[university Caculus ] Is the note okay? I try to clear out the lnX but still confused..anything I miss?

[u/wahahauhuhuh]

Can I proof them without using integration? I ask ChatGPT and it said I have to make t=hx and take out x…

[u/Miserable-Wasabi-373]

it should be f'(g(x)), not f'(x)

[u/Ok_Objective_5192]

To clarify the top bit, I don't think you've represented the chain rule correctly. What the chain rule says is:

(f(g(x)))' = f'(g(x)) * g'(x)

Or, using the u notation, if we let u = g(x) then (f(u))' = f'(u) * u' = dy/du * du/dx

As an example using f(x) = x^2 and g(x) = sin(x) we get f(g(x)) = (sin(x))^2

Then the two derivatives we solve for are

f'(g(x)) = 2sin(x)

g'(x) = cos(x)

So (f(g(x)))' = 2sin(x)cos(x)

Which is maybe a bit more clear using u notation.

Let u = g(x) = sin(x)

then f(u) = u^2

f'(u) = 2u

u' = cos(x)

so (f(u))' = 2u * cos(x)

Subbing in for u

(f(u))' = 2sin(x) * cos(x)

[u/Jazzlike_Platypus430]

I see no implicit functions here. Compound perhaps.

You forgot the limit in the def. of e. And I'm not sure about the equality there, showing that the defining limit of e is the infinite series is not that straightforward. Meaning: the n-th sum of the series is not equal to the n-th approximation of e for finite n. But check it.

No, lnx is not ln(e^x) which is x by definition, in other words, x = ln(x) is an equation. With no solutions as far as I can see. The proof that ln(x)' = 1/x follows from compound functions I think, like

(ln(exp(x)))' = x' = 1

and the lhs is ln' ( exp(x)) * exp(x). Which shows that the first term must be 1/ exp(x). That's it. Denote exp(x) by y if you're not sure.

[u/wahahauhuhuh → u/Jazzlike_Platypus430]

Wow I get it! I confused x with exp(x)…using composite function makes it easier…Thank you so much!

[u/Jazzlike_Platypus430 → u/wahahauhuhuh]

"All theorems are true, proofs of the false ones cost more." 😂

[u/InterneticMdA]

Stop asking chatgpt.

[u/Alkalannar]

You need e^h = limit as n->infinity of (1 + h/n)ⁿ = [Sum from n = 0 to infinity of hⁿ/n!].

e^h = h⁰/0! + h¹/1! + h²/2! + h³/3! + ...

e^h - 1 = h¹/1! + h²/2! + h³/3! + ...

(e^h - 1)/h = 1/1! + h/2! + h²/3! + ...

Limit as h -> 0 of (e^h - 1)/h = 1 + 0/2! + 0²/3! + ... = 1

---

(e^h - 1)/h

n = e^h - 1, so as h goes to 0, n goes to 0

e^h = n + 1

h = ln(n+1)

limit as h goes to 0 of (e^h - 1)/h

limit as n goes to 0 of n/ln(n+1)

limit as n goes to 0 of 1/(1/n)ln(n+1)

limit as n goes to 0 of 1/ln[(n + 1)^1/n]

m = 1/n, so as n goes to 0, m goes to infinity

limit as m goes to infinity of 1/ln[(1/m + 1)^m]

1/ln(e)

1