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Jul 18, 2022 at 10:04 vote accept Aminos
Jul 17, 2022 at 20:57 answer added Victor Souza timeline score: 2
Jul 16, 2022 at 22:54 comment added Тyma Gaidash It alludes to how $\sin(t)$ gives the arc length of a circle given an angle $t$, $\arcsin(x)$ gives the angle given the arc length x, and $\arcsin’(x)$ is the integrand. $\arcsin(x)$ can be put in terms of $\ln(x)$ while $\sin(x)$ can be put in terms of $e^x$. As for the intuitive explanation, $\int x^{-1} dx=\lim_{p\to0} \frac{x^p}p=\ln(x)$
Jul 16, 2022 at 22:41 answer added Jean-Armand Moroni timeline score: 2
S Jul 16, 2022 at 19:50 history suggested bobeyt6 CC BY-SA 4.0
edited for latex
Jul 16, 2022 at 19:43 comment added user2661923 So, letting $y = e^x$, you have that $g(y) = x.$ Then $g'(y) = \dfrac{1}{f'(x)} = \dfrac{1}{f(x)} = \dfrac{1}{y}.$ Thus, the derivative of the function $g(y) = \ln(y)$ is $g'(y) = \dfrac{1}{y}.$
Jul 16, 2022 at 19:40 comment added user2661923 Posting this as a comment, rather than an answer, because my response is very informal, and not backed up by a solid understanding of the topic. The function $f(x) = e^x$ has the property that it is equal to its own derivative. That is, $f'(x) = f(x).$ Let $g(x) = \ln(x)$. Then $g(x)$ is the inverse function to $f(x)$. There is a principle in Real Analysis that when any $g(x)$ is the inverse of any $f(x)$, and when $f'(x_0) \neq 0$, then $g'(x_0) = \dfrac{1}{f'(x_0)}.$ ...see next comment
Jul 16, 2022 at 19:13 comment added bobeyt6 Please use LaTeX and Mathjax formatting for future posts.
Jul 16, 2022 at 19:13 review Suggested edits
S Jul 16, 2022 at 19:50
Jul 16, 2022 at 19:07 review Close votes
Jul 17, 2022 at 6:05
Jul 16, 2022 at 19:02 history edited Aminos CC BY-SA 4.0
added 10 characters in body
S Jul 16, 2022 at 18:42 review First questions
Jul 16, 2022 at 18:43
S Jul 16, 2022 at 18:42 history asked Aminos CC BY-SA 4.0