Simple differentiation help please

[mystic aurora]

Q5 , the answer is b apparently, but I keep getting the ln y as -infinity over -1 which is +infinity if I'm doing simple first grader logic here, but if I do that then y will be e^infinity , which is (c), gpt keeps tryna switch the infinity sign and I'm not convinced as to why, thank u in advance

[cerulean pulsar]

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[lapis lake]

just

differentiate

and

this is

L

[mystic aurora]

I did?

[lapis lake]

hopital

[honest plover]

is this one

just asking

[odd sand]

okay.

[mystic aurora]

Yea

[odd sand]

kinda hard to explain here.

[lapis lake]

differentiation of cos x

say

1

2

[honest plover]

-senx

[lapis lake]

sen x?

okay

[honest plover]

but negative, (cosx)' = -senx

[lapis lake]

now

[mystic aurora]

I did the work up there

[lapis lake]

tan x

and then put

90

[mystic aurora]

And gpt did the same exact work and reached the infinity

[honest plover]

it can be sec^2 (x) or 1+ tan^2(x)

you choose bc both are the same

[mystic aurora]

Wait hold up

[lapis lake]

@mystic aurora you can just solve the derivatives for the trig functions
and then subsitute

the pi/2

[mystic aurora]

Ok ok

But

I want to know

If I worked the way I did

Why didnt I reach

The answer

[lapis lake]

where is the working

[honest plover]

it doesnt exist tan (90) if we talk about degrees

[mystic aurora]

In the photos

[honest plover]

ok so

[mystic aurora]

Then where to explain 🥹

[lapis lake]

oh shit ya right

[honest plover]

bro,

[odd sand]

i will try my best

[honest plover]

l'hopital cant be used on this problem

[odd sand]

recall that x=exp(ln(x))

[mystic aurora]

It's a question in the l'hôpital sheet wdym

[lapis lake]

wdym

[mystic aurora]

Yes

[honest plover]

i'm so fucking confused right now

[gritty karma]

Ok

[odd sand]

so the limit turns into $\lim_{x\to\frac{\pi}{2}}\exp(\tan(x)\ln(\cos(x)))$

[polar trellis]

dark matter

[lapis lake]

hi femboy

[odd sand]

can you try to do that?

[mystic aurora]

Wait

[odd sand]

yes?

[mystic aurora]

Why did u put e as the base

[odd sand]

because x=exp(ln(x))

also so it’s easier to differentiate

[mystic aurora]

Mm

Ok I'll try holup

[odd sand]

(e^f(x))’=e^(f(x))*f’(x)

[honest plover]

chain rule yea

but here i got one question

bc it made me confused

l'hopital isnt focused on solving terminations of 0/0 or infinite / infinite?

[odd sand]

(you should really try with a regular limit though)

[honest plover]

like it's exclusive for fractions of algebraic expressions

[odd sand]

and note that x->pi/2 from the left

[odd sand]

no.

[honest plover]

give me this wisdom please

[odd sand]

this is a 0^infinity case

[mystic aurora]

Bro how is this any easier

[honest plover]

yea, and why l'hopital can be used here?

[odd sand]

take $\lim_{n\to\infty} (1+\frac{1}{n})^n$

[polar trellis]

dark matter

[honest plover]

THAS E

[mystic aurora]

The exp will stay as I keep differentiating

[odd sand]

yes.

[mystic aurora]

And the exp is to the power of undeterminate

[odd sand]

which is why i said to evaluate the original limit with exp.

[mystic aurora]

What

Evaluate exp of the whtvr?

That's still undeterminate

[odd sand]

evaluate this without L’H

[odd sand]

no, it isn’t.

[mystic aurora]

That's e to the power of infinity times negative infinity

[odd sand]

make a new help post for this, i’ll explain later.

[odd sand]

and what is infinity*infinity?

[mystic aurora]

Undeterminate

[odd sand]

it’s not an indeterminate form.

[mystic aurora]

It isn't?

[odd sand]

no.

infinity-infinity is.

[mystic aurora]

Wow

[odd sand]

but times isn’t

[mystic aurora]

Mind blowing

Ok

So

[odd sand]

yes, it’s kind of unclear where an indeterminate occurs

i highly suggest you create a different post to address this, but we’ll answer this first

so what is -infinity*ifninity

[mystic aurora]

I just knew the fact that smth exists called undeterminate last Thursday so pls excuse my ignorance on this

[odd sand]

it’s alright

[mystic aurora]

Um negative infinity prob

[odd sand]

even it is unclear for me sometimes, and i’ve been doing this for a few years now

[odd sand]

yes

and what is e^(-infinity)?

[mystic aurora]

Ok cool

I get it this w approach

[odd sand]

yes.

[mystic aurora]

But why is the other approach not working?

[odd sand]

you mean, taking the ln?

[mystic aurora]

Yea

L'hôpital

[odd sand]

$\ln(y)=\ln(\lim_{x\to\frac{\pi}{2}}(\cos(x))^{\tan(x)})$

[polar trellis]

dark matter

[mystic aurora]

Yes

[odd sand]

this.

it works literally the same.

[mystic aurora]

I mean like after putting the ln inside the lim

[odd sand]

$\ln(y)=\lim_{x\to\frac{\pi}{2}}\tan(x)\ln(\cos(x))$

[polar trellis]

dark matter

[odd sand]

don’t see the issue here.

[mystic aurora]

Oh yeah

Same shi

Sorry sorry

[odd sand]

yes.

[mystic aurora]

Thanks alot

Is there a thank u thing here?

[odd sand]

it’s alright, we can discuss this further either here or in a new help post

it’s up to you

yes, there is

typing +close will prompt you to thank the helper and close the post

[mystic aurora]

No I got the gist of it , I was just missing an important fact that's why

[mystic aurora]

Alrr

[odd sand]

👍

[mystic aurora]

+close

[rocky cove]

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[rocky cove]

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@mystic aurora

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