precalculus

[restive bridge]

Desmos I thought was just for graphing

good to know there's more to it

[willow bear]

$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

[obsidian monolith]

Ann

[willow bear]

also consider $(\cos(\theta) + \sin(\theta))^2$

[obsidian monolith]

Ann

[willow bear]

@viscid thistle

[ancient rampart]

are 2 vectors considered parllel if they are like on top of each other

for example <2, -1> and <4, -2>

[mild swan]

I would argue yes

[sick steppe]

parallel if they're scalar multiples

[finite birch]

Hi!! I was wondering if I could have help with my Pre Calc parametric a review!! I’ll send a photo, but i’m really confused and was hoping for some help on these 3 problems:)

[tardy ridge]

How do I represent 2 equations into a parametric equation

like -4x-6y+3z-6=0

and 3x+2y+4z+2=0

nevermind I got it

[frigid sonnet]

can some one break this down please

[uncut mulch]

its notation for function compositions

eg.
(fog)(x) = f(g(x))

which you should know how to do with a basic understanding of function notation

[visual brook]

Aren't the vectors called "anti parallel" if the scalar is negative?

[tardy ridge]

wut

[rigid beacon]

I've never heard that term before

I guess you could use it but no one does

if vectors are scalar multiples of each other, people will say parallel regardless of what that scalar is

[stable crown]

You can switch that equation to an imaginary form, which would be easier to figure out I think.

That's my approach to find theta following trigonometry properties
\theta = $(\pi n+tan^{-1} (\frac{1}{5}(3-\sqrt(14)))$ ; $n \me \mathbb{Z}$

[obsidian monolith]

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

[willow bear]

$\theta = \pi n + \tan^{-1}\left(\frac{1}{5}(3 - \sqrt{14})\right), n \in \mathbb{Z}$

[obsidian monolith]

Ann

[willow bear]

@stable crown fixed that tex for you

but also you don't need this lol

[barren hatch]

Could I get some help?

[willow bear]

with what?

[visual brook]

Just do $\sine{x}+\cosine{x}=\sqrt{2}\sine{\left(x+\frac{\pi}{4}\right)}$

[obsidian monolith]

September22nd

[barren hatch]

Vectors

@willow bear

I have a pic sending

[visual brook]

Can you send your q instead of saying you need help

Okay

[barren hatch]

See it??

[visual brook]

When you multiply a vector by a scalar you have to multiply each component by the scalar

[barren hatch]

So it would be [15,-6]?

[visual brook]

Yes

But use the <> notation

[barren hatch]

Ahhh

Thank you

It was the notations all along

I had a problem before that and I couldn’t figure it out so I clicked get a similar problem to try a different one and I kept running into the same problem so I started kinda guessing

Thanks Pali

Papi

[visual brook]

Uhhh

No problem

[barren hatch]

Have a goodnight sleep tight so that the bed bugs don’t bite

[willow bear]

that's kind of vague

weird way of phrasing it, but ok...

go a meters, turn around, go b meters, turn around again, go c meters

this is a - (b - c)

go a meters, go b meters backward, then go c meters

this is a - b + c

go a forward, then b backward, then c backward

"turn around" is more like -(...)

than just -

you could say that

thats a little trickier to get a visual for, i'm afraid

[fickle steeple]

Can someone please help in solving this question

Feel free to DM if you like

[tranquil lava]

can you send a pic of the entire question

[fickle steeple]

Sure

Q.no.86

[stable crown]

I would like to know the name of the book pls.

[fickle steeple]

Calculus of a single variable,10th edition

[stable crown]

Thanks

[tranquil lava]

what you can do is get an expression for $$a_{n+1} - a_n$$

[obsidian monolith]

indolentbrains

[tranquil lava]

for part a

or an expression for $$\frac{a_{n+1}}{a_n}$$

[obsidian monolith]

indolentbrains

[tranquil lava]

this will imply that {a_n} is a Cauchy sequence

[near epoch]

having a bit of trouble here....

[shell flicker]

Hey if someone is available Friday to help me out with some hw pls DM me

[vivid moth]

@shell flicker I can help.

[shell flicker]

DM

[outer inlet]

I got (0,5) and this being a point of inflection, is this correct?

[willow bear]

no

the point itself is correct but your judgment of its type is not

[outer inlet]

how would I determine whether its a point of max, min or inflection

[willow bear]

well how did you get that it's a point of inflection?

was it a guess or did something make you conclude this?

[outer inlet]

I don't really know to be honest

[willow bear]

so it was a guess?

[outer inlet]

sort of

[willow bear]

there is no "sort of". if you can't explain what led you to the answer, it is as good as a guess.

in any case, you would look at the second derivative of your function at the stationary point.

f''(0), in your case.

[outer inlet]

so 0?

[willow bear]

no, f''(0) ≠ 0 for your function.

[outer inlet]

is't f'(x) = -2x

[willow bear]

that's the first derivative.

[outer inlet]

-2x+h

[willow bear]

what is this?

[outer inlet]

idk

I am lost

[willow bear]

refer to this message.

do you understand what a second derivative is?

[outer inlet]

no

[willow bear]

did that never come up in your class?

did the concept of "take the derivative of the derivative" never earn so much as a passing mention by your teacher?

[outer inlet]

not sure, is there an equation for it?

[willow bear]

.......

take the derivative of the derivative

[outer inlet]

the examples look like this

[willow bear]

oh, so you were never told about it.

[outer inlet]

no

[willow bear]

in that case, i suppose you are forced to consider the values of the first derivative at points near your stationary point, and see whether it is positive or negative

[outer inlet]

wait is it a max?

[willow bear]

if it's the same sign on both sides, you have a point of inflection.
if it's negative to the left and positive to the right, you have a minimum.
if it's the other way around, you have a maximum.

yes, in your case you have a maximum.

[outer inlet]

oh ok now I see it

I drew the wrong diagram on my paper

[edgy blaze]

Can cos(x) and cos(2x) be the same value?

yeah so I had a tan(x) = -sqrt(3) from pi/2 <x <pi

so doing math I found

sin(x) = sqrt(3)/2
cos(x) = -1/2

using cos(2x) = cos^(2)(x) - sin^(2)(x)

cos(2x) = (-1/2)^(2) - (sqrt(3)/2)^2

cos(2x) = 1/4 - 3/4
cos(2x) = -1/2

[tardy ridge]

yes. In this case cos(120) = cos(240)

[viscid thistle]

I need help with this problem

[molten garnet]

sonmeone help me with functions

I hv no idea what a refers to

[quick bluff]

"a" is just another variable

m(a) means we substitute "a" in the place of x

m(a) = 2 is nothing but (a+1)/3 = 2

[quick bluff]

you can use this to solve for "a", same method for other sums

[viscid thistle]

I can help you with it

[viscid thistle]

opt a

[molten garnet]

aight man thanks guys

[brazen marsh]

it was an exercise u cant just give an answer like that

[brazen marsh]

ok see, what this wants u to do is find values of x for which the function is equal to whatever is given,
like m(a)=2 basically means for which value of a
(a+1)/3=2
or in case of the other options, u will have to basically change the right side of the equation to whatever is given and solve it for a

[viscid thistle]

.

Ok @brazen marsh ?

[brazen marsh]

Oh i didnt read that

sorry

mb

[viscid thistle]

np

[plush copper]

hi

[wary mulch]

idk how to do this sum

if anyone is willing to help me pls mention me

[trim hemlock]

express sec in terms of cos

and then do common denominator for the expression in the numerator

and you should apply sum to prod formula

[wary mulch]

thanks i got the ans

[tranquil haven]

Hey is proving things just trying valid operations and seeing if you're getting closer ?

Or is there some logic behind it

Also

I can't seem to prove this one, could someone maybe help ?

[blissful ridge]

Convert all the cosine terms to sine

[tardy ridge]

you could do it the teacher way with LHS and RHS

or change the statement to $sin^2x(cos^2y-1)+sin^2y(1-cos^2x) = 0$

[obsidian monolith]

Xiangli

[tardy ridge]

would would be easier to prove.

[visual birch]

can anyone help me with precal?

cashapp is an option, i have no idea how to do it rn

[tardy ridge]

why would you pay them

just ask the question here

[stuck lark]

@visual birch don't offer cash for hw help

[rough carbon]

I would start with the left
Write cos ^ 2 y as 1- sin^2 y
Same for the cos^2 x
So now it’s sin^2x (1-sin^2y) - (1-sin^2x)(sin^2y)
So LHS= sin^2x -sin^2x sin^2y -sin^2y + sin^2xsin^2y

The sin^2xsin^y terms cancel

Leaving you with the RHS

[mental wedge]

How do I solve the 7 and 8 sums??

[patent beacon]

"The distance between |x - 1| and 2 is greater than 2"

Which means that |x - 1| < 0 or |x - 1| > 4

[mental wedge]

Ye i got till this much

Wats not??

Nxt*

[patent beacon]

@mental wedge
We can repeat the same logic. On the first equation:
"The distance between x and 1 is less than 0"
That's obviously impossible, there's no solutions there. What does it look like when applied to the second equation?

[mental wedge]

X, is either greater than 5 or lesser than minus 3??

[patent beacon]

Nailed it

[mental wedge]

Ig, IDK this is all new for me....

Thnx

[mental wedge]

@patent beacon

Wat abt the second one??

The eighth question

[patent beacon]

Note that you can rewrite it as:
||x - 1| - (-2)| < 1

[tranquil haven]

Is it allowed to rewrite a/b/c as a/bc ? If so, why ?

Nevermind I understand it

[willow bear]

what do you mean by a/b/c?

do you mean (a/b)/c or a/(b/c)? these are not the same

[mental wedge]

Oh... OK.

[raven fulcrum]

Hey, I found f'(x) for x ≠ 0 but I'm not sure how I would proceed to find f'(0) if it exists

[tranquil haven]

(a/b)/c

= a/bc

[willow bear]

a/(bc) but yes

[willow bear]

using the limit definition

[raven fulcrum]

okay I'll try that thanks!

[simple parcel]

how would you solve this with the cauchy schwarz inequality?

[willow bear]

let u = a+b, v = a+c, w = b+c, then a+b+c = 1/2 (u+v+w)

apply cauchy-schwarz to the vectors [sqrt(u), sqrt(v), sqrt(w)] and [1/sqrt(u), 1/sqrt(v), 1/sqrt(w)]

[simple parcel]

thanks!

[flint bloom]

,rotate

[flint bloom]

I know part b is correct, but i’m unsure about c, it just doesn’t feel right i guess

[tardy ridge]

why are you not using woflam alpha.

or you could just check it with your calculator

substitute B = arctan(40/9) + 180 and A = 180 - arcsin(7/25). Use your calculator to see if you got the angle additions correct

[tardy ridge]

@flint bloom

[flint bloom]

dk why i didn’t think of that, i was treating A as 7/25 instead of the arcsin of that since A is an angle, it worked out correctly

[viscid thistle]

Can someone help me out with basic pre-calculus? Pls ping m

[flint bloom]

just ask

[tardy ridge]

you should and you are right, but I meant to check your answer, you would do it the arcsin way.

[flint bloom]

@tardy ridge

using the same way to check, the answer should be the positive version of this

i’m not sure why it’s coming out negative though

does it have to do with me checking it with arcsin

bc i checked with arccos and it comes out negative instead

[dire lintel]

Can someone help me out with calculus?

like the basic

[flint bloom]

ask your question

[dire lintel]

what is the menaing of dx/d?

[flint bloom]

do you mean d/dx

[dire lintel]

ahh yess

[flint bloom]

it’s the notation used which means to differentiate with respect to the variable, in this case, x

[dire lintel]

what is the meaning of this ∫ symbol?

[flint bloom]

to integrate, the opposite of a derivative

anti derivative

used to find areas or volumes

[dire lintel]

what are the variable we write on top and botton of ∫ ?

[flint bloom]

they can be variables but are mostly numbers

they are the bounds

[dire lintel]

Why do we write those?

[flint bloom]

to show which section we want to take the area or volume of

it like tells us which region we are working with

[dire lintel]

Are those mandatory ?

[flint bloom]

no

we can have ∫ without anything above and below, it just means to take the anti derivative of the integrand

but with numbers we are gonna end up with a number at the end of it all

[dire lintel]

Can you give me an example on how to represent a derivative and an intergral?

[flint bloom]

put d/dx infront of a function of x

put a function of x after the integral symbol, and then a dx after the function

[dire lintel]

Oh k

thank you

[merry chasm]

using k as the constant of proportionality, what does it mean if z is jointly proportional with w and u and is inversely proportional with p?

this was a question I was given, and I wasn't sure how to write out the equation.

[cyan glacier]

should i read a book about precalc or no? i use khan academy is that enough?

[willow bear]

z = kwu/p

[merry chasm]

ok, that makes sense

thanks for the help :)

[quick bluff]

For the sandwich theorem, is it correct to say that if the left and right sides of the inequality (in this case the functions f(x) and h(x)) give different values when applying the limit, then the limit for g(x) does not exist?

[willow bear]

no, it's not.

if $f(x) \leq g(x) \leq h(x)$ but $\lim_{x \to c} f(x) \neq \lim_{x \to c} h(x)$, you cannot say $\lim_{x \to c} g(x)$ doesn't exist. after all, maybe it DOES exist, and you just picked a bad pair of bounding functions!

[obsidian monolith]

Ann

[willow bear]

the squeeze theorem simply does not apply.

[quick bluff]

makes sense

wait i have another doubt

one sec

[willow bear]

yes?

[quick bluff]

suppose i have limit (x tends to 1+) -(1+x) . cos(1/(1-x))

can I just directly say that the limit does not exist as I can't simplify it further?

[willow bear]

$\lim_{x \to 1^+} -(x+1) \cos(\frac{1}{1-x})$

[obsidian monolith]

Ann

[willow bear]

this?

[quick bluff]

yes

[willow bear]

"as I can't simplify it further" is not good reason for the nonexistence of a limit.

[quick bluff]

true

so could I say it doesn't exist because I get 0 x cos (infinity) ?

I'm not sure how I can say the limit doesnt exist

[willow bear]

you will not get 0 * cos(∞)

[quick bluff]

I incorrectly used the sandwich theorem to "prove" it prior to this

[quick bluff]

oh

[willow bear]

anyway, to establish nonexistence, you could construct two sequences $x_n, y_n$ such that $x_n \to 1^+$ and $y_n \to 1^+$ but $\lim f(x_n) \neq \lim f(y_n)$

[obsidian monolith]

Ann

[willow bear]

where f(x) is the stuff in your limit

[quick bluff]

ohh

reminds me of the method to prove whether a function is one-one or not

but i haven't been taught how to do that yet

[quick bluff]

by "that" i mean this

[willow bear]

then i cannot help you

[quick bluff]

oh you have already helped me SO much

thanks a lot Ann

[tardy ridge]

the negative answer is right.

are you sure you're using A = 180 - arcsin(7/25)

the angle is in the second quadrant.

do not use A = sin(7/25), that's the angle in the first quadrant.

[trim hemlock]

What have you tried?

[warm forum]

$\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{g^2}$

[obsidian monolith]

Muzan Jackson

[mortal iron]

Does anyone know about the proof of this statement?

[warm forum]

https://www.zweigmedia.com/RealWorld/proofs/quotientruleproof.html

[mortal iron]

ty

[shell flicker]

https://cdn.discordapp.com/attachments/490557019623915520/856698083035447337/image0.jpg

Anyone know how to do this?

[merry chasm]

@shell flickerso here was my thought process.

Assuming there isn't a horizontal shift in the entire function, I know that sin of zero is zero, meaning the sin function crosses the origin. Because this graph does, it's a sin function.

I also know that because the slope from x = [-pi,pi] is negative, that equation is going to have a negative in it.

This is where I combined luck and knowledge. So I know that B increases/decreases the period of a function. If you understand the unit circle or memorize how a sin wave moves in the real world, from the origin, it moves down and up to make a minimum, then goes up and passes the x axis and comes back down eventually to make a maximum. When it crosses the x axis again, it makes a full revolution (2pi).

This is longer than what the graph shows, as all is happening in the graph above is it's making a local maximum and passes 2pi at the x axis. The part where I got lucky was I set B to 1/2 and I imitated the graph shown on the screen. If you're not as lucky, there is a formula that would show you what B would be, but I don't remember that anymore.

Taking all those elements, the equation I got was -2sin.5(x)

The reason why it's 2 is because that's the amplitude, which is what the extrema are.

How do I graph a circle using this equation? I know the center radius one with h,k and r, but this isn't as obvious to me. I had a problem like this on a test I did a while back, and I didn't understand it at the time. I couldn't find many resources for this ether.

[uncut mulch]

there should be plenty of resources for this

look up completing the square

and converting equation of a circle from general to standard form

[merry chasm]

I think I know how to complete the square. Is the standard form the name of the equation above?

[uncut mulch]

the eqn you posted would be general form

[merry chasm]

🤔

this shit is giving me mixed messages

@uncut mulch

[uncut mulch]

different locations use the words general and standard differently

[merry chasm]

ah ok

[uncut mulch]

you could call that center (-radius) form if you want

[merry chasm]

ok, I found some resources while finding out what the name was too. thx for the help :)

[near epoch]

could someone please help me

no idea what I'm doing

also don't understand how its possible to evaluate trig functions with values greater than 90 lol

sry I'm stupid

so yeah

[shell flicker]

Thx

[merry chasm]

does this jog your mind?

[near epoch]

@merry chasm thx but someone helped me

already

[fleet yew]

Still need help with this

[merry chasm]

Are you asking me for help or if I need help?

[fleet yew]

Do you need help with the circle eq

[merry chasm]

Nah I think I got it

Just do completing the square twice to turn it to the other equation

[fleet yew]

K. Gg

[cunning hollow]

Is this demonstration for the limit of x*sin(1/x) using the squeeze theorem accurate? I dont understand why you can multiply by x in the 1st line and maintain the inequality. Doesn't x have to be positive for that?

[stuck lark]

it holds for x>0; we can use it to squeeze the right side lim. we can write a similar inequality for x<0 that lets us squeeze the left side lim. then we have the 2 sided lim

[cunning hollow]

oooooh thanks!

[viscid thistle]

For #20 how do you set that up

[trim hemlock]

you just have to integrate twice

[merry chasm]

why does the sign flip in this problem?

[willow bear]

are you talking about the bit i highlighted in red?

[merry chasm]

yea

@willow bear

[willow bear]

well they tell you right here

they divide both sides of the ineq by -9, a negative number

[merry chasm]

so whenever you divide by a negative, you change the sign?

[willow bear]

yes. whenever you divide both sides of an inequality by a negative number, the sign gets flipped.

[merry chasm]

ah ok, I didn't know that, thx for the help :)

[vapid torrent]

i have a super simple problem that im not understanding why im getting the wrong answer

wouldnt we just be getting

e^ (delta v / ve) = ma/mf for the answer

i thought thatsd what it was, but using the given in #hint 1

this is wrong

[willow bear]

$e^{\Delta v/v_e} = \frac{m_0}{m_f}$ is a step towards the solution but you aren't done yet

[obsidian monolith]

Ann

[vapid torrent]

wym, dont i want that ratio

[willow bear]

no, the proportion of fuel is $\frac{m_0 - m_f}{m_0}$

[obsidian monolith]

Ann

[willow bear]

m0 = total, mf = payload

[vapid torrent]

ohhhhh

how would i get to that equation from m0/mf

we arent given their individual values

@willow bear

[willow bear]

consider: $\frac{m_0 - m_f}{m_0} = 1 - \frac{m_f}{m_0} = 1 - \paren{\frac{m_0}{m_f}}^{-1}$

[obsidian monolith]

Ann

[vapid torrent]

thank you

[viscid thistle]

can anyone help with question b

[trim hemlock]

what is y in part a?

[quick bluff]

Once you found y in part a it's simple

Find the second derivative

Multiple both sides by x^2

Try to bring y in the right hand side

Your y should be x^2 + (3/x)

The second derivative should be 2 + (6/x^3)

[viscid thistle]

ok @quick bluff thanks.

[bleak cradle]

Having trouble applying compound interest formula. P = 10000 two different rates 3.5% monthly vs 6.5% quarterly for 10 years. I get 610,634 for the 3.5 and 114,160.75 for the 6.5. What am I doing wrong?

[willow bear]

what you're doing wrong is having the entire annual rate get applied every month and quarter, respectively.

you probably calculated the first investment as $$10000 \cdot (1 + 0.035)^{12 \cdot 10},$$ didn't you?

[obsidian monolith]

Ann

[willow bear]

@bleak cradle

[bleak cradle]

I did. What should I have done?

[willow bear]

you should've done $10000 \cdot \paren{1 + \frac{0.035}{12}}^{12 \cdot 10}$

[obsidian monolith]

Ann

[bleak cradle]

Ahh. Let me try it again. Thank you.

[willow bear]

when compounding several times a year, the annual interest rate gets spread out evenly over your compounding periods

[bleak cradle]

Ok. I think I follow.

[willow bear]

this applies to both your problems

[bleak cradle]

Thank you so much.

[jaunty thistle]

Omg

hilarious

[past meadow]

<@&268886789983436800>

[worldly badger]

wow

[obsidian monolith]

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

[steep sinew]

Huh

[crimson magnet]

Bruh

[worldly badger]

<@&268886789983436800> help plz ^^"

[rain tide]

wow

[pastel jay]

seriously just what is the point

[white thicket]

lol

[jagged violet]

i love having a name that starts with the letter A

[proper pendant]

Wtf is going on

[severe verge]

[viscid thistle]

What a weirdo

[orchid nova]

christ

[rain tide]

that was terrible

[stuck lark]

ty

[jaunty thistle]

Why

[jagged violet]

nothing quite like it

[digital kestrel]

smh~my head

[rain tide]

the world is ending

[twin escarp]

kek

[crimson magnet]

Mans def hates people with the letter A

[keen parrot]

who pinged

[pure light]

ping

[frank spruce]

at least the chat is finally active

[karmic burrow]

ayo fuck u

[twin escarp]

glizzy time

[rain tide]

pang

[digital kestrel]

oh itachi's here

[tender wigeon]

hm?

[misty bay]

??

[crimson magnet]

It’s chi

[digital kestrel]

chat came back from the dead

[tender wigeon]

lols im not even in precalc

[pure vine]

whats up fellow As

[wet remnant]

who ping

[white thicket]

lol all the A letter people have been united

[knotty spear]

precalc gang

[steep sinew]

Random user used: Summon

[wet oar]

huh

[golden flume]

https://media.discordapp.net/attachments/451168261007081523/852008483754606592/image0-3.gif

[pale geyser]

sup!

[lament radish]

Euler was wrong

[lilac marsh]

someone pinged me

[frail steppe]

amogus

[pure light]

bruh

[whole flint]

[frank spruce]

time to share our math memes and reminisce high school

[viscid thistle]

I don't ave letter A

[raven burrow]

[tender wigeon]

is this a smart people chat i might have to leave

[lilac marsh]

and i dont even chat here

[pure vine]

now that we have all been gathered here

[stuck lark]

everyone please chill

[wet remnant]

A GANG UNITE!!!!

[worldly badger]

i was chilling, then ping guy arrives

[tender wigeon]

LMFAOO

[worldly badger]

lol

[wet oar]

pre calc gang

[lime tusk]

Illegal

[pastel jay]

i dont even know pre calculus smh

[karmic burrow]

Letter A mfs unite to destroy the pinger

[digital kestrel]

y'all

[tender wigeon]

wait what's goin on

[plush birch]

packin

[digital kestrel]

the hell

[lilac marsh]

ikr

[viscid thistle]

im just here for the emotes

[tender wigeon]

me neither lmfaoo

[knotty spear]

ok #chill moment 😌

[worldly badger]

massive ping

[viscid thistle]

<@&268886789983436800>

[rain tide]

@dark plank

[knotty spear]

begone

[tender wigeon]

damn he got ass

[stuck lark]

ok we're on slowmode

[lilac marsh]

th ot

[rare nebula]

Sigh

[steep sinew]

Back to my hole I go

[plush birch]

dam

[karmic burrow]

shitpost hours

[wet oar]

YOoooooo

[lament radish]

One is prime change my mind

[tender wigeon]

for..?
also why slowmode if ping

[white thicket]

ok this was nice cya

[plush birch]

nah u wrong as hell

[wet oar]

TF

[viscid thistle]

how come i have pings turned off for this channel but i still got it

[stuck lark]

everyone stop shitposting

[sturdy relic]

is this precalc??

[pastel jay]

whats with the shitposting at least make them good shitposts

[worldly badger]

for nothing, someone wanted to ping everyone haha

[opal veldt]

[lime tusk]

magic

[viscid thistle]

1÷0=♾

[lilac marsh]

damn yall really taking the chance]

[maiden smelt]

who pinged me?

[marsh sail]

https://tenor.com/view/discord-ping-pingus-bingus-cat-gif-18905873

[tulip prism]

[frail steppe]

egg

[pastel jay]

yay its over

[worldly badger]

a spammer

[tender wigeon]

cringe. maybe ping in general next time

[viscid thistle]

Change my mind @tender wigeon

who pinged

[cursive fiber]

who pinged me lmao

[static reef]

Didn't seem to work

[frail steppe]

🦦

[steep sinew]

@stuck lark just turn chatting in this channel off for 5-10 minutes

[viscid thistle]

6 hr slowmode

[lime tusk]

Sorry

[frail steppe]

pornography

[orchid nova]

kek

[pastel jay]

6h slowmode petition

[tender wigeon]

bruh im not smart
im not even in precalc:dead-4:

[boreal pumice]

Lmao is some cunt with no life masspinging

[orchid nova]

aww it's not actually 6 hours

[viscid thistle]

shamrock can you help me with vector fields

[knotty bane]

#discussion

[viscid thistle]

Manjaro sucks btw

[orchid nova]

#multivariable-calculus

[tender wigeon]

wtf

oh nvm
btw staff here should change member perms lol?

[boreal pumice]

Blocked

[pastel jay]

your os sucks btw

[viscid thistle]

I'm freaking out over this ping, I can't continue my life, it's blocking my daily life man

[stuck lark]

anyone who wants to chill go to #chill

[frail steppe]

i mildly dislike anime

[stuck lark]

any non precalc msgs will be met with mutes

[viscid thistle]

Manjaro is based off Arch Linux and Arch Linux is the really thing

[lapis girder]

who tagged

[pastel jay]

thanks for your valued opinion

[stuck lark]

to anyone wondering why they were pinged, there was a masspinging troll

[boreal pumice]

Ok at least u like arch and not like macos or some shit

[viscid thistle]

Yea build your own arch lmao

[stuck lark]

first mute

[viscid thistle]

bruh I used to use my Hackintosh

[pastel jay]

guys move to another channel

[stuck lark]

2nd mute

3rd mute

[frail steppe]

i could say something in this channel, but if i got muted it would be a mute point

[stuck lark]

4th

[pastel jay]

heck i just wanted to continue this os talk in a different channel...

[stuck lark]

i'll say again. offtopic convos should be moved to #chill

[lime tusk]

Is math analysis the same as precalculus if it is the name of the course you do right before calculus at my school?

[sick steppe]

you'd need to share parts of the curr

[ivory needle]

I was pinged yes

[raw orbit]

Who hath pinged me

[ivory needle]

I think some guy decided to just ping everyone alphabetically

[raw orbit]

[raw orbit]

What a legend

[sick steppe]

yeah looks like it's just precalc

w/ first bit of regular calc 1

[lime tusk]

hmmm
Should I be terrified?

[grizzled orchid]

Did i get pinged

[lime tusk]

I did algebra 2/trig last year and it was mostly fine

[grizzled orchid]

Was it a mass ping

[sick steppe]

idk you as a math student

[lime tusk]

see the pins

[opaque belfry]

bruh

[grizzled orchid]

Wacky

[charred frigate]

lot of people starting with 'a' and people with the honorable role. probably pinged everyone on the user list from top to bottom

[viscid thistle]

Somehow i got pinged too

[toxic karma]

ah man I got pinged too and I thought someone finally solved the problem I asked like 2 months ago 😔

[ivory ravine]

wtf ping?

[versed escarp]

:ping:

[proud sparrow]

omg whoa okay that explains it

[viscid thistle]

ping

[viscid thistle]

lol i mean dont u like the little burst of dopamine, when you see that you got pinged because that means someone is paying attention to you and thinking of you, even though its an online stranger most likely? isnt that the reason you so readily came to this channel but then was dissapointed to find nobody of the sort?

[ivory ravine]

i like not being pinged unless its important?

[quick bluff]

The 7th question, the answer given is option A

But according to my steps doesn't "a" have to be 4?

[past meadow]

,rotate

[past meadow]

:(

[golden latch]

you're checking differentiability at a point so you will be taking the LHD and RHD at x=0 so you should replace x with 0 in your expression

[quick bluff]

OHH

Thanks again

[crimson steppe]

cue the 10000000 clones of macos' ui done in different linux distros

[misty hemlock]

why ping

[viscid thistle]

who pung me

[past meadow]

there was a mass pinger earlier

[agile nest]

Can someone differentiate this?

[golden latch]

you would want to simplify this to an algebraic expression in x using the trigonometric identity 1+tan^2x=sec^2x , so you would want to do the substitution theta=arctan(x) and then proceed

[agile nest]

The final simplification part was a bit confusing

But I think I solved it now

Thanks for your support :-)

[golden latch]

sec(x)+tan(x)=tan(pi/4+x/2) iirc

[agile nest]

I solved till this

[golden latch]

I mean this is just chain rule , which is not incorrect but I think turning this problem into algebraic is more efficient and pretty standard , this works tho

[agile nest]

Yep thanks for the help

[viscid thistle]

not sure how to tackle this

[sick steppe]

find f^(-1) for that f(x)

then equate them and determine (a,b)

[viscid thistle]

f^-1(x) would be x - b/ a, correct?

[sick steppe]

no

$x=ay+b$

[obsidian monolith]

Mosh

[viscid thistle]

oh right

oops

wait but then

when you isolate y

it's x - b/ a right?

[sick steppe]

no

$y=\frac{1}{a}(x-b)=\frac{1}{a}x-\frac{b}{a}$

[obsidian monolith]

Mosh

[viscid thistle]

excuse me but wouldn't that be the same thing

[sick steppe]

no..

1/a clearly isnt 1

[viscid thistle]

(1/a) times x equals x/a

[sick steppe]

yeah.. not what you wrote

[viscid thistle]

x/a - b/a because they have the same denominator would be x - b/ a

[sick steppe]

$x-\frac{b}{a}$ is what you wrote

[obsidian monolith]

Mosh

[viscid thistle]

oh nono

I meant

(x-b)/a

[sick steppe]

then yes

[viscid thistle]

alr

so then what do I do next to solve the problem

[sick steppe]

.

$ax+b=\frac{1}{a}x-\frac{b}{a}$

[obsidian monolith]

Mosh

[sick steppe]

so clearly you need a=1/a and b=-b/a

[viscid thistle]

oh got it

that makes sense thank you

[quick bluff]

Part (A)

The answer is given as 4

My logic was the function becomes non differentiable everytime the mod switches signs

That is every time tan x = cot x after which the function changes

But I'm pretty sure it's wrong

I drew the graph of tan and cot and I see only three places where | tan x - cot x| would open with different signs

OH

WAIT

One sec

Lmao the graph is wrong

Oh wait

[trim hemlock]

Technically all you need is to find the spot where f(x) is undefined and f'(x)is undefined

Because

1. If the function is already undefined, how cna there be a rate of change anyway?
2. Rate of change is undefined

[quick bluff]

Wait why does it need to be undefined?

[trim hemlock]

Reasons are above

Not differentiable implies that the rate of change is undefined at some points

Or the limits dne

[quick bluff]

Isn't it not differentiable even if there's just a sharp turn?

That is, when the function changes?

[trim hemlock]

Yes that too

Usually that would be when abs sign are present

[quick bluff]

Yes

[trim hemlock]

Actually, i was wrong, scratch the first point

[quick bluff]

Okok

[trim hemlock]

Oh wait lemme see

[quick bluff]

Waler could you verify one thing for me?

[trim hemlock]

I think im confusing kyself with another fact here

[quick bluff]

Isn't this graph incorrect?

[trim hemlock]

Oh yeah nvm, this is right

[quick bluff]

The dark lines are tan x and lighter ones are cot x

[trim hemlock]

The function has to be defined first for it to be differentiable

[trim hemlock]

Ill check

[quick bluff]

Pretty sure this is correct

I think I got it

Because tan x and cot x intersect 4 times between 0 and 2 pi

After which | tan x - cot x | will open with different signs

[quick bluff]

This changes the entire function giving it a sharp edge pretty sure

Am I right?

[trim hemlock]

Ermm no

Wait

[quick bluff]

Yeah my brain is shitting itself lmfao

[trim hemlock]

Is it the graph of |tanx -cotx| or graph of the given function

[quick bluff]

No it's the graphs of tan x and cot x

Imposed on one another

[trim hemlock]

Ah

[quick bluff]

The dark lines are for tan x and the lighter ones are for cot x

[trim hemlock]

Then yeah, seems good

[quick bluff]

Nono, actually...

Even that seems wrong

Because there are like 6 regions where tan x and cot x switch in their magnitudes?

One sec

[trim hemlock]

Im not sure what you mean

But the graph is correct

The two graphs should intersect at kpi/2 + pi/4

[quick bluff]

Split it into 8 regions

In the first region, cot x > tan x

2nd region, tan x > cot x

3rd region, cot x > tan x

And so on

[trim hemlock]

Yep

[quick bluff]

So shouldn't it be 8 then?

[trim hemlock]

8 what?

[quick bluff]

8 points where | tan x - cot x | opens differently

Thereby changing the entire function

Giving it 8 points of non differentiability?

[trim hemlock]

Since its an oscillating function so yes

[quick bluff]

^

[trim hemlock]

That was a different problem

[quick bluff]

This discussion was entirely for part A of this question right?

[trim hemlock]

Oh wait qctually

[quick bluff]

When the mod opens with +ve, the function is cot x, when it opens with -ve, the function is tan x

[quick bluff]

The opening of the mod changes only when tan x > cot x switches to cot x > tan x or vice versa

[quick bluff]

From here I'm able to see 8 places where the opening of the mod changes

[trim hemlock]

No no, i have mistkaen

[quick bluff]

Oh

[trim hemlock]

The sign only changes when tanx -cotx equates 0

[quick bluff]

Is this wrong?

[trim hemlock]

Lemme just draw it out

[quick bluff]

Okok

[trim hemlock]

Disregard this

This is the graph of the asb value function

The red is the graph

Dotted red is the mirror of the bpue portion

[quick bluff]

Ok

[trim hemlock]

Hmm wait a min

[quick bluff]

Kk

[trim hemlock]

Im not sure how to explane this

[quick bluff]

My issue right now is by my logic the answer should be 8

But it's flawed somehow and I don't know where

[trim hemlock]

Ok so basically

[quick bluff]

Ye

[trim hemlock]

lets take it easy, and consider tanx - cotx

[quick bluff]

Kk

[trim hemlock]

This is the graph of tanx and cotx

[quick bluff]

Yes

[trim hemlock]

You will see that they have x ints every kpi/2 + pi/4

(Basically where tanx and cotx intersect)

[quick bluff]

Yes

[trim hemlock]

And is negative in portion 1, 3, 5 and 7

[quick bluff]

Yup

Because cot x is greater

And it's positive in regions 2, 4, 6 and 8 right?

[trim hemlock]

So their graph is something like

[quick bluff]

Oh

[trim hemlock]

That is what we should be having

Btw my sketch was wrong

[quick bluff]

This is the graph of tan x - cot x?

[trim hemlock]

Yep

[quick bluff]

Oh kk

[trim hemlock]

So now when applying the absolute value to it

We want to mirror the negative portion of the graph

[quick bluff]

Yes

[trim hemlock]

To the positive y axis

Which results in

Sorry not very steady hands

[quick bluff]

Yes

[quick bluff]

Lmao np

[trim hemlock]

Basically now you notice that these "sharp turns" exists at points where |f(x)|=0

[quick bluff]

Yes

[trim hemlock]

Or simply put f(x)=0

[quick bluff]

Yup

[trim hemlock]

Which there exists to be 4

[quick bluff]

[trim hemlock]

Huh, i just realised that, the sharp turns are always at the x-int

Alright thats cool

New knowledge obtained

[quick bluff]

Alright, now there's this one final doubt

when i actually graph it out

OHH NVM

NVM

lmao

[trim hemlock]

Yeah seems logical

[quick bluff]

the question states points where they are "continuous and undifferentiable"

Thanks a lot dude

[trim hemlock]

👍

[unborn blade]

is $\frac{1-\sqrt{x}+x\sqrt{x}}{x}$ the same as $\sqrt{x}$?

[obsidian monolith]

Pahul's Uncle

[unborn blade]

clearly it's not since the graphs are different when i check on desmos but when I try simplifying they are the same

im not sure why

im probably simplifying wrong

but the textbook answer key says the continous interval is x>0 which led me to believe that it was indeed the same as sqrt(x)

[simple parcel]

i think i know why you got $\sqrt{x}$, but the numerator simplifies as $\sqrt{x}(x-1)+1$, not $\sqrt{x}(x-1+1)$

[obsidian monolith]

mchen10

[unborn blade]

my algebra skills are wack

can you explain a bit more

maybe the steps

i cant get it to simplify to $\sqrt{x}(x-1)+1$

[obsidian monolith]

Pahul's Uncle

[simple parcel]

just the numerator

not the whole fraction

[unborn blade]

ohh

im trolling

ok yeah i got it now

that makes more sense

[simple parcel]

ok

[unborn blade]

thanks

[simple parcel]

np

[unborn blade]

Can someone point me in the right direction with this question

[willow bear]

not every function continuous only on the positive axis is sqrt(x)

[unborn blade]

yeah but i figured I was on the right track

obviously I wasnt

[somber sphinx]

s l o p e

also this might help you;

as r->0, ab becomes perpendicular to y axis

so center-c becomes (inf, 0)

it basically diverges to nigh

oh well bye

[quick bluff]

Is this correct?

[rancid viper]

Looks right to me

[quick bluff]

Thank you

[rancid viper]

If you want to just derive the functions, which I think you do

[quick bluff]

Yes

Thought that we couldn't directly derive each function within individually for some reason

Thanks again

The 12th question

I'm able to say f(x) is continuous by using the sandwich theorem

But I'm not sure how I can tell anything about f(x)'s differentiability at x=0

[grand lotus]

Take f(x) =x^2

[viscid thistle]

What does |f(x)| imply? Does it imply we can access the negative range values as well?

The answer should be option 3, if my earlier statement holds true. We get two graphs, and origin act as a mirror for x^2.

[quick bluff]

Yes, the answer given is option 3

But how can we take f(x) as x^2 itself?

Is it becausef(x) and x^2 more or less become the same when tending to 0?

[grand lotus]

X^2 full fill all the conditions

[viscid thistle]

Can you tell me when does a function become non-differentiable?

[quick bluff]

When it's not continuous or when the derivative is infinite or when the left and right hand derivatives don't match up?

[viscid thistle]

Sure. You are right! If the function is discontinuous at point P, it will be non-differentiable at P(--temporary--). And, at sharp corners, it is non-differentiable too. That sentence is an underestimation. We should rather say, at sharp corners, the function becomes enormously differentiable. So much so that at one specific point, we get multiple distinct values of the derivative.

[quick bluff]

true

But what does that have to do with taking f(x) as x^2 ?

[quick bluff]

Is this statement incorrect?

[viscid thistle]

I am not sure what you mean, but probably you are misinterpreting it. Think of x^2 as a function (which it is) and it is = f(x). They both affect each other. Guess the value of x and you will get f(x). At 0, yes, they converge to a specific number which is 0. But you will notice that |f(x)| is not a function, which I guess you can make sense of (or feel free to ask).

Also, remember this - Tangent is not the same thing as derivative. If f(x) is a function then at point x0 its derivative is f'(x0), which is the slope of the tangent line to f(x) at point P. At a sharp point, we get multiple tangents. It implies there have to be multiple slopes, which in turn means distinct derivates.

[viscid thistle]

Having additional knowledge doesn't hurt. I was trying to give a conceptual and clear understanding,

[quick bluff]

Oh ok

I thought that was the end of the explanation and I missed something

My bad

[quick bluff]

Ah yes, I keep confusing derivatives as the tangents themselves instead of the slope of the tangents

[quick bluff]

Suppose the inequality was just < and not <=, would I still be able to take |f(x)| as x^2?

[viscid thistle]

Nope!

The whole function will become discontinuous as a result.

[quick bluff]

Oh yes

We can't use the sandwich theorem if it's just <

[viscid thistle]

That's a naive statement. We should never just say a function is continuous or discontinuous. We MUST always mention the bounds (or the domain) while mentioning continuity or discontinuity. Ex - Every discontinuous function can be continuous for some ∂≈0.

[viscid thistle]

It is not about seeing the applicability of some theorems. One should get a sense of whatever is going on.

[quick bluff]

So when I remove the = from <= in the question, I can't tell anything about it's continuity since f(x) might be close to the upper and lower bounds (x^2 and -x^2) or f(x) might not be near them at all?

Oh wait

When I reread what I wrote it makes no sense

Yes, it doesn't make sense because I didn't mention for which domain I'm checking the continuity for?

[viscid thistle]

One thing that helps me a lot in mathematics is to frame a literature sense of the equation. I would recommend you do that as well. Ex: y < x. Mathematically : y < x. Literature wise - "So, it is saying we have to find all those x's that are bigger than y. But where can Y = X? oh, it's that y = x line. Below it, x > all y's and above it y > all x's." That's one way to solve things in math. Step by step from the ground up.

[quick bluff]

Noted, I'll try my best to implement this into my thinking process

OH WAIT

I think I got something

f is from R to R

So every input x HAS to give an output

When u graph out the upper and lower bounds

That is x^2 and -x^2

They have a common tangent at x=0

And since f(x) lies between x^2 and -x^2

[quick bluff]

It has to pass through this point with the same slope?

I think the mistake I have been making is trying to look at f(x)'s behaviour at all R instead of just around x=0

[viscid thistle]

Exactly! But a little correction in your above statement. It should be f(x) is = x^2 and f(x) = -x^2 simultaneously. It would lie between them when there's just "< or >" inequality.

[quick bluff]

Oh yes

[viscid thistle]

And, since |f(X)| has two different simultaneous outputs (x^2, and -x^2), that is why it is NOT a function.

[quick bluff]

Makes sense

[quick bluff]

Yeah I think I get what you and learn4math were telling me now

Thanks a lot!

[viscid thistle]

JEE is doable. All you need is an enthusiastic teacher (who can extend equations to real life applications) and a passion ingrained in you.

[quick bluff]

I have the enthusiastic teachers, searching for the passion

[ionic vessel]

This is dumb but how can I simplify 4^2^x

[sick steppe]

unclear

$(4^2)^x \ 4^{2^x}$

[obsidian monolith]

Mosh

[ionic vessel]

The second one sorry

[sick steppe]

it cant be as far as I'm aware

[ionic vessel]

Actually like 2^(2^(2x+1))

[sick steppe]

ok yeah, $2^{2\cdot 2^x}=2^{2^{x+1}}$

[obsidian monolith]

Mosh

[ionic vessel]

Is there anything else that can be done

[sick steppe]

not really

[quick bluff]

The 14th sum

Answer is given as option 2

Sending some steps, one sec

I'm getting the answer but I'm pretty sure my reasoning is incorrect

[quick bluff]

From the first statement

I substituted x = 0

f'(0) give -1 and 1 respectively

f'(f(0)) give values -cos(ln2) and cos(ln2) respectively

Can I multiply the pairs of values of f'(0) and f'(f(0)) and conclude that since both of them give the value cos(ln2), g'(0) is cos(ln2) ?

[viscid thistle]

i mean that's okay, but when finding the piecewise function, as you are asked at x=0, you can perfectly plug it into ln(2)≥sin(x) and ln(2)<sin(x) and see which one is true, and then go with that one instead of doing both values

[quick bluff]

OHH

So at x = 0 only ln(2) > sin(0) is true so I can ignore the function in the interval ln(2)<sin(x) ?

[viscid thistle]

pretty much, yeah. piecewise functions are made so that when working with a certain point, you'd want the part of the piecewise functions that corresponds to that interval.

[quick bluff]

Thanks a lot

[viscid thistle]

The way you wrote f'(x) after f'(f(x)) is quite astonishing.

[quick bluff]

Wait does it look weird?

[willow bear]

is it? all they did was apply the chain rule

[quick bluff]

Or did I make a mistake of some sort?

[viscid thistle]

It is not weird. Rather a notational abuse I would say. f(x) is not differentiable(at a lot of points), so I am not sure what you are going to get by f'(x).

[quick bluff]

But isn't it differentiable around 0?

The options make it clear that they want us to check the differentiability at 0 I'm pretty sure

[viscid thistle]

It sure is differentiable around 0. I am speaking in general and I have seen people taking such a path with the composition rule of differentiation (chain rule) where they might differentiate the inner function without realizing whether it is differentiable or not.

[quick bluff]

Oh

I'll keep that in mind then

But is what I did incorrect then?

[viscid thistle]

I mean we have a rule(chain rule), and that is fine. But we must understand the kind of functions(and domains) we are dealing with. It is such a subtle thing I guess.

[viscid thistle]

In this case? No. Can it be wrong in othe case? Maybe.

[willow bear]

abovewater i think you're being a little pedantic here

[viscid thistle]

Ironically, limit is about being precise with the degree of approximations we are making. I am just trying to give him a different lens to look through.

[quick bluff]

A new word has been added to my vocabulary

[quick bluff]

Is the reasoning correct? The question was to find K which is the set of all points where f(x) is not differentiable

[viscid thistle]

Paradon me but I am having trouble reading your handwriting. From what I can barely make, it seems correct.

[dusk edge]

What is this font lmao

[dark gyro]

Find the derivative of

is this solution correct ?

[simple parcel]

no, because youre taking the derivative with x in the exponent, so the power rule doesn't apply

[dark gyro]

ah shit wait

the derivative would be the same wouldnt it

the same as e^x/2

[simple parcel]

the derivative of e^x is e^x, but you need to use the chain rule

[dark gyro]

no not here

unless the book is messing with me

[simple parcel]

$\frac{d}{dx}e^{\frac{x}{2}}=\frac{1}{2}e^{\frac{x}{2}}$

[obsidian monolith]

mchen10

[dark gyro]

im confused

the book didnt give me any tips

ffs

like it gave me some formulas, i thought i was supposed to follow them

the book gave me this

i see nothing here that can help with this scenario besides the e^x

do you?

[simple parcel]

it says that the derivative of $e^{kx}$ is $ke^{kx}$

[obsidian monolith]

mchen10

[dark gyro]

sorry had to go

but can you explain that ?

a bit confused how you got there

[simple parcel]

the derivative of e^kx is in your book, but basically the derivative of f(g(x)) is f'(g(x)) * g'(x), which is also called the chain rule

[dark gyro]

Well here is the weird thing

the book teaches the chain rule a bit further up

where im at

its before chain rule

which is weird why the book would expect me to know that

[simple parcel]

i guess they just want you to use the formula first

[dark gyro]

ok so i should learn the chain rule then get back to the question?

so you didnt use the e^kx to solve the question?

[simple parcel]

i did. the book says $\frac{d}{dx}e^{kx}= ke^{kx}$, so i just plugged in k=1/2 to get the answer

[obsidian monolith]

mchen10

[dark gyro]

wdym plugged in k = 1/2

just randomly ?

or was there a reason

[simple parcel]

you wanted to find the derivative of $e^{\frac{x}{2}}$, so that's just what i did

[obsidian monolith]

mchen10

[dark gyro]

omg xD

why did u go with 1/2 ?

its x/2

[simple parcel]

$\frac{x}{2}=\frac{1}{2}x$

[obsidian monolith]

mchen10

[dark gyro]

i see

ok

[molten garnet]

How do u get rid of the log thing

Do u expand on the right or smth?

[willow bear]

are these logs meant to be decimal logs or natural logs?

[molten garnet]

just natural

[willow bear]

natural, ok

note that any solution to this must be greater than 21 (otherwise log(x-21) is undefined!)

and after doing so, take e^(both sides)

[molten garnet]

then I simplify and find the 2 x values

[willow bear]

what you should end up with is an equation without logarithms, which you will then solve as normal

and then, once you've got your x values, you check them against the condition x > 21

[molten garnet]

ok

thanks

[hoary grail]

Because it's a perfect square?

[willow bear]

yuo

yup*

[hoary grail]

ty

[mild swan]

How do we know that?

[hoary grail]

(ax - b)^2 is (ax - b)(ax - b)

[mild swan]

How does this immediately lead to a discriminant of 0?

How do I not see this?

OHHHH

Only one solution for x

If f(x) = 0

[hoary grail]

yep

[quick bluff]

Not sure how I can proceed from here or if there's a simpler method

[quick bluff]

The question actually asks for y'(1)

[viscid thistle]

before heading onto that, the derivative of x^(2x) and x^x aren't correct, you can't differentiate these like you would with x^2 for example.

[quick bluff]

I should use the chain rule right?

[viscid thistle]

uh yeah, but still that's not how you'd differentiate these.

you can consider e^ln(x)=x for example.

[quick bluff]

e^ln(x) . (1/x) = 1 ?

Is this correct?

Oh wait

[viscid thistle]

so that $\overbrace{{\color{green}{x}}^{x}}^{{\color{green}{x}}=e^{\ln( {\color{green}{x}} )}}= \left(e^{\ln( {\color{green}{x}} )}\right)^x=e^{x\ln(x)}$

[viscid thistle]

uh yeah.

let me fix it.

[quick bluff]

Ok

[obsidian monolith]

Al𝟛dium

[quick bluff]

so basically I try to convert these functions into known functions by taking ln?

[viscid thistle]

kind of yeah, you are essentially manipulaying x^x so that it becomes an easier function to differentiate.

[quick bluff]

Would it be correct if I take both negative terms to the RHS and take ln on both sides and then differentiate?

[quick bluff]

Ah I see

[viscid thistle]

do you mean from the very beginning?

[quick bluff]

Yes

[viscid thistle]

yeah you perfectly could.

[quick bluff]

Nice

[viscid thistle]

so that's up to you

[quick bluff]

I'll give it a shot then

Thanks a lot

[viscid thistle]

alright, you can ping me if you get stuck.

yw