert ivy]

nvm

idk which one is wrong then

is it c?

[copper kestrel]

Yeah

They swapped dy and dx

[covert ivy]

ok

can you help me with more pls

[copper kestrel]

Gtg sorry

[covert ivy]

.close

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[wet osprey]

Don’t guess

[pearl pond]

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[midnight reef]

From the information provided in each case, find the coordinates (x, y) of the point on the unit circle:

a) The x coordinate of the point is 4/5 and the y coordinate is positive.
b) The y coordinate of the point is -1/3 and the x coordinate is positive
c) We have that t = π/2

[midnight reef]

It's been a long time since I used the unit circle, how do one find the corresponding coordinates?

[opal moss]

unit circle = every point on the circle is distance 1 from the origin

[midnight reef]

I know that, but what is the proceed one do to find the a coordinate if we were given one of them?

[opal moss]

construct an equation using that fact then the question tells you which quadrant the point is supposed to be in

[midnight reef]

I don't remember much about the unit circle honestly

Should I use the Pythagoras Theorem? Or just replace, for instance in a), x = 4/5 in the equation of the circle: x² + y² = 1?

Since this is pre-university

[pearl pond]

@midnight reef Has your question been resolved?

[opal moss]

ye that’s it

[pearl pond]

@midnight reef Has your question been resolved?

[midnight reef]

For c) how would it be?

[opal moss]

some nonsense
jk, probably use cos^2 θ + sin^2 θ =1

[midnight reef]

replace theta = π/2?

or you mean (sin^2 θ, cos^2 θ)?

since its asking for an (x, y)

[opal moss]

yes, otherwise the question is nonsensical

[pearl pond]

@midnight reef Has your question been resolved?

[midnight reef]

but should it be (sin^2 θ, cos^2 θ) or (sinθ, cosθ)?

[gusty prism]

.close

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@midnight reef Has your question been resolved?

[pearl pond]

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[toxic summit]

what is the second method he uses called?https://youtu.be/2k030RuNyOE?si=8qiFc8bCooEkNBp7

[candid spire]

I’m not sure if it has a name exactly

It’s pretty standard though

[toxic summit]

man i thought that everyone uses the first one

so iwas so confused

its so much easier than the first one though

thanks

[candid spire]

What is the first one exactly

[toxic summit]

i think its like

actaully

idk

but i learnt that one

and i didnt get taught the second one

anyways

thanks

.close

[pearl pond]

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[candid spire]

Yeah

Second one I mean

Learn why it works though

Just stare at this:
(x - r)(x - s)

Expand it out

What do you get?

[toxic summit]

okay

um

[candid spire]

(Do you know how to expand it?)

[toxic summit]

yeah yeah

wait so

wait isnt like

a^2 + ab + b^2 = (a+b)^2

so therefore its

[candid spire]

No no not at all

Hold on

Let’s just talk about how to expand it

Are you aware of the distributive property

[toxic summit]

uh no

[candid spire]

3(a + b) = a+b + a+b + a+b

[toxic summit]

oh

[candid spire]

= 3a + 3b

[toxic summit]

i see

[candid spire]

Yeah

[toxic summit]

yeah

[candid spire]

So actually

(a + b)(c + d)

Consider “(a+b)” as a single term

[pearl pond]

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[toxic summit]

ac ad bc bd?

[candid spire]

And distribute it to c and d individually

So you get

(a+b)c + (a+b)d

Following?

[toxic summit]

ohhh

yeah

[candid spire]

And now you can apply it again

and what do you get

[toxic summit]

wait apply to what?

[candid spire]

ac + bc + ad + bd

[toxic summit]

oh

[candid spire]

You see?

[toxic summit]

yeh

yueahj

yeah

[candid spire]

So now you have a nice rule:
(a + b)(c + d) = ac + bc + ad + bd

Okay

So now do this with (x - r)(x - s)

What do you get?

[toxic summit]

x square + -sx + -rx + rs

[candid spire]

yep

And you can factor out an x from the middle two terms

[toxic summit]

ohhh

[candid spire]

x^2 - (r + s)x + rs

Following?

[toxic summit]

yeah

[candid spire]

Okay cool

So now we have this:
(x - r)(x - s) = x^2 - (r + s)x + rs

Now, when you’re “solving a quadratic”, you’re searching for roots

ie, numbers for x which when you plug in it gives you zero

[toxic summit]

yep

[candid spire]

Here, clearly r and s are the roots

Plug them in, you get 0

So this is actually very useful

Because now you can look at a quadratic like

x^2 + 5x + 6

If this has two roots, call them r and s, then we must have

5 = -(r + s)
6 = rs

Following?

[toxic summit]

yeah

[candid spire]

Okay cool

So this actually justifies what that guy was doing in the video

We’re searching for two numbers, r and s, which add to -5 and multiply to 6

And then once you get r and s, the factorization will be (x - r)(x - s)

That’s it

[toxic summit]

yeah i do get monics

but i dont really understand non monics

[candid spire]

Ohh ok ok

Yeah so that’s a good question

hm

[toxic summit]

its like why do they multiply

[candid spire]

do you mean the (2x - 3)(x - 4) or whatever

Like how do you get the 2

Hmmm

Lemme think

okay this is gonna be very ugly but this is what I recommend

Let’s generalize our previous r,s idea

[toxic summit]

wait no no

i mean like

[candid spire]

What

[toxic summit]

why do you multiply the co efficient with the product

[candid spire]

What coefficient

[toxic summit]

so like

2x^2 - 13x +21

why do you have to multiply the

2 with 21

[candid spire]

Ohhhh

I see what you’re saying

Yeah let me think about that

(ax - r)(bx - s) = abx^2 - rbx -sax + rs
= abx^2 - (rb + sa)x + rs

goddamn this is a lot of variables

2 = ab
-13 = -(rb + sa)
21 = rs

[toxic summit]

uhhh what

[candid spire]

I’m trying to find my own reason for this method

[toxic summit]

oh ok

[candid spire]

This is a non-monic generalization

See if you can follow

[candid spire]

This is for the quadratic you gave me, 2x^2 - 13x + 21

Hmmmm

[toxic summit]

sorrya nd thanks a lot but igtg

bye

thansk a lto

[candid spire]

Take this for the road

This is how I learned how to factor

It’s a bit ridiculous and even the authors admit it

Honestly idk why the video method works but

Yeah gl man

[pearl pond]

@toxic summit Has your question been resolved?

[pearl pond]

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[midnight haven]

When the 3-digit number ABC is divided by the two-digit natural number BA, the quotient is 20, and the remainder is the two-digit natural number BC. Therefore, what is A+B?

[midnight haven]

I m new to these problems

[candid spire]

Do you understand what number ABC is

It’s 100A + 10B + C

For example, 352 = 3*100 + 5*10 + 2

[midnight haven]

Oh yes

Ok

[candid spire]

Okay

And when you divide with remainder

For example, dividing 20 by 7 gives you a quotient of 2 with remainder 6

[midnight haven]

Yes i know

[candid spire]

Oh ok ok

So literally translating this problem it is
ABC = 20(BA) + BC

100A + 10B + C = 20(10B + A) + 10B + C

100A + 10B + C = 200B + 20A + 10B + C

[midnight haven]

Ok

[candid spire]

Can you simplify this

[midnight haven]

Yes

[candid spire]

To what

[midnight haven]

80A=200B

?

[candid spire]

Yep

So the problem is asking what is A + B

Thus C is irrelevant

So this is good

Also, A and B are digits, ie they are <= 9

B can be 0, but A must be >=1

(If A = 0, then ABC will not be a three digit number)

Let’s simplify again

[midnight haven]

Where do we simplify

[candid spire]

80A = 200B
8A = 20B

[midnight haven]

Yes

[candid spire]

What’s next

You can divide out even further

[midnight haven]

English is not my main language

And this is math

[candid spire]

Oh no worries

8A = 20B

2A = 5B

Right?

[midnight haven]

Yes

[candid spire]

And what could A and B be?

Remember, they must be single digits

[midnight haven]

2 and 5

[candid spire]

Yep exactly

A = 5, B = 2

That’s it

[midnight haven]

Ok thanks

[candid spire]

So A+B = 7, and that’s your answer

Np

[midnight haven]

İll check out again

From start

[candid spire]

Ok 👍

[midnight haven]

Cause i gotta know this

.close

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[shut sierra]

how did we go from that fractional exponential onto them multiplying with negative exponents? If this is a rule could someone please explain it to me.

[pallid leaf]

(1/2x)^r is same as

(1/2)^r × (1/x)^r

[calm wing]

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

[jolly parrot]

artemetra

[pallid leaf]

Same as 2^-r x x^-r

[calm wing]

then split

[shut sierra]

ohh, I understand, we just took reciprocal inside of the brackets which caused both 2 and x to be multiplied with -1, which then multiplied with r.

Thank you!

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[cedar scarab]

How do I go about this? This was the one class that I missed and I have an exam on this in 8 hours.

[cedar scarab]

It's probably extremely obvious, I am just not sure which function on my calculator this is.

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[woven briar]

someone help mw with this

[midnight haven]

What is the question?

[woven briar]

algebraic fraction

im at 4 rn

ik im supposed to change 7 int fraction

done that

[sudden pumice]

?

[sudden pumice]

yep

[woven briar]

ive done that

[sudden pumice]

what was the result

[woven briar]

3/a + 7/1

[sudden pumice]

u have to make the denominator same

[woven briar]

change a into 1?

i was tryna find the LCM

[sudden pumice]

The lcm would be' a' since the lcm of 1 and a is a

[woven briar]

oh ok

so do 3/a + 7/a

and then 10/a

[sudden pumice]

yes

noooo

[woven briar]

ty

wat

[sudden pumice]

U gotta multiply 7 by lcm

[woven briar]

7 x a

[sudden pumice]

yep

[woven briar]

and whats that

7?

cus a = 1

@sudden pumice ?

[sudden pumice]

Why is a=1

[pearl pond]

@woven briar Has your question been resolved?

[pearl pond]

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[hollow willow]

hello may anyone please help me with this last question of my hw, i have tried everything i could and it is still not correct

[potent steeple]

!status

[pearl pond]

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

[hollow willow]

3

[potent steeple]

Okay, would you mind explaining your steps, and the answer you got?

[hollow willow]

i have used a calculator to work out the final result and give it to 1 dp and it is still not correct

i have been trying to look for help

[potent steeple]

okay, what is the formula for the circumference of a circle

[hollow willow]

multiply the diameter by pi or 3.14, then round it to 1 dp

[potent steeple]

So the Formula for the circumference is:
C = 2(pi)(r)

Like you said

[hollow willow]

yh

[potent steeple]

But we need to substitute pi for 3.142

So its:
C = (3.142)(1.8)

^ Could you solve that?

[hollow willow]

gimme a sec please

[potent steeple]

np

[hollow willow]

yh i got it correct thank u

now i can submit

[potent steeple]

Awesome

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[hollow willow]

alr

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[violet sparrow]

How would you differentiate this?

[midnight haven]

diffrenciate the term in the prenthacies

then use f`g+fg`

[violet sparrow]

ohhhhh

[violet sparrow]

thank you! Have a great day

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[river aurora]

I'm a bit stuck on this question:

Approximate f by a Taylor polynomial with degree n at the number a.

f(x)=sin(x), a=π/6, n=4, from x=0 to x=3.

All of our previous questions have been Maclaurin series, and so I'm not quite sure how to go about putting in the a value.

[pearl pond]

@river aurora Has your question been resolved?

[river aurora]

<@&286206848099549185>

[stray sequoia]

a taylor polynomial isnt that different from a mclaurin one

how do you tranlate a function along the x axis

[river aurora]

So do I change the original function, to f(x)=sin(x-π/6)? Or do I input it into the sum after I get the Maclaurin for sin(x)?

[stray sequoia]

well it wanted a taylor series centred at a

how would you modify a mclaurin series to make the centre a instead of 0

[river aurora]

My question exactly 😵‍💫

You subtract a, right? Or do you add it? I'm lost

[stray sequoia]

right so call this mclaurin series M(x)

how could you transform the function

[river aurora]

(x-C)^n?

Oh no wait

You put it in the f(n)(C)?

[stray sequoia]

almost!
you now just need to account for the centre of the nth derivative

[stray sequoia]

thats the second piece of the puzzle

good job

[river aurora]

So with the sin function, where do I put it? Which part is the nth derivative?

The coefficient is (-1)^n/(2n+1)!, right?

[stray sequoia]

how was this simplified form obtained from the general one

[river aurora]

Deriving sin(x)?

I think I know where to go now, thanks for your help

[stray sequoia]

youre welcome :)

[river aurora]

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[midnight haven]

hi guys

[midnight haven]

how do i solve this

in the notes it says X_t ~ Poi(lambda t)

so i was thinking U ~ Poi (2 * lambda)

but that is not one of the choices

i think V ~ exp (60/ lambda)

[fluid axle]

yeah the question seems screwed up

your reasoning is fine

[midnight haven]

oh alright

thanks so mcuh

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[silver imp]

Hi

[silver imp]

What is the derivative of $\frac{d}{dy} {ln(y+1)}$?

the inner is 1

ln(y)= 1/y

[jolly parrot]

maizz01

[silver imp]

Whats the solution?

1/y * (1+y) + 1?

[fluid mortar]

It’s 1/(y+1)

[silver imp]

yes how

ln(y)=1/y

Y+1= 1

derivatives

[fluid mortar]

Y is y^1, using the power rule, you get 1(y^0) which is one

For the derivative of ln it’s the derivative of the whats inside ln over 1

[silver imp]

I cannot follow like this. Can you write it in the Texit bot? It would be much easier to follow

<@&286206848099549185>

[buoyant stump]

You just have to use the chain rule. Are you familiar with that?

[silver imp]

Actually yes but I‘m stupid rn and do not get the answer I want

[buoyant stump]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[silver imp]

u' is wrong ig

but I also tried without that

without (y+1) at u'

Solution is 1/y+1

[buoyant stump]

I am not quite sure what you have done as i can't tell apart between u and v. 😶

Anyway, you don't even need u and v. You only need one.

Take u = y+1

Now, use this.
$\frac{\dd{(ln(u))}}{\dd{y}} = \frac{\dd{(ln(u))}}{\dd{u}} \cdot \frac{\dd{u}}{\dd{y}}$

[silver imp]

u=ln(y)
v= 1+y
u'= 1/y
v'=1

[buoyant stump]

Why are you even assigning u to be ln(y)?

[silver imp]

wasnt it:

[jolly parrot]

Enemagneto

[buoyant stump]

Edited.

[buoyant stump]

Yes.

[silver imp]

okey lets take this

[buoyant stump]

What is your u' ?

[silver imp]

1/y

[buoyant stump]

You are naming variables in a confusing way.

u(v(x)) = ln(v(x). Okay?

So u' = 1/v(x)

And v(x) = 1+y

Gosh. Naming of variables is so bad. Lol

[silver imp]

no because u is my outer function. For chain rule i need inner and outer function. ln(y) is the outer function and if you derivative it=> 1/y

Just a moment!

[buoyant stump]

You can't give (1+y) the name "y".

It's confusing.

[silver imp]

So just 1+y?

[buoyant stump]

Give it some other name. u, v, x, t, w, a, b, or whatever.

Anything but y.

[silver imp]

okey x

[buoyant stump]

Fine. So, x = (1+y)

[silver imp]

Yes

[buoyant stump]

So expression that you have to differentiate is now ln(x). Okay?

[silver imp]

Yes

[buoyant stump]

We can't differentiate ln(x) with respect to dy.

So we use chain rule.

We differentiate it w.r.t. x and then differentiate x w.r.t. y.

i.e.

$$\frac{\dd{(ln(x))}}{\dd{y}} = \frac{\dd{(ln(x))}}{\dd{x}} \cdot \frac{\dd{x}}{\dd{y}}$$
$$\frac{\dd{(ln(x))}}{\dd{y}} = \frac{1}{x} \cdot \frac{\dd{x}}{\dd{y}}$$
We know that $x = 1+y$
$$\frac{\dd{(ln(x))}}{\dd{y}} = \frac{1}{1 + y} \cdot \frac{\dd{(1+y)}}{\dd{y}}$$
$$\frac{\dd{(ln(1+y))}}{\dd{y}} = \frac{1}{1 + y} \cdot 1$$
$$\frac{\dd{(ln(1+y))}}{\dd{y}} = \frac{1}{1 + y}$$

Okay so far?

[silver imp]

What is w.r.t.?

[buoyant stump]

With respect to

[silver imp]

ah

[silver imp]

Go on

[buoyant stump]

Now, What is derivative of ln(x) ?

[silver imp]

1/x

[buoyant stump]

Good

[silver imp]

Lemmie show short

[buoyant stump]

Now, remember what x was ?

x = (y+1)

We substitute x.

[silver imp]

the problem came from here. Its a clue why ln(x) is 1/x

[silver imp]

Or no. Wait

[buoyant stump]

Now finish it.

yes

[silver imp]

Yes go ob

[buoyant stump]

Do the last part.

What is derivative of (1+y) w.r.t y?

[silver imp]

1

[buoyant stump]

YES

[silver imp]

uff. Would it work if we do not substitute?

[jolly parrot]

Enemagneto

[silver imp]

Actually I do it almost every day at home but today It didn’t work

I do it by my own

[buoyant stump]

I don't think there is another way. It is very simple though. At one point, you would not need to do explicitly all this. You assume it to be some variable and do it all in your head.

[silver imp]

Its a little difficult alone

Some steps to understand

[silver imp]

Okey

[buoyant stump]

Umm... I think that as long as you don't name variables in a confusing manner and do some practice, you should be good.

[silver imp]

👍

Thank you

No one helped me really except you

:)

[buoyant stump]

Happy to be of help.

Once you are done, close the channel.

[silver imp]

.close

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[main pollen]

can someone tell a 6th grader how to do calculus in the most simplest way possible

[vivid pilot]

um, how?

calculus is comprised of many different topics

also why does a 6th grader need calculus

[old marsh]

‘How to do calculus’ wdym by that

[old marsh]

Why not, if they’re interested, let them pursue it

[vivid pilot]

alright no stopping them

[main pollen]

ok

[vivid pilot]

but this is still a very vague question

[old marsh]

The most esteemed king of math Sir Tao supposedly did calculus at the tender age of 7

[main pollen]

hoq

[vivid pilot]

sir tao is built different

[main pollen]

how

[old marsh]

That’s how

[main pollen]

dont you need to know algebra 1-3

[old marsh]

He was probably doing algebra in the womb homez

[main pollen]

ok i brb

[vivid pilot]

and precalc before he could talk

[feral leaf]

https://www.youtube.com/watch?v=TzDhdvVg9_c&ab_channel=LukeyB.ThePhysicsG
It's not an explanation but something I found that might help

[main pollen]

ok

isn't calculus just solving math problems like if you were to drive somewhere and you drove 40 mph and it takes 1 hour it's simple, But doesn't calculus add how many times you stop at traffic lights, etc

[pearl pond]

@main pollen Has your question been resolved?

[pearl pond]

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[sterile crystal]

its a definite integral

[dire snow]

is it just without the U?

[sterile crystal]

i dont know why youre adding C

[dire snow]

Idk I thought u always do that when integrating

[sterile crystal]

were evaluating

[near haven]

this whole thing is wrong

[sterile crystal]

rip

use summation

[dire snow]

.close

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[broken burrow]

I'm learning about trigonometric ratios and ive come up with this questions but idk how to solve it.
Can someone help me

[broken burrow]

I can calculate YR

but im having a confusion with how should I calculate the hypotenuse in the xrz triangle when im only given with one side of it

yeah that's fine

so?

yeah i know this one i just dont know how to use the trig functions

yeah

tangent?

ohh so based on the sides i know i can determine that which trig functions i should use

wait a second, let me just try to process this, its new for me

ohh yeah, like how to express it in terms of the adjacent

ill start doing it and see whether ill be stuck or not

so thats o over t

yep

i think so let me just go thru it and see

I got t right

thanks very much for helping

i appreciate it

@wind salmon what do these sign mean?

okay, and what do the signs around rpt mean?

okay got it thanks

[pearl pond]

@broken burrow Has your question been resolved?

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[humble tinsel]

Using fermats little theorem, find the inverse of 3 mod 17

[humble tinsel]

So, i think the inverse using fermats little theorem =
3^15 mod 17

But whats next?

Calculate 3^15?

How>

[shrewd basin]

I mean that’s not hard to do you can do (3^3)^5 for example

[humble tinsel]

Ah yes 27^5 very easy

[shrewd basin]

What’s 3^3 mod 17

[humble tinsel]

Oh i can do mod here

okay

10^5

Hmm

[humble tinsel]

okay so i have (10^5) mod 17 now

cant rly see what the next step is

[hallow ice]

You can keep doing the same thing where you break the number up and then reduce it mod 17 to make it easier

I.e you observed that 3^15=(3^3)^5=10^5

now you can break 10^5 up into smaller chunks

[shrewd basin]

What’s so hard about 100000 / 17

If you really want to (10)^5 = 2^5 • 5^5 = -2 • 3125

Now just calculate 3125 mod 17

[humble tinsel]

Bruh.....

I just kept splitting it like Layne said

Way easier than doing all that

[shrewd basin]

Okay…

[humble tinsel]

😭😭😭

[humble tinsel]

you realise i dont have a calculator right

💀

[shrewd basin]

Do you not know how to do division?

[humble tinsel]

No

[shrewd basin]

Okay good luck

[frozen lantern]

💀

[shrewd basin]

5•5 = 25 = 8 mod 17
5^5 = 64 • 5 = -20 mod 17

This good enough?

[humble tinsel]

Yes

I did it like this, but yours works too indeed

[pearl pond]

@humble tinsel Has your question been resolved?

[pearl pond]

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[waxen gate]

The question itself is larger, but technically it just says this, if N is a natural number that is <1000 then how many values for N confirm that N(N+12)/N+3 is irreducible

[waxen gate]

I tried adding +27 -27 to complete a reduced form so it has N+3 as a factor to then use it and see if the results are irreducible for any value that N takes, but it ended up giving me 334 values that are reducible, which is 666 values for N that make it irreducible, however, there is no option for 666

[pearl pond]

@waxen gate Has your question been resolved?

[pearl pond]

@waxen gate Has your question been resolved?

[waxen gate]

<@&286206848099549185>

[pearl pond]

@waxen gate Has your question been resolved?

[midnight haven]

@waxen gate I am doing this type of question for the first time but If N is a natural number that is <1000 then how many values for N confirm that N(N+12)/N+3 is irreducible then n can be all prime natural number except 3 and it's multiple and all multiple of 2 which are not a multiple of 3.

[waxen gate]

does it need to be necessarily prime numbers?

because if so, then the answer would be less than 200 (I think) which also doesn't appear as an option

options are a)669 b) 667 c) 333 d) 334 e) 668

[midnight haven]

No all prime numbers except 3 and it's multiple. AND ALL multiples of 2

@waxen gate

[waxen gate]

that would end up being 666 I think?

maybe I'm doing some calculations wrong

but that is the answer I get

[woven matrix]

N(N+12)/(N+3) is irreducible iff gcd(N(N+12), N+3) = 1
N²+12N = N²+3N+9N = (N+3)N+9N, so by your +27-27 trick you found that gcd divides 27 = 3^3
so either N = 0 mod 3, in which case both N+3 and N(N+12) are 0 mod 3 -> not irreducible and vice versa
so you're counting the integers in [[1, 999]] that are not divisible by 3
there are 999 numbers in total and the multiples of 3 are 3, 6, ... , 999, there are 333 of them
999-333=666, so the answer is 666
so I see 3 possibilities:

• you're in a country that includes 0 in natural integers
• the question is <= 1000 and not < 1000
• the question is wrong

[waxen gate]

ok so, I think the question is wrong, because all the other possibilities aren't, my school doesn't use 0 as a natural number and it says on paper that is <, so yeah, I'll just say to the others that 666 is the answer, thanks

.close

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[steep walrus]

how does this factor into that?

[steep walrus]

when I multiply it out I get it but moreso wondering how I'm supposed to quickly recognize that

<@&286206848099549185>

[grizzled dust]

!15m

[pearl pond]

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[plush bramble]

Multiply it out and see

[grizzled dust]

$4t^2+4+\frac{1}{t^2}=\frac{t^2}{t^2}(4t^2+4+\frac{1}{t^2})=\frac{1}{t^2}(4t^4+4t^2+1)\
=\frac{1}{t^2}[4(t^2)^2+4(t^2)+1]$

[jolly parrot]

Soosh

[grizzled dust]

maybe its easier to see as a square if you think of it that way

and after factoring you multiply 1/t into each individual factor

and as far as how you would recognize that...well i think the coefficients of 4 4 1 kinda hint at you and the powers of x decreasing with even steps: 2, 0, -2. my instinct would be to get rid of the denominator by multiplying everything with t^2 as i showed, so to do that i would need also a 1/t^2 to cancel out and that's how i'd come up with t^2/t^2

[steep walrus]

I already said I did that, I'm moreso wondering how I'm supposed to quickly recognize that..

[steep walrus]

o shoot that makes a lot of sense

[steep walrus]

wait maybe I'm being slow rn but when I get to that last line where I have 1/t^2 (4(t^2)^2 + 4t^2 + 1), how would I factor that lol

[grizzled dust]

how would you factor 4u^2+4u+1

it's just same thing u = t^2

[steep walrus]

so I would get (2u+1)^2 which would become (2t^2+1)^2

oh and then I'm dividing that by t^2 oops I forgot ab that

lmao that's y I was confused my bad

[grizzled dust]

yeah and give 1/t to each factor

since you have 2 factors

[pearl pond]

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[pearl pond]

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[devout totem]

Can someone help me asap?

[inland lantern]

[pearl pond]

@devout totem Has your question been resolved?

[pearl pond]

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[hexed path]

Guys help hahahahaha

[hexed path]

I really dont know how.....

[cinder flower]

Bruhh

[hexed path]

Dont mind that hahahahaha

[cinder flower]

haha

[hexed path]

Guys how hahahaha

[cinder flower]

have you tried writing n+1 choose k = (n+1)!/(k!*(n+1-k)!) etc and just using algebra

[hexed path]

?

[cinder flower]

$$\frac{(n+1)!}{k!(n+1-k)!} = \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-(k-1))!}$$

[jolly parrot]

layla

[hexed path]

I dont get it

[cinder flower]

what i wrote is the same thing as what you wrote

[hexed path]

Oh

[cinder flower]

do you know ${n \choose k} = \frac{n!}{k!(n-k)!}$

[jolly parrot]

layla

[hexed path]

Oh ok thank you

I forgot to close hahahahaa

.close

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[cursive mantle]

anyone can help me with these geometry questions its just 3 ill appreica te it ill send one by one as anyone helps me

[glass salmon]

shoelace

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[vivid pilot]

how can i answer parts 3 and 4? not sure what to do

[vivid pilot]

also seems like i got part 2 wrong

[warm current]

3. Population is increasing when dP/dt>0

[vivid pilot]

right but how do i find when it is >0

[warm current]

You are given dP/dt in terms of P. So solve for P.

[vivid pilot]

in terms of dP/dt?

also do you know why part 2 is wrong

@warm current

.close

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[lethal plover]

hello

[lethal plover]

when you divide an equation on both sides by a variable, do you lose an answer?

[robust oar]

like what specifically

[pearl pond]

@lethal plover Has your question been resolved?

[warm ledge]

sometimes yes
for example
x = x^2
if I divide both sides by x I get x = 1
but x = 0 is a solution too!

its better to do x^2 - x = 0 which gives x(x-1) = 0 which gives me x =0 and x = 1

when you divide by something, you're asserting it is non-zero but its possible that something being 0 is a solution that you just got rid of!

[pearl pond]

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[lethal plover]

ohhh, thank youu

[lethal plover]

but how about if you subtract both sides by a variable? the equation wouldnt lose an answer right

[karmic fern]

nope

the reason why multiplication/division can remove/add solutions is because you may be multiplying or dividing by zero without knowing it

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[pulsar umbra]

im trying to get the domain for this but i keep getting {xer|x≠-pi/4 + pi/2(k), kei}. the answer should be x≠-3pi/4 + pi/2(k). what am i doing wrong 😔

[pulsar umbra]

if anyone figures it out pls ping me...

nvm guys im just so smart i figured it out by myself

im so awesome

.close

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[lethal plover]

i just asked htis question but i dont think i fully get it

[lethal plover]

i understand that the bases are all powers of 2, but i dont understand how you solve it after switching the bases to from 16-> 2^4 and 4-> 2^2

[hasty crater]

log(a^n, x) = (1/n)*log(a, x)

[lethal plover]

is that another log rule? i think my teacher only wants us to use the multi/div/power/change of base rules to solve problems

[hasty crater]

that is the power rule. no?

[lethal plover]

but i thought the power rule only works on the argument, which is x?

[hasty crater]

there are 2 power rules
log(a, x^n) = n*log(a, x)
log(a^n, x) = 1/n*log(a, x)

[hasty crater]

works both ways

[hazy crypt]

have u tried using change of base to prove the "second power rule" tbh I've never heard of it called that

[lethal plover]

ohh i see, is there any other possible way of solving it without using the other power rule

[hazy crypt]

put everything in terms of log 2 using the change of base

[lethal plover]

okay let me try rn

[hasty crater]

idts. you wont be able to simplify it otherwise

man, i have no idea wut each of these stuffs are called. i just know that these exist

names are confusing and too hard to memorize

focus on the main stuff

[hazy crypt]

let me know once u got smth

I have the answer

too

[lethal plover]

idk my teacher only gave us those rules so i assume its solvable? and i dont think he'll allow the seocdn rule?

[hazy crypt]

like the change of base part

ur allowed change of base right

?

[lethal plover]

im stuck, do you mind telling me

[lethal plover]

yupp

[hazy crypt]

ok

I'll give u the first part

and then u can do the rest

[lethal plover]

sounds good

[hazy crypt]

$$\frac{log_2(x)}{log_2(16)}$$

[jolly parrot]

WOWWA

[hazy crypt]

is the first term

does that make sense?

I used change of base

[lethal plover]

yes, i understand that

[hazy crypt]

ok

and what is that denominator

what is log_2(16)

[lethal plover]

4

[hazy crypt]

exactly

so it simplifies to log_2(x)/4

do the same for the rest

[lethal plover]

ohh i got it

so it would be x/4 + x/2 + x/1 =7?

[hazy crypt]

log_2(x)

instead of x

for what u have

make sure u get why

[lethal plover]

right, and then would i factor out the log_2 (x)

[hazy crypt]

yeah sure

u can do like that

[lethal plover]

wowow, thank you, i got the answer

[lethal plover]

is there another way of doing it w/o factoring?

[hazy crypt]

nah it's like same thing

i just think abt it differently

if I have
(x+1)/ 4 + (x+1)/2 + (x+1) I just add the fractions in my head which give 7(x+1)/4

it's just a in my head thing

it's basically me skipping the factoring part

no difference

[hazy crypt]

surely you could also just multiply everything by 4

and then do something like that

lots of ways

[lethal plover]

got it 👍 thank you so much

if you are still available, i have one more question

[lethal plover]

this one is a bit tricky because the exponent is around the whole log thing

[hazy crypt]

ya

@lethal plover first look at the second term

on the lhs

what can u do

[lethal plover]

i can use the power rule?

[hazy crypt]

ya

so what happens then

[lethal plover]

idk ...? it looks like a quadratic so i can subsitute log x with a?

[hazy crypt]

bingo

[lethal plover]

WOW

THANK YOU

AGAIN

[hazy crypt]

lol that one was basically all u

[lethal plover]

not rly LOL, i needed the walking through to get it