precalculus

[jovial coral]

But for the y we have to take the derivative of y with the standard notion that we are derivating with x

I'm not really good at teaching I'm sorry if you don't understand

[stiff depot]

Instead of checking the tangent line at x you do it at the y

[winter comet]

?

[proven night]

Hello guys

I always manage to fail when it comes to trig equations with restricted range of solutions, for example cos(x) = 0 but x € [-pi;pi]

Am i right if i just get the general solution (pi/2 + pi*k) and then i just substitute k with random integers (for example, -1, 0, 1 etc)?

And then i just check whether the solution is within given range

[summer ruin]

yes

but it's obvious from the trig circle which x solve this particular equation

[foggy valley]

I recently just got in pre calculus class and was talking about circles. I've been stuck for 2 hours in this assignment I have which is

"Find the equation of the circle that passing through p1 (-3,6), p2 (-5,2), and p3 (3,-6). (General equation)" I watched a tutorial video with each having different methods, I had done the equations 1,2,3 and so on. Doing the elimination method. But I just don't seem to know if I'm doing it correctly

P.s I even used some ai like chatgpt and IT MADE MY UNDERSTANDING WORSE

[uncut mulch]

don't use gpt, show what you did

[frank spindle]

Dont use chatgpt with that stuff

It spits out nonsense 99% of the time regarding math

[foggy valley]

So I followed this video tutorial BUT IDK I'M NOT SURE IF I'M DOING IT RIGHT

I did the equations and subtract the equation 1-2 which you'll get the equation 4 and 1-3 which would be your equation 5, after that you'll subtract it again

I DONT KNOW

"Find the equation of the circle that passing through p1 (-3,6), p2 (-5,2), and p3 (3,-6). (General equation)"

[frank spindle]

The equation of a circle is (x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center

Thats the general equation if i recalll

You can easily solve for r with the information given

[uncut mulch]

its more or less the same, you end up with a system of 3 equations

one thing that stands out is the notation issues

when squaring stuff like -3 and -5
you MUST have () around them

${\red{(\black{-3})}}^2$

[foggy valley]

I don't know if I'm even doing it right anymore:( I'm lost

[obsidian monolith]

ℝαμΩℕωⅤ

[uncut mulch]

can't check every single detail
but you seemed to have reached a system of two equations
try solving that to get the values of D and E

[foggy valley]

I don't even know, like can I get the value of D or E in this equation "(16 + 2D + 4E) - (-6D + 12E)"

What did I do wrong T-T

[uncut mulch]

if you want to approach it with elimination,
you'd want the magnitudes of one of the coefficients to be the same before you add/subtract, depending on personal preference

which can be achieved by multipling one or both equations by an appropriate value

[foggy valley]

If I want to eliminate D for example

I'll multiply

16 + 2D + 4E to 6

while I'll multiply -6D + 12E to 2?

[foggy valley]

ooOoOooOoOo

[uncut mulch]

you can be more efficient considering lcms

considering that the lcm of 2 and 6 is 6,
its sufficient to multiply that first equation by 3 or -3

[foggy valley]

IDK IF IM DOING IT RIGHT

[uncut mulch]

show your attempt?

this is going on in two places, choose one,
here or #help-32

[uncut mulch]

<@&268886789983436800> political shitpost

[modest vine]

Solve for the ctr of the circle by solving for the intersection of the perpendicular bisectors of two pairs of points

At least i think that should work idk

Yep that works

[daring tapir]

Whats a fucking GPT Im interested

[winter comet]

maybe it is what it sounds like...

i don't think we wanna know

💀

[daring tapir]

I wish it is what it sounds like lmfao

[daring tapir]

Real

[winter comet]

🗿 💀

[muted sun]

Hello can anyone help me understand this?

[willow skiff]

tan theta represents the slope of the red line

If you change the angle by 180 degrees, the point will become directly opposite and thus have the same slope

Same for 360 deg ofc

[muted sun]

ty a lot

I appareciate

[trim isle]

hello,can anyone help me? I don't fully understand what I'm doing

[somber void]

that's a cool problem i reckon

So ok

1x - (-1)y = -3
and
7x - (1)y = 7

We need to do 2 things:

1. We need to find the intersection point, because bisector has to pass through that

2. We need to find the slope of the bisector

For 1 we can just solve the system of equations to get

2 is cool i think

because

1x - (-1)y means that for every 1 unit to the right the line goes 1 unit down.
7x - 1y means that for every 1 unit to the right it goes 7 units up

it would be nice if there was some system that represents rotation with pairs of numbers...

oh wait

it's the complex numbers

so
-1 + i and 1 + 7i represent the angles

we can multiply them together to get double the rotation
(-1 + i)(1 + 7i) = -1 -7i +i - 7 = -8 - 6i

but the problem is that we actually need their average

which means that we need to find the square root of that somehow

[hasty lily]

Angle bisector?

Wouldn’t basic trig suffice

[somber void]

i mean yeah but that's boring

(a+bi)² = (a²-b²) + (2ab)i
(a+b)(a-b) = -8
2ab = -6
a = 1
b = -3
works

[hasty lily]

Complex numbers are just trigs

Actually they’re based on e^x and Taylor series

[somber void]

1-3i

[hasty lily]

What’s with my enter key today

[somber void]

it was better when it just said "actually they're based"

[modest vine]

Same thing anyway

My god I just wanted the image

But the trig part is from that right

Oh nevermind it's just vectors 😭

[somber void]

you get
x = 1/2
y = -7/2
and then from part 2
3x+y = c
3(1/2) + (-7/2) = c
c = -2
3x+y=-2 is the line

on one hand we got lucky that the sqrt is nice

on the other hand we got unlucky that we had to do this by hand

[willow skiff]

https://www.desmos.com/calculator/jclvmbhiow

that is unfortunately the wrong line: you have bisected the obtuse angle

luckily the fix for this is to make one of the sides negative

the correct version of the distance to a line formula has absolute value signs: for example, if we have |x| = |y| and x = y doesn't hold, we must thus have x = -y

then it's literally just:

[trim isle]

I had a feeling It was, I was watching a obtuse angle solving question. I thought it was the same thing for acute, turns out not to be correct. 🤔

[willow skiff]

I have another approach using the angle bisector theorem actually:
the point (0, -7) lies on the blue line and the point (4, -7) lies on the red line, and their intersection is at (0.5, -3.5)

so using the angle bisector theorem
our bisector must pass through $\left(4 \cdot \frac{ \sqrt{0.5^2+3.5^2} }{\sqrt{0.5^2+3.5^2} + \sqrt{3.5^2+3.5^2} }, -7 \right)$

(https://en.wikipedia.org/wiki/Angle_bisector_theorem)

wait that's not right, I checked on Desmos

[willow skiff]

WAIT WHY IS 2 YOUR DENOMINATOR

that explains everything

[trim isle]

i did say I didn't know what I was doing

[willow skiff]

okay that makes more sense

[trim isle]

yeah

[obsidian monolith]

south

[willow skiff]

simplifies down to (5/3, -7)

[trim isle]

question: how to know if I've bisected the obtuse angle or the acute angle.. ( teacher hasn't taught us yet )

[willow skiff]

if you calculate the slopes of the two original lines, you get -1 and 7

from slope = -a/b

so that means the bisector has to have negative slope, if you make a sketch

[willow skiff]

now, you can check the coefficients of x and y in here
if 0 is on the RHS, you get y/sqrt(2) + y/sqrt(50) so positive
and x/sqrt(2) - 7x/sqrt(50) so also negative

so the slope -a/b will be negative * negative/positive = positive

so that means that if you choose the positive sign on both sides of the equation, that's the wrong bisector

you need to make one of the sides negative

[trim isle]

hmmmmm, ok, i can comprehend what you just said. I'll do the rest, Thank you c:

[willow skiff]

no worries!

if there's a better method it's probably more slick

but you need to figure out which of the angles is acute which depends on the two lines so it can't be that short I think

nw

[timid belfry]

Can I get some help with this one pls... it's expaded to standard equation of a circle. I used the method my teacher used for the later question but my classmates all have different answers on no.4. I genuinely don't know what I'm doing...

I've been up the entire night trying to figure out which part I got wrong...

[still tendon]

Hi im not rlly sure where to ask this but i just started pre calc (only half way through functions unit) and am kinda annoyed there r still word problems theyre actually the bane of my existence. I was wondering if they get easier in the future with calc n uni math

[winter comet]

it might get easier in the sense that you will be already familiar with word problems

how hard your problems are really depends on your teacher tho

it might get "easier", might get "harder"

[uncut mulch]

when you are familiar with the terminology, there shouldn't be too much of an issue setting up the equations
whether you have difficulty solving those is a separate issue

[modest vine]

32/4=8

You made an error when factorising

[timid belfry]

OMG THANK YOU 😭

[still tendon]

I see thank you!

[still tendon]

Yeah its just setting them up

[oak bluff]

Uh Im kind of confused because it looks like none of the options are correct

[cyan berry]

Im pretty sure its the top one

[oak bluff]

it was

[cyan berry]

Do you understand why?

[oak bluff]

but thanks anyway

yeah

[cyan berry]

All goodie then

[oak bluff]

(2^2)^t/2

2^t

[cyan berry]

Yeah thats basically it : )

[oak bluff]

: D

[unreal dawn]

yeah for this one you have the intiial set up correct

you can just check both lines and see which does have the positive slope as someone lse said

[brittle geode]

help!

i did tan^-1(9)= 83.6

and then rounded to 83.7

[winter comet]

but are there any other angles that satisfy tan(theta) = 9 between 0 and 180 deg?

[brittle geode]

oh

so 180-83.6

= 96.3 ?

when i added 180 to 83.6 i got (263.7)

which i discard because is over 180 degrees

[winter comet]

nah wait what

there aren't any other angles what am i on

i have no idea why thats not correct 💀

[brittle geode]

okay

thank for your time

[winter comet]

have you tried putting in 83.6 just in case they round wrong?

or u only have so many tries 💀

[brittle geode]

i have 2 tries

i used one

so its my last chance

[winter comet]

yeah ok so you gotta be sure 💀

uhh

[winter comet]

can you ask your teacher?

[brittle geode]

i emailed her on friday but no response

[winter comet]

dang

[brittle geode]

I checked multiple times

and notice that for that specific problem i had to put the degree sign

which didnt really make sense because the answer was telling me to put in degrees already

but thank you for your time and patience!

[winter comet]

bruh

how do you even put a degree symbol 💀

is part of the program?

[brittle geode]

no 💀

LOL

[winter comet]

💀

[nocturne cave]

I found someone saying that it was possible to solve a linear system using matrices, and he showed this equation. Can someone tell me what the name of this method is and if it really works? I would like to learn more about it.

rx + sy = t

[summer ruin]

the inverse matrix

[hushed sphinx]

(That was also asked in #prealg-and-algebra, with more answers there).

[flat shuttle]

Would a quadratic model have small residuals and the residuals plot show no pattern?

[brittle geode]

hi,

i just wanted to double check this problem

my answer is -sqrt3

is that right?

🙂

[desert void]

yep

[modest vine]

Uh let's say there's a type of sequence that is defined by two seed values
F(1) and F(2) and F(n+1)=F(n)+F(n-1) basically the fibonnaci sequence but for any two seed values

How can I determine if two pairs of numbers belong to the same sequence

Like for example for the normal fibonacci sequence both 2,3 and 1,1 belong to the same sequence

I could just continue the sequence but that'd take a bit long in some cases

[somber void]

If F is increasing you can do a binary search

You can also maybe do mod, and see if it appears in the same sequence

there's also F(1)f(n) + F(2)f(n+1) = F(n+2)
where f(1) = f(2) = 1 and f(n+2) = f(n+1) + f(n), and F is your sequence

[somber void]

There's also some obvious cases where it won't be. For example if your sequence is 2, 2, 4, 6, ... then 7382921 is never in the sequence

[normal shadow]

how can i use a calculator to get coordinates on a unit circle

i have memorized the method to remember radians and degrees on the unit circle

but i dont know how to figure out the coordinates / memorize them using some kind of rule

so how can i get them with a calculator

[placid bone]

well I don't think there's a traditional calculator that can give you the coordinates in (x,y) form

but you know that the x coordinate on a unit circle is cosine of the angle, and the y coordinate on a unit circle is sin of the angle

so if you have the radians/degrees memorized, you can find the coordinates pretty easily off of that principle

but it's honestly 10x better to just memorize the coordinates, makes trig problems so much smoother

[modest vine]

Alright ig

[somber void]

idk what that means. Just take the number of turns you want around the circle and multiply it by τ (or 2π) this is the angle in radians. Or multiply it by 360 to get it in degrees. use cos(angle) to get the x coordinate and sin(angle) to get the y coordinate. What are you trying to memorize here?

you want 1/6 of a turn? that's τ/6 radians and 360/6 degrees. Put that into calculator and you have your x and y. Don't have calculator? Realize that 60° is the angle inside the equilateral triangle. Drop height and see that cos(60°) = 1/2. Use pythagoras to get sin(60°) = sqrt(1² - (1/2)²) = sqrt(3/4) = sqrt(3)/2 that's something like 0.85 i think

[normal shadow]

when solving for inverse values of trigonometric functions, do I have to graph them and/or look at the unit circle to find their values, or is there some kind of equation that can be used

[river drift]

not really any way to find values of trig or inverse trig functions other than known values

[lyric sleet]

anyone help me with this?
300.
A toroid is a rotational body where a plane figure rotates around a straight line that does not intersect the plane figure. For example, the holed wooden ball from the previous problem is a toroid. If the rotating plane figure is a circle, it forms a torus, which is a special case of a toroid.

a) Calculate the volume of a bicycle's inner tube when its inner radius is 25.0 cm and the outer radius is 29.0 cm. The cross-section of the inner tube is a circle.

b) Calculate the volume and mass of a gold ring when the cross-section of the ring is a semicircle, the inner diameter is 18.0 mm and the outer diameter is 19.5 mm. The density of the gold alloy used in the ring is 17.3 kg/dm³.

I'm currently doing the **b) **, and been stuck getting answers like 49.96 or 52 when I should get 51.7

The second picture is is the a), because the two are similar so I think you solve them quite the with the same technique.

The radius of the inner circle is 9 (18/2) and outer circle 9,75 (19,5/2). So the width of the ring should be 0.75. Because you can think it as half of a round ring, it should 9 should be the distance from the origon to the center of the ring (if it was whole).
x^2-y^2=r^2 so then I think you could just (x-9)^-y^2=0.75^2 so x=9+-sqrt(0,75^2-y^2).
Then you plug it in the integral from -0,75 to 0,75 and put (pi/2)*f(y)^2=(pi/2)(9+-sqrt(0,75^2-y^2))^2 as the function. There then comes the problem as I get the answer 49,964.

[uncut kestrel]

Find the concavity of the functions in diff intervals $f(x)=\frac{x-2}{(x^2-x+1)^2}$

[obsidian monolith]

TᖇᗩᑎᔕᑭᗩᖇEᑎT ᔕᕼᗩᗪOᗯ

[willow skiff]

set f''(x) = 0 then make a sign diagram

like this, check the first derivativ

[uncut kestrel]

Find the concavity of the functions in diff intervals $f(x)=\frac{x-2}{(x^2-x+1)^2}$

[obsidian monolith]

TᖇᗩᑎᔕᑭᗩᖇEᑎT ᔕᕼᗩᗪOᗯ

[uncut kestrel]

I got the second derivative and got

And I got $x =\frac{4 \pm \sqrt{6}}{2}$ and $x=0$

[obsidian monolith]

TᖇᗩᑎᔕᑭᗩᖇEᑎT ᔕᕼᗩᗪOᗯ

[uncut kestrel]

I represented them on the number line

[willow skiff]

,w second derivative of (x - 2)/(x^2 - x + 1)^2 = 0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by sooouth.

[uncut kestrel]

Isn't $\left(\frac{4 + \sqrt{6}}{2} ,+\infty\right) \approx \left(3.2 ,+\infty\right)$ ?

[obsidian monolith]

TᖇᗩᑎᔕᑭᗩᖇEᑎT ᔕᕼᗩᗪOᗯ

[willow skiff]

cool that's correct (there's also 2 + sqrt(3/2) which got cut out)

[uncut kestrel]

If I put 5 and above values I get f"(x) to be positive implying tat at those intervals it's concaved up

[willow skiff]

cause 2 + sqrt(6/4) = 2 + sqrt(3/2)

[uncut kestrel]

But when I put 4 it becomes negative

[willow skiff]

wait

oh right you are finding concavity, continue

[uncut kestrel]

And 4 is literally under tat interval I mentioned

[willow skiff]

f''(4) is still positive

[uncut kestrel]

No way Ig i did my second derivative wrong 💀

I got $x(2x^2-8x-5) = 0$

[obsidian monolith]

TᖇᗩᑎᔕᑭᗩᖇEᑎT ᔕᕼᗩᗪOᗯ

[uncut kestrel]

I'm equating the second derivative to zero

[willow skiff]

Wolfram Alpha tells me you should instead have

6x(2x^2 - 8x + 5) = 0

[uncut kestrel]

Bruhhhhh
I saw jn 💀
It was +5

Not -5
My brain

I hate quotient rule 💀

Anyways tysm 🥹😭@willow skiff

[willow skiff]

no worries

you can literally use Wolfram Alpha and Desmos/GeoGebra to check yourself

[uncut kestrel]

30 mins wasted 🤣

[willow skiff]

next time

[uncut kestrel]

Fr 💀

I'll do some other problems now
I'll be back if I need help ✨💫

[mystic sluice]

I’m looking for question a

How do I find the asymptotes of this function and there is 2 horizontal

[willow skiff]

the horizontal asymptotes are 1 and -1

as x tends to -infinity, you get -x^3 / (-x^3) = 1

as x tends to infinity, you get x^3 / (-x^3) = -1

because $|x| = -x$ if $x < 0$

[obsidian monolith]

south

[willow skiff]

|-5| = -(-5) = 5 for example

[young gust]

Hello

Precalculus is awesome XD

[keen fox]

@young gust hey which uni do u study at?

[mystic sluice]

Thx

[ember hemlock]

Hey, anyone from India??

I need help with the whole syllabus 💀

[modest vine]

I'm making a graphing calculator

How should I handle asymptotes

I use lines to connect points

So the asymptotes kinda have this vertical line connecting the two ends

How can I detect asymptotes

[daring tapir]

That depends a lot on what you are using

And a server for that particular language/library will be way more help than here

[modest vine]

Uh idk I'm just using raylib

Anyway im essentially just putting pixels

So it shouldn't really matter that much

[daring tapir]

Yea try some c++ server in that case

[modest vine]

Never mind then

The people there told me to ask a math server😭

That's why I came here

[daring tapir]

Rip- but i mean you need some way to get the derivative and then setting it equal to zero so its more of a language dependent thing, i dont really have enough experience with c++ to advice you directly on this

[modest vine]

No I don't need it in c++ I just need a general algorithm

[daring tapir]

Cuz writing the algorithm for finding the derivative is gonna be too lengthy and im pretty sure there is already some library for that

[modest vine]

Whatever ig I'll just search for libs 😭

[daring tapir]

Hmmm

[viscid thistle]

Bros being played like a volleyball 😭

[modest vine]

Recursive refferals 😭

[viscid thistle]

You gotta teach ai the calculations 💀

[daring tapir]

Well doable

But not recommended

I mean he could use the openai api

Which is doable but will probably be more expensive than coding it himself

[winter comet]

how to solve for t with just a calculator?

😐

oh they probably want you to graph it ._.

[jagged nova]

I had a question about partial fraction decomposition. If you are trying to split say 3/(x^2+1)(x-2), would it be impossible to split this fraction if you are limited to the real numbers? My reasoning is that typically to solve for the numerator of either fraction you need to set each polynomial equal to zero, but x^2+1 has no real solutions.

Sorry if this isn’t the right place for this question I didn’t know quite where to put it

[summer ruin]

x-2 has real roots, so it is possible

[viscid thistle]

It's free

For some time

[daring tapir]

It gave you some free credits the last time i checked

[winter comet]

you can set each polynomial equal to 0, but thats just one way

[viscid thistle]

Yea

[daring tapir]

Well i guess some is not enuff for a calculator

[viscid thistle]

I'll try

[jagged nova]

Yes, so I guess my question is if it doesn’t have real roots, does that imply it’s impossible. I know that setting each of the roots equal to zero is only one way. Is there another way that avoids this problem? I am only familiar with that method

[winter comet]

you can distribute it out, combine like terms. then use the fact that the equality is only true if the constants are equal to the constants on the other side, the x terms are only equal to the x terms on the other side, the x^2 terms are only equal to the x^2 terms on the other side, ect.

then you can create a system of equations to solve

[summer ruin]

,w partial fractions 3/((x^2+1)(x-2))

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by transparent_elemental

[jagged nova]

Interesting, thank you thank you. I am curious then if you can ever not separate the fraction, and if so what the necessary and sufficient conditions are for that to be the case.

[dapper rune]

Can somebody tell me how to prove that there are inf prime numbers?

[jagged nova]

For the sake of contradiction assume there are finitely many prime numbers, say n of them. Then consider the product of all of those prime numbers and add one to that:

$$p_1 p_2 p_3…p_n +1$$

Now you should be able to reach a contradiction from here. Does that make sense?

[obsidian monolith]

Squirtle Squad Gang Affliate #1

[jagged nova]

There are two cases, one where $$p_1 p_2…p_n +1$$ is prime, and one where it is not prime

[obsidian monolith]

Squirtle Squad Gang Affliate #1

[full mirage]

when finding how many complex zeros a polynomial can have, can you just graph the equation, count the zeros normally, then mulitply the number of zeros by 2?

[somber void]

counterexample:
x² + x + 1

has 2 complex zeroes, but 0 real ones

[river drift]

the fundamental theorem of algebra says that the number of complex (incl. real) zeros (counting multiplicity) = the degree of the polynomial

[uncut mulch]

the number of effort points needed to ping <@&268886789983436800>
(trivial addition troll)

[heavy merlin]

not gonna lie I'm just starting pre calc in school and I'm probably gonna need some help later on but hopefully I don't.

[turbid breach]

(𝑥 + 1)² + (𝑦 + 1)² = √26 how do I get the radius of the circle if it's already in a radical? Do I get the fourth root?

[turbid breach]

Same

[safe basin]

yeah as its written thats what youd do although i'd just doublecheck the problem again to make sure they didn't mean for 26 to be the right hand side and root 26 to be the radius

[turbid breach]

Ohhh okay, thanks!

[zinc compass]

On Sept 9th I'll have to take an exam about differential calculus, subjects are limits, derivatives, numeric successions, functions and so on.
Would you say Khan Academy is a good resource to learn these things?

I'm currently learning trig on Khan Academy, since I lack trig and as I understand it's definitely something to learn before heading to more advanced calculus concepts

[uncut mulch]

its a good place to start

[turbid breach]

Aren't all of these circles?

[summer ruin]

no

[uncut mulch]

how are you reaching that conclusion?

[turbid breach]

All of these are A = B. A being the coefficient of x2 and B being the coefficient of y2

I saw it from some video

Is it wrong?

Teacher hasn't really taught us anything, just told us to answer this

[uncut mulch]

its flawed

consider
x^2 + y^2 = -100
would you consider this to be an equation of a circle?

[hushed sphinx]

That teacher should be forced to take their own horribly mistypeset exam...

[uncut mulch]

also ^

[hushed sphinx]

^ indeed.

[turbid breach]

Ohhh

No

Ok I think(?) I get it

[uncut mulch]

so here, you'd want to do something like express these equations in standard form

[turbid breach]

Ah, thank you so much

[hushed sphinx]

We can shortcut (6) and (7), because once we find even one point where the LHS is <= 0, the equation must make a circle somewhere

[narrow briar]

Hi! One question: can I use Lambert’s W function for solving this equation? 2^x = -x^2 -3x +44

[hushed sphinx]

Looks unlikely.

The typical way to check is to type the equation into Wolfram Alpha and see wether it produces an expression that involves W.

[winter comet]

v and w and v+w are referring directly to vectors, not magnitudes, right?

💀

i feel kinda stupid i thought they were talking about magnitudes and i was like erm no 🤓

yea i think its just the vectors themselves 💀

[modest vine]

Yea they're talking about the vectors not their magnitudes

Classic tail to tip vector addition

Physics class ahh addition

[winter comet]

yup

[winter comet]

wwwut 💀

[modest vine]

Do you wanna try a graphing calculator

Which is totally goated

[winter comet]

wdym?

[modest vine]

https://github.com/denzio231/GraphingCalculatorRayLib

[acoustic spoke]

Can someone help me out! I don't understand how to do arcsine, arccostine, and arctangent problems like this without a calc

[willow skiff]

you don't need to at all

show the options of the question

but to get the principal angle, you just keep adding pi until you are more than 0, cause the period of tan is pi

[acoustic spoke]

Ok thats an old question so heres a newer one

[willow skiff]

oh I think you do need a calculator for this

you can figure it out though without one

[acoustic spoke]

Yeah... I was wondering tho is there a process to get it without one

[willow skiff]

from the options

[willow skiff]

the range of arccos is [0, pi]

because cos x passes through (0, 1) and (pi, -1)

so when you take the inverse, those points are now (1, 0) and (-1, pi)

we have chosen those two points cause that's the largest domain where cos x is one to one

[-pi, 0] also works as a domain for the inverse, but that's a bit weird to have only negative numbers

[acoustic spoke]

ok thanks i figured out how to do it manually lol

[willow skiff]

nwnw

[spice vault]

Yo how to do I solve

( 16x² + 96 + 2,304) + (16y² - 40y +400) =315

Idk how to solve cause the x has a number

Ps. Pls tag me if u know the answer

[willow skiff]

there are no solutions if x, y are real numbers

what are the conditions on x, y?

that's weird

,w (16x² + 96 + 2304) + (16y² - 40y +400) =315

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by sooouth.

[willow skiff]

you see that -4x^2 - 615 < 0

[modest vine]

I'm pretty sure this is just a circle

[willow skiff]

oh should be 96x

nope discriminant still negative

,calc 96^2 - 16 * 2304

[obsidian monolith]

Result:

-27648

[willow skiff]

it is but in the complex plane

[modest vine]

Yea the radius has a negative sqrt

[trim isle]

hello! can someone help me understand this?

[elfin cargo]

This was actually a quite challenging problem

Do you know that the intersection of angle bisectors of a triangle is an incenter?

[elfin cargo]

The only method I can think of is to find the equation of 3 angle bisectors and then find their intersection which is the center of the circle inscribed

Actually we only need to find 2 angle bisectors

Wait we still need 3 to locate incenter exactly

Even though, this method can still lead to correct answer but it is quite long so I don't know if anyone has a better method

[willow skiff]

well notice your triangle is right angled

x + 2y - 5 = 0 and 2x - y - 10 = 0 are perpendicular

https://www.youtube.com/watch?v=ISA6LGOXmAM

from the above diagram we first solve x + 2y - 5 = 0 and 2x - y - 10 = 0 to get (5, 0)

then solve the other two equations similarly: the sides are 3sqrt5, 4sqrt5, 5sqrt5

so r = sqrt5

[willow skiff]

now note the square in the diagram: the vector (-1, -2) is perpendicular to the line x + 2y - 5 = 0 and the vector (-2, 1) is perpendicular to 2x - y - 10 = 0

both vectors already have length sqrt(5), which matches the radius

so the centre is just (5, 0) + (-1, -2) + (-2, 1) = (2, -1)

[elfin cargo]

Yeah, I also got that answer

[somber void]

there's a way to solve this that uses stupid amounts of linear algebra (use cramer's to solve for vertices, cross product to solve for sines of angles (or you can just find side lengths, that's easier), then use barycentric coordinates to solve for incenter)

[raw hill]

Angle bisector theorem + partioning a line segment moment:

[brittle geode]

hi, could someone help me with this ?

[pulsar oxide]

This isn't a function right?

[uncut mulch]

what's your reasoning

[sharp yarrow]

Each value on the horizontal axis (independent variable) corresponds with only one value in the vertical axis (dependant variable) right?

[mellow sorrel]

you know why I dont like cos and sin and stuff like that

because I have to use calculators for them and it is depressing to not be able to calculate them

[chrome ether]

not everything is a nice rational value

best get used to that now rather than later

[arctic dagger]

Depends on the problem. You don't always have to use them

[arctic dagger]

Have u learned about trigonometric identities?

[arctic dagger]

I think it's a function

Because a vertical line only intersects the function once for every point on the function

[pulsar oxide]

What about these points though?

[arctic dagger]

Do u know what those points say?

It's like, imagine a vertical line intersecting those filled and empty circles, and x=a on that vertical line. The function must only have one value f(a) on x=a, and that value is the filled dark circle. The empty one is like open ended, it approaches a but the true value at a is the dark circle

And pls don't confuse a and f(a). a is the value on x-axis, f(a) is the value on y-axis

[pulsar oxide]

OJHHHH

My teacher didn't really explain that

Thank you so much

[arctic dagger]

Damn. Your teacher should. Or maybe you can ask ur teacher when something is confusing

But also sometimes it's hard to put smth in words things that u don't even know about

[pulsar oxide]

Yeah

Also

We're in online class

And there's no way to contact the teacher

So 😭

[arctic dagger]

I see

[pulsar oxide]

Thanks for the help!

[arctic dagger]

Yw

[pulsar oxide]

I treat this server more like my teacher than my actual one

[lunar hemlock]

I think circle with no dark is means a value in under it is not included

And circle with dark us where a value ends like 5 > x, dark circle goes forever

You can just write it as arrow

[lunar hemlock]

I see

[little wigeon]

Anyone able to VC

Real fast to explain a problem to me

Good karma awaits you

Or just DM

[cyan canopy]

I need a little help

[willow skiff]

for example, 6.

Desmos and GeoGebra can actually do q8 for you

just type in {0 < x} for example

for the domain restrictions

[cyan canopy]

oh

But is this right?

[willow skiff]

8a is correct

[cyan canopy]

9?

How it’s not completely to scale tho

[willow skiff]

wait

sorry brainfart

[cyan canopy]

All good

[willow skiff]

yeah that's correct

[cyan canopy]

What about the characteristics

[willow skiff]

just state what you observe

[cyan canopy]

Like is there a VA for 8a?

but 0 has an end point

[willow skiff]

so there are asymptotes for 8 at x = 0 and y = 0, for x < 0 only

[willow skiff]

yeah

[cyan canopy]

9 is easy

[willow skiff]

that the graph is decreasing for x < 0 but increasing for x >= 0 etc

yeah you get the idea

[cyan canopy]

Ok thanks man

[willow skiff]

oh right after x = 7 draw the graph a bit higher than y = -5 ofc

or you can make the slope at x = 7 increasing, that also works

[cyan canopy]

thanks you so much

[willow skiff]

np

it's testing your attention to detail

[cyan canopy]

Thanks again

[native smelt]

i need help

[arctic dagger]

What is the standard form of the equation of the circle

[native smelt]

what formula am i gonna use?

[arctic dagger]

The circle's standard equation

[slender vault]

Wooo

Lol

[viscid thistle]

Is this equation

The one that goes like (x-a)^2-(y-b)^2

[little wigeon]

Looking for a pre calc tutor to help me review for final. Willing to pay hourly rate

Please contact ASAP

[pulsar oxide]

How should I go about substituting this?

Specifically the f(x+h)

I don't know where to put the +h

[simple tundra]

f(something) is a function that has an input of (something) and an output of 3(something) - 1

So whatever that something is, whether x or x+h, you just put it where the (something) should go

[pulsar oxide]

So would 3(x+h) - 1 be correct?

Also, what does it mean by "each replacement value of x"?

[crude solstice]

Yes

each replacement value is referring to x+h and x in f(x+h) and f(x)

for f(x) though it is obvious

[pulsar oxide]

Okay, I get it now

Thank youuuu

Can I do anything more with b?

[river drift]

you can multiply numerator and denominator by the conjugate

[tardy ridge]

I'm doing precalc this year, any tips?

[arctic dagger]

Close. (x-h)² + (y-k)² = r², where (h, k) is the center of the circle and r is the radius

And i think that equation comes from the distance from the center (h, k) to every point in the circle. So it's basically a distance formula. As all conic sections are, i think

[steel dew]

how do i graph 86

[willow skiff]

that's the fractional part of x

so for x, you take off the whole number part, and so your range is 0 <= f(x) < 1

and there's a repeating pattern every 1 unit

[arctic dagger]

Graph crosses x axis: odd roots
Graph touches x axis but doesn't cross: even roots

And mind the sign

[twilit ridge]

i need help with conic sections 😭 i don’t understand how standard and general form works

[willow skiff]

from the general form, you can always complete the square for both x and y to get into standard form

[arctic dagger]

If i still remember, the standard forms always involve the center point

(h, k)

No that's only for circle

[willow skiff]

(if there are no cross terms, like xy)

[twilit ridge]

how can you find radius though in this problem

[willow skiff]

for example

[willow skiff]

where?

[willow skiff]

easyyyyyyy

x^2 + y^2 = 2^2, radius is 2

[twilit ridge]

ohhhh

[willow skiff]

the full form is $(x - h)^2 + (y - k)^2 = r^2$, so you have $h = k = 0$ and the origin is the centre

[obsidian monolith]

south

[willow skiff]

long story short, it's minus h, minus k
because x - 2 makes the graph go right by 2 units
(x - 2)^2 makes the parabola go right by 2 units, so it's the exact same with the circle

[twilit ridge]

ohh okay ,, thanks

[twilit ridge]

is this the general form ?

[willow skiff]

no

the general form of any conic is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

[obsidian monolith]

south

[spice vault]

Idk how to solve, I need help

[willow skiff]

find the perpendicular bisector of AB and of BC

where the two bisectors meet is the centre

[spice vault]

Bro ngl, I'm very slow at math

[willow skiff]

no worries

[spice vault]

So how? Could u pls give me an explanation can't really understand on how to solve

[willow skiff]

1. find the gradients AB and BC: $m = \frac{y_2 - y_1}{x_2 - x_1}$

2. take the negative reciprocal, -1/over, the gradients AB and BC: those are the slopes of the perpendicular bisectors

3. find the midpoints of AB and BC: the midpoint of (a, b) and (c, d) is ((a + c)/2, (b + d)/2)

4. y = mx + c for both lines, where m are the gradients in step 2: sub in the x- and y-values of the midpoint to obtain c for both lines

5. solve both lines simultaneously: either elimination or substitution works
   now you have the centre

6. use Pythagoras to get the radius squared: with point C for example, $r^2 = (7 - p)^2 + (2, q)^2$ if (p, q) is the centre

7. now you can write (x - p)^2 + (y - q)^2 = r^2

[obsidian monolith]

south
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[spice vault]

TY, I just gotta understand it now, Thank you very much for the help

[willow skiff]

no worries

[small shard]

@little wigeon i got some messages to you

[fiery crest]

Im starting college and id like to get to calculus but I have no clue where to start. I've never completed a full math course before but I did ok in the parts I tried. Should I go straight for precalc or is it too far of a jump from not understanding algebra 2?

[warped wharf]

Can anyone help me with conic sections

[pseudo osprey]

If you didn’t understand algebra 2, I would recommend first re doing it at your earliest convenience and then do precal after. If you do this, you will have a much easier time in your calculus journey.

[fiery crest]

What even was the quadratic formula for? I kept switching schools so I never understood

[pseudo osprey]

It’s a way to factor quadratic equations. Decomposing polynomials into factors is one of the key problem solving skills that will help you in classes like precal and calculus.

At the end of the day, your success in these classes is dependent on how much time you take to understand each concept as it is given. Math may seem an insurmountable obelisk currently, but with enough concerted time and effort, that once-obelisk will morph and change, revealing layers of simplicity that make up the most complicated equations.

You too, can see math as others do—just as they once saw math the same way you do currently.

You can be successful.

[fiery crest]

Yeah I think I can do ok I just constantly changed schools so I never was able to learn anything

I don't really wanna have to redo a class

Especially cause I had no clue what was going on in algebra 2

[pseudo osprey]

It’ll suck but if you really think it’s the best thing to do for your understanding, you should.

[pseudo osprey]

Circumstance also plays a huge factor. I am sympathetic.

[fiery crest]

I'm not sure if I should go farther back than algebra 2 cause I missed the last half of algebra 1. I did really good in the first half

[pseudo osprey]

Look into the course material and see what it is you’re missing. You could either teach yourself from there or use it to guide your decisions, ultimately.

[fiery crest]

I really appreciate the advice

[pseudo osprey]

Np man. I wish you luck.

Math is extremely rewarding and challenging—however, I’m sure you’ll find your peace with it eventually.

[cedar drift]

for precalc do i round this to 0

[sinful halo]

whats the question asking for

generally, you should leave it in scientific notation and round to 3 decimal places

[gentle totem]

probably

Looks like just a floating point error

[real briar]

should i learn algebra 2 before precalc

[cedar drift]

yes

[uncut mulch]

its more or less the same

[warped wharf]

Here a quick question 1^x=2 solve x is a complex number

[drowsy swan]

Anyone know quickly what my teacher means by this

[somber void]

if you know what newton's method is it should be clear

[fading monolith]

Also that is not precalculus

[winter comet]

everything under the sun is precalculus atp 😔

[dreamy pike]

can someone explain this? i got it right from guessing the second answer but why is the identity suddenly (-180-theta)

[summer ruin]

a typo

[viscid thistle]

Calculus seems pretty hard

A lot of people says it is

Is this really the case?

[boreal remnant]

nah

[winter comet]

i don't believe it is that hard, no

[boreal remnant]

maybe unintuitive at first

[winter comet]

of course it depends on how hard the teacher makes it

but like the core concepts are pretty easy

[winter comet]

but yea

[viscid thistle]

Fr?

[winter comet]

yeah

[viscid thistle]

My older friends struggle

[winter comet]

it's kind of overrated imo lol

[viscid thistle]

I just rlly want to get it

[winter comet]

the difficulty

[viscid thistle]

Would you recommend prelearning

Calculus

[winter comet]

yes

definitely

[viscid thistle]

Okay

Thank you

[boreal remnant]

im prelearning basically all of highschool, if you enjoy it then why not lol

[winter comet]

as long as you put in the work you should do fine

[viscid thistle]

I haven’t been taught it yet in school

[viscid thistle]

Thanks

[storm lodge]

holy hell

crab

[viscid thistle]

SHH

I’m trying okay

[winter comet]

wait its a crab :O

[viscid thistle]

Bubo

[boreal remnant]

why tf yall talking to eachother in 5 different channels 😭

[winter comet]

a talking crab

[boreal remnant]

just chose 1

[storm lodge]

hes trying to get active

[viscid thistle]

I am a miracle yes

[boreal remnant]

lmfao

[winter comet]

🙏

[viscid thistle]

Uhhh

No im not

[boreal remnant]

give ppl help in one of the help channels

[viscid thistle]

[viscid thistle]

I’m the one that needs the help tbf

[winter comet]

everyone is jealous of your active role

[viscid thistle]

Evidently so

[boreal remnant]

ah

just pretend like you know what you're talking about

they'll listen

💀

[viscid thistle]

But thanks for the calculus help guys that I actually needed

[winter comet]

literally

😂

[winter comet]

if you ever have questions you can go to #calculus

[storm lodge]

help channels are the bane of my existence

i dont have time to have an hour long conversation

[boreal remnant]

i wake up every day with 40 notifications

[viscid thistle]

LOL I just reply to the question with random comments like ‘hmm.. that seems difficult’

[boreal remnant]

lmfaooo

[winter comet]

I wake up every day

most days

[viscid thistle]

Thanks

[storm lodge]

doesnt matter

bc at the end of the day

its night

[boreal remnant]

i mean somedays just dont sleep and then u dont wake up so

[winter comet]

at the end of the day, i sleep

usually

[viscid thistle]

First funny bubo moment

[boreal remnant]

😭

[winter comet]

or just sleep and don't wake up

[storm lodge]

https://media.discordapp.net/attachments/1233150338509438996/1241247013253808188/sussss.gif?=&width=2230&height=807

[boreal remnant]

mood

[viscid thistle]

Let’s save this for discussy

[boreal remnant]

discussy 😋

[viscid thistle]

Mhm

[winter comet]

crab 😋

[viscid thistle]

OI

crab is not a food pls

It’s been wrongfully labelled as one

[winter comet]

on a scale from 1 to 10 how cannibalistic are you

[boreal remnant]

https://tenor.com/view/image-perm-bad-piggies-drip-gif-23469521

@viscid thistle

[hushed sphinx]

Kindly stop the random shitposting here.

[storm lodge]

isnt images allowed for everyone in this channel

[viscid thistle]

My bad tropo

[boreal remnant]

my bad

[viscid thistle]

I’ll take it to discussy

[viscid thistle]

Is this website DFM?

lol

[fossil cypress]

hi

i need hlep

[winter comet]

nice

[fossil cypress]

like

[viscid thistle]

bro i dont remember anything about conics 💀

lemme search it up

[winter comet]

i'm bad at conics

c = sqrt(a^2 - b^2) is for ellipse i believe

lol

[viscid thistle]

yeah and the -8 in the x term means its shifted to the right by 8

[winter comet]

yea

[viscid thistle]

@fossil cypress

[fossil cypress]

o

[winter comet]

we're just doing the problem for him 😂

[viscid thistle]

lmao

[fossil cypress]

wait

[winter comet]

do you know the center?

[fossil cypress]

so

yes you divide the thing

by 2

hm

[winter comet]

..what

[fossil cypress]

um nvm

[fossil cypress]

(8,2)?

[viscid thistle]

the fact that its shifted to the right means you find the vertices as if it wernt and then just add 8 to the x values of what you got

[proven night]

Hi

If i am quite good at precalculus (i need to get better only at logarithmic and trig equations), and i already know about derivatives, can find extrema of function, have basic understanding of continuity, limits, but i cannot solve harder limits, like i know how they work, but i didnt master solving them, so with this knowledge, can i start learning integrals?

[proven night]

We skipped limits at school and started calculus with derivatives

[timber blade]

i did the same as you and i would say - yes, you can start solving integrals, integrals require variety of different techniques and most of them have nothing to do with what u learned in limits

[proven night]

Okay

[lime lotus]

How do you put this in standard form:

In like this

When I tried to manipulate

this happens

[summer ruin]

what happens

[modest vine]

I guess we'll never know

[lime lotus]

how do I put it in a form like this?

[modest vine]

140y^2/35=(y^2)/(35/140)

[summer ruin]

and replace $y = \hat{x}, x = \hat{y}$

[obsidian monolith]

Transparent Elemental

[lime lotus]

can you show the algebra?

[modest vine]

Divide by 140 on both numerator and denominator

[odd agate]

Shouldn’t the 4y^2 be negative?

[lime lotus]

I multiplied -1/35

on both sides

[summer ruin]

you're replying to the comment that shows the algebra

140/35 = 1/(35/140)

[lime lotus]

So this is the standard form?

[odd agate]

Then why doesn’t the third step have the (140y^2)/35 positive?

[odd agate]

Yes that is in standard form

[lime lotus]

Thanks!

[willow skiff]

conics are beautiful

[viscid thistle]

except they arent

[willow skiff]

shuttuponlyhelpfulshavespeakingrights

[winter comet]

keep that line in it

[willow skiff]

actually the correct statement is quadratic forms are beautiful

but studying them is hell

[viscid thistle]

good

[viscid thistle]

I had a cancer

reading this

[viscid thistle]

????

wtf

[lime lotus]

how do you put this in equation

The product of the distances of P(x,y) from the coordinate axes is 5.

[winter comet]

how far away is P(x,y) from the x axis

[hushed sphinx]

Ahem, no, that's not what the Helpful role is supposed to indicate.

[willow skiff]

~~okay, did you mean that only mods have speaking rights? ~~

[lime lotus]

is this right?

[winter comet]

not quite

the distance from P(x,y) to the x axis is that, but its not equal to 5

[odd agate]

How do I write subscripts?

[winter comet]

you mean type them here?

i do like a_n

[odd agate]

Ye

[winter comet]

a sub n

[odd agate]

1_2

No like with the math bot thingy

[winter comet]

oh

idk

💀

[lime lotus]

Can you show me the algebra?

[willow skiff]

(x, y) is y units above/below the x-axis

so how many units is (x, y) left/right of the y-axis?

[winter comet]

to the right of/to the left of

💀

[willow skiff]

corrected haha

[odd agate]

Is that not just
y=5

[odd agate]

Wouldn’t it just be
x_p * y_p = 5

I think either way it really depends on what you want the equation to represent

You could do an inequality to represent the area that contains all valid points

[willow skiff]

yes

xy = 5

actually, |x| |y| = 5

[odd agate]

Unless P represents something beside a point, it’s just
|xy| = 5

[odd agate]

Good catch

[willow skiff]

like (0, -3) is still 3 units away from the x-axis

so basically |xy| = 5

that's the same

[odd agate]

You right

[surreal crescent]

Is directed distance the same as r?

[wise ferry]

depends on the circle

[surreal crescent]

Talking about polar coordinates

[winter comet]

what do you mean

if directed distance is like it can be positive or negative

then r can be positive or negative so it is A directed distance

but r is not defined as a directed distance and vice versa

i guess

i'm not really sure what you mean by are they the same

[slim jasper]

Is there anything I should familiarize myself with regarding Precalculus Algebra? I haven't had a math class in over a year, and old concepts have kind of slipped my memory.

[odd agate]

For what class?

[slim jasper]

Class names “Precalculus Algebra”

[foggy plover]

ik factoring is a big thing in calc 1 and precalc so you def wnat to get that down

[dreamy pike]

but the rest of the answer guide uses the "typo"

[dreamy pike]

whats that

[obsidian monolith]

ඞඞඞ

[stiff sluice]

please help

[lime lotus]

Thanks!

Am I right?

or wrong?

[arctic dagger]

Right

I mean the computation is correct

Shouldn't be the 2 on the other side? Not sure

[unborn saffron]

could someone help me out with this somebody else asked me for help with it and i don't get it 💀

[stiff sluice]

Let $A = {(x,y) \in \mathbb{R}^2 \mid -2 \leq x \leq 3; , 1 \leq y \leq 4}$ and $B = {(x,y) \in \mathbb{R}^2 \mid 0 \leq x + y \leq 5}$, and let $P = (1,-1)$, $Q = (-2,2)$, and $R = (-1,3)$.
\begin{enumerate}
\item Determine which of the points $P, Q, R$ belong to $A \cap B$ and which belong to $A \cup B$.
\item Let $C = {(x,y) \in \mathbb{R}^2 \mid y = 2x}$. Find $A \cap C$ and determine if $A \cap C \subseteq B$.
\end{enumerate}

help

[obsidian monolith]

ඞඞඞ

[trim cobalt]

Jesus

[somber void]

Can someone explain

,w d/de (e^x)

[obsidian monolith]

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[somber void]

why would it be 0?

[elfin cargo]

,w find the derivative of e^x wrt to e

[obsidian monolith]

Wolfram Alpha doesn't understand your query!
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Requested by closervd1k

[elfin cargo]

,w find the derivative of e^x with respect to e

[obsidian monolith]

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[arctic dagger]

Do u mean d/dx ?

[somber void]

no i mean d/de

maybe i don't know what e is and i have some iterative method to find it

and i want to know the derivative with respect to e

i mean it should be xe^(x-1), but why does wolfram alpha say 0

[arctic dagger]

It could be that it's treating the 2nd e as the euler's number and as a result, e^x is treated as a constant

[somber void]

oh

[arctic dagger]

And the derivative of a constant is 0

[somber void]

yeah that's probably what it is

[arctic dagger]

Otherwise just choose a different variable aside from e

[arctic dagger]

This is the expected answer?

Ah yea that makes sense

,w d/df f^x

[obsidian monolith]

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[arctic dagger]

^here it gives the expected result. Just replace f with e. I conclude it treats the 2nd e as euler's number

[elfin cargo]

I don't know why Wolfram Alpha here won't work

You can get that by using Wolfram Alpha website

[arctic dagger]

Oh i see

[viscid thistle]

youre asking it the derivative of e to the x?

i think its confused

between the power rule

and the e to the power x thing

[elfin cargo]

The guy above wanted to find the derivative of e^x wrt to e so I showed him

[viscid thistle]

💀

why is wolfram confused

[modern carbon]

💀

[storm lodge]

its supposed to be d/dx @elfin cargo

[elfin cargo]

It was for the guy above he wanted to find d/de of e^x 😂

[storm lodge]

lmao

[shadow ether]

Ur using partial derivative with respect to e

Not to x

It treats x as any random constant

And e as a variable

[winter comet]

he meant to do that

he was saying it didnt work on discord but it worked on the website

[shadow ether]

Oh

My failt

Fault

Misunderstanding

[winter comet]

its confusing anyway XD

[arctic dagger]

Is it possible that discord uses a different version of wolfram alpha?

[shadow ether]

There are different versions?

Sorry I know I'm living in the past

[slender vault]

Lol

Y'all still going at it

[shadow ether]

Ur name literally q

Who u talking to

[slender vault]

I've been reading from the past day Ig

[hasty lily]

No

[stiff sluice]

wassup how is it going guys

I killed the chat mb

[slender vault]

Why is it not correct? X is just a constant here so won't this be it

[stiff sluice]

yeah differential of e^x with respect to e is 0

anyways

[hasty lily]

notice the difference in the d

[slender vault]

Just say that it's d

[hasty lily]

Maybe they send different inputs to WA?

[slender vault]

Might've been the case

[shadow ether]

Wher

[slender vault]

e is a constant, and can you differentiate something with respect to a constant

I think it should be a variable

[hasty lily]

Not really. this follows from the limit definition

[stiff sluice]

lets use the definition of derivative as a limit shall we

,w d/de (e^x)

[obsidian monolith]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[stiff sluice]

,align y &= e^x \ \frac{dy}{de} &= \lim_{h \to 0} \frac{f(e+h) - f(e)}{h} \ \frac{d}{de} \left[e^{x} \right] &= \lim_{h \to 0} \frac{f(e+h) - f(e)}{h}

[obsidian monolith]

ඞඞඞ

[stiff sluice]

.

[winter comet]

why does it say 0

should be xe^(x-1)?

[stiff sluice]

I thought it was zero

can you show how did you differentiated

[winter comet]

the variable is probably just whats confusing

lets call e -> y, and x -> a

d/dy (y^a) right?

then you just use the power rule

[stiff sluice]

okaay sure

then?

[winter comet]

the power rule gives ay^(a-1)

do you want me to prove it with limit definition or?

[stiff sluice]

oh okay

[winter comet]

lol

[stiff sluice]

but catgod said limit def

wtf

[winter comet]

limit definition of derivative

it gives the same result as using derivative rules

[stiff sluice]

.

[winter comet]

oh he was saying taking the derivative of something with respect to a constant

in this case, we were treating e as a variable

[stiff sluice]

ye is constant

my bad I didnt said

[winter comet]

so ur saying why can't you differentiate with respect to e

huh

[stiff sluice]

ye

[stiff sluice]

.

[winter comet]

i don't know how it follows from the 'limit definition' of derivative

but it does make sense with rates of change

the derivative just represents the rate of change

dy/dx means you look at how y changes as x changes

[stiff sluice]

ohh okay

[winter comet]

so dy/de means you look at how y changes as e changes

except if e is a constant, it doesn't change

[stiff sluice]

exactly, correct

[winter comet]

i'm pretty sure that just means dy/de doesn't exist

talking about how it changes makes no sense

[stiff sluice]

thank u for the explanation sir, yeah I was confused why = 0

also catgod used the derivative as limit definition yeah

[winter comet]

very unrigorously, i believe de would be 0 because its not changing

and dividing by 0 makes no sense

lol

[stiff sluice]

oh okay

[winter comet]

i think this better though

[winter comet]

this is kind of iffy explanation lol

[winter comet]

yea

[stiff sluice]

yeah when I saw that it was = 0 I shit the bed

[winter comet]

i don't know how you would do that, there is no e axis XD

[winter comet]

💀

[stiff sluice]

ye

[winter comet]

wait a minute

it is 0?

hmm

[stiff sluice]

ye

[winter comet]

why is it 0

💀

[stiff sluice]

because rate of change mf

[winter comet]

i would think taking the derivative with respect to a constant doesn't make sense

[stiff sluice]

e is constant so no rate of change

[winter comet]

e doesn't change

its a constant

[stiff sluice]

lmaoo

[winter comet]

dy/de means you look at how y changes as e changes

e doesn't change

so this shouldn't be 0 it should be dne

it would be 0 if y doesn't change but e changes

😔

[stiff sluice]

if its dne why WA say its 0 doe

u sure bro

,w d/de (e^x)

[obsidian monolith]

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[winter comet]

that's what im tryna figure out

[stiff sluice]

u see its 0

[winter comet]

i see its 0

ah you know what

it might be thinking of different e

taking the derivative with respect to the variable e

of e^x, this one is constant e

[stiff sluice]

is looking to differentiate with respect to e
and taking (e^x) as constant

[winter comet]

cuz the e font is different

[stiff sluice]

e is variable

[winter comet]

yea like

its taking the derivative with respect to e of e^x

notice the different e

e is variable, e is constant

[stiff sluice]

exatly

[winter comet]

yeah

and x is constant

so is 0

[stiff sluice]

lmaoo

ye

[winter comet]

it just misinterpreted 😂

but yeah taking the derivative w.r.t a constant makes no sense

[stiff sluice]

😂

[winter comet]

im pretty sure

[winter comet]

funny 🤣

[shadow ether]

It's 0 cuz

X is a constant

So e^x

Is also a constant

Remember constant rule brother

Wait dumbass

[river drift]

note that the e in e^x is bolded, meaning it's euler's number, but the e in d/de is not, meaning it's a variable

[shadow ether]

It took e as a constant

And ur differentiating variable e

[winter comet]

👍

thats what it did

thats where the confusion came from 💀

[shadow ether]

Bro I swear every time I enter this channel

Yall idiots always pondering abt smth

When the answer

Right In front of you