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If you let $y=\ln(x)$, then $x=e^y$. Performing implicit differentiation on $x=e^y$ to find $y'$ will give you $1=y'e^y$. Since $x=e^y$, we can substitute $x$ for $e^y$ to give us $1=xy'$. Solving for $y'$ will give $y'=(1/x)$

Is this a sound proof for $(\ln(x))'$?

user1551
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X Stanton
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    It is a sound proof, provided that you have shown that the function $\ln$ is differentiable in the first place. Depending on the level of rigour of the calculus course you are taking, this might be taken for granted. If not, then you should be aware of a proof of the inverse function theorem. And you should also state how you are defining $\ln$. – Joe Jun 23 '21 at 16:01
  • For $x \gt 0$, the function $ln(x)$ exists and is continuous. Since $ln(x)$ is continuous for $x \gt 0$ and there are no sharp turns in the graph, $ln(x)$ is differentiable for $x \gt 0$. I haven't taken a course on rigorous proofs yet and the calculus course I am taking isn't really rigorous. – X Stanton Jun 23 '21 at 16:05
  • It's true that $\ln$ is continuous and has no sharp turns in the graph. But that's not a proof that $\ln$ is differentiable. For one thing, there is no rigorous definition of "sharp turn". – Joe Jun 23 '21 at 16:07
  • How would you rigorously prove that ln(x) is differentiable? Can you give a hint? – X Stanton Jun 23 '21 at 16:08
  • That's not at all an easy task unless you have built up a lot of other theorems. To begin with, what do you mean by $\ln(x)$? Judging by your question, you seem to have defined it as the inverse of the function $x \mapsto e^x$. If so, the question becomes "what do you mean by $e^x$"? – Joe Jun 23 '21 at 16:10
  • Is this more talked about in real analysis, which is more like pure math than calculus? – X Stanton Jun 23 '21 at 16:11
  • Yes, it is. If you are interested in studying real analysis, then you could read Calculus by Michael Spivak (which, despite its name, is really about real analysis). This would give you a rigorous proof that $\ln$ is differentiable. Every University course in pure mathematics would also have a real analysis module. But if you are just studying calculus in a school course, then I would assume that your proof would be considered perfectly valid. – Joe Jun 23 '21 at 16:16
  • (Remember to begin your comment with @Joe if you want to ping me a message.) – Joe Jun 23 '21 at 16:17
  • @Joe. I'm in college right now, but I'm still in freshman year. I'll be taking far more math courses in sophomore year, which should hopefully be less rough with COVID restrictions being lifted. – X Stanton Jun 23 '21 at 16:20
  • @XStanton you may use definition of derivatives and evaluate limits – Lalit Tolani Jun 23 '21 at 16:45
  • @XStanton: See Bernard's answer for a proof that $\ln$ is differentiable. If you have any questions regarding his answer, then you ask me or him. – Joe Jun 23 '21 at 16:53
  • @XStanton: I just realised that you were the same user as the one who asked about using the chain rule to compute the derivative of an inverse function. If you look at the theorem that I wrote about in my answer at the bottom (it's called the inverse function theorem), then you can confirm that $\ln$ is indeed differentiable. – Joe Jun 23 '21 at 17:12

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