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93/1000, why is the derivative 2^x equal to 2^x*ln(2)

• bprp calculus b… 3일 전 2026.08.08 16:00 인기
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@nic-u6fnic 　3일 전

horrible explanation at the end

2026-08-09 07:30

[ytcid:UgwD4pGPnwPLT3YTwLx4AaABAg] horrible explanation at the end



@Eric-jx3lb 　3일 전

You basically need to define what e is.  \(e\) as the upper limit of \((1 + \frac{1}{n})^n\). Then we talk!

2026-08-09 02:05

[ytcid:Ugy04nmYUDujjIxR0fh4AaABAg] You basically need to define what e is. \(e\) as the upper limit of \((1 + \frac{1}{n})^n\). Then we talk!



@HowardThompson… 　3일 전

Bruh, that’s cheating.

2026-08-08 22:58

[ytcid:Ugzl32gb9PGGYu3rkxF4AaABAg] Bruh, that’s cheating.



@brainfellow514… 　3일 전

Something's missing,  he jumped a step somewhere to conclude ln(2).  Feels incomplete...

2026-08-08 21:26

[ytcid:UgxsQtaxEpcJLxdJRwx4AaABAg] Something's missing, he jumped a step somewhere to conclude ln(2). Feels incomplete...



@David_Math_Bib… 　3일 전

Nobody is tryna use the definition. Tedious you know. Just set 2^x equal to y, ln both sides, and proceed from there

2026-08-08 21:19

[ytcid:Ugw1-Eo9oZ_BnI12adJ4AaABAg] Nobody is tryna use the definition. Tedious you know. Just set 2^x equal to y, ln both sides, and proceed from there



@EHJ-V 　3일 전

Please use 94/1000 to show how you do the last part. This was... unsatisfactory.

2026-08-08 21:06

[ytcid:Ugyug7uvALrPLYZY5vN4AaABAg] Please use 94/1000 to show how you do the last part. This was... unsatisfactory.



@Iomhar 　3일 전

Using the calculator is cheating.
Your 93 is invalid.

2026-08-08 20:24

[ytcid:UgyGWPOrftDd8KneuQd4AaABAg] Using the calculator is cheating. Your 93 is invalid.



@Φόβος729 　3일 전

So we could define
ln(X) = lim_h->0 of (x^h - 1 )/ h?

2026-08-08 19:19

[ytcid:UgwpryPOm8cX8y2Q1GN4AaABAg] So we could define ln(X) = lim_h->0 of (x^h - 1 )/ h?



@xeoneo6841 　3일 전

Without using any other derivatives or the definition:

2^x*lim h->0 2^h-1 / h
*note 2^h = e^(h*ln2)

Now take the series expansion
e^(h*ln2) = 1 + hln2 + (hln2)^2/2!…

Now look back at the original limit the 1’s cancel divide through by h and your left with ln2 + h(ln 2)^2 + h^2(ln3)^3…
Taking the limit as h -> 0 gives you just ln2. Now remember you had the 2^x outside of the limit so…
d/dx 2^x = ln2*2^x

2026-08-08 17:53

[ytcid:UgzCwpJGhPqEoTO9v5p4AaABAg] Without using any other derivatives or the definition: 2^x*lim h->0 2^h-1 / h *note 2^h = e^(h*ln2) Now take the series expansion e^(h*ln2) = 1 + hln2 + (hln2)^2/2!… Now look back at the original limit the 1’s cancel divide through by h and your left with ln2 + h(ln 2)^2 + h^2(ln3)^3… Taking the limit as h -> 0 gives you just ln2. Now remember you had the 2^x outside of the limit so… d/dx 2^x = ln2*2^x



@rohangt1 　3일 전

Is there any proof for the result ln(2) coming from it's limit?

2026-08-08 17:18

[ytcid:Ugx9kEAhLhhx-rCxYrN4AaABAg] Is there any proof for the result ln(2) coming from it's limit?



@עולםמיינקראפטש… 　3일 전

That not profe that is realy ln(2)

2026-08-08 17:15

[ytcid:UgzRQwp3kj4H4br_RlR4AaABAg] That not profe that is realy ln(2)



@vikramrawat526… 　3일 전

Day 93

2026-08-08 16:56

[ytcid:UgyIXiJDlRWR4BoX4vh4AaABAg] Day 93



@mohbomran7087 　3일 전

Is there really no other way to do that???

2026-08-08 16:12

[ytcid:Ugx7BtChOSJxvsO1Pap4AaABAg] Is there really no other way to do that???



@cjfool5489 　3일 전

Without using the definition .
Let y= 2^x
lny = ln(2^x)
lny = x.ln2
(1/y)dy/dx = ln2.dx/dx
(1/y)dy/dx = ln2
dy/dx = y.ln2
Substituting the value of y we get,
d(2^x)/dx = 2^xln2

2026-08-08 16:12

[ytcid:UgyHtDbWfx1mf5psegp4AaABAg] Without using the definition . Let y= 2^x lny = ln(2^x) lny = x.ln2 (1/y)dy/dx = ln2.dx/dx (1/y)dy/dx = ln2 dy/dx = y.ln2 Substituting the value of y we get, d(2^x)/dx = 2^xln2



@BSky2025 　3일 전

Can you do some integrals by substitution please?

2026-08-08 16:05

[ytcid:Ugwl_PnzEJigDYS94z14AaABAg] Can you do some integrals by substitution please?



@OfficialSuper_… 　3일 전

Oh,my fist

2026-08-08 16:05

[ytcid:UgxCZaG6Bjjj-F7ysNB4AaABAg] Oh,my fist



@marksandsmith6… 　2일 전

Everybody unhappy with last step ?

2026-08-09 18:15

[ytcid:UgzsRcqe2uO1pC2-Lad4AaABAg] Everybody unhappy with last step ?



@wannabeactuary… 　2일 전

2^h  =  h ln 2 = h exp(ln (2) =  h (1 + ln 2 +  (ln2)^2! +....
(2^h - 1)/h = hln2  + o((ln2)^2)

2026-08-10 06:17

[ytcid:UgwU9AXzULecaws5o-14AaABAg] 2^h = h ln 2 = h exp(ln (2) = h (1 + ln 2 + (ln2)^2! +.... (2^h - 1)/h = hln2 + o((ln2)^2)



@russelltaylor7… 　2일 전

y=2^x
lny=xln2
y'/y=ln2
y'=yln2=2^xln2

2026-08-10 05:11

[ytcid:Ugy8BDLHLsNdWy59eGZ4AaABAg] y=2^x lny=xln2 y'/y=ln2 y'=yln2=2^xln2



@Mediterranean8… 　1일 전

L = lim h→0 (2ʰ-1)/h , let t = 2ʰ-1 , h = ln(1+t)/ln 2 , L = ln 2 × lim t→0 t/ln(1+t) = ln 2/ lim t→0 ln(1+t)ᵗ = ln 2/ln e = ln 2

2026-08-10 21:10

[ytcid:UgzcPtNv9bwYng4Xds54AaABAg] L = lim h→0 (2ʰ-1)/h , let t = 2ʰ-1 , h = ln(1+t)/ln 2 , L = ln 2 × lim t→0 t/ln(1+t) = ln 2/ lim t→0 ln(1+t)ᵗ = ln 2/ln e = ln 2