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I would like to read the demonstration that proves :

$\frac{\mathrm{d} ln(y)}{\mathrm{d}y}=\frac{1}{y}$

where y=g(x)

Do you know where can i find it ?

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  • $\begingroup$ This is not true. The derivative of $\ln{g(x)}$ is $g'(x)/g(x)$ using the chain rule $\endgroup$ – pwerth Feb 11 at 17:41
  • $\begingroup$ What does $\frac{d}{dg(x)}$ means? $\endgroup$ – AlessioDV Feb 11 at 17:42
  • $\begingroup$ I miss writed. Edited $\endgroup$ – Jhdoe Feb 11 at 17:44
  • $\begingroup$ I edited again, is it more clear ? $\endgroup$ – Jhdoe Feb 11 at 19:48
  • $\begingroup$ @pwerth Your answer of $g'(x)/g(x)$ is correct for taking the derivative wrt $x$, but the question asks for the derivative wrt $y$. $\endgroup$ – John Omielan Feb 11 at 21:10
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There are many places you can find a proof for

$$\cfrac{d\left(\ln\left(y\right)\right)}{dy} = \cfrac{1}{y} \tag{1}\label{eq1}$$

One such place is the MSE Proof of the derivative of ln(x). As for \eqref{eq1} where $y = g\left(x\right)$, note this is true regardless of how you represent $y$. To see why, consider what the LHS of \eqref{eq1} represents. It is asking for the rate of change of the function $f\left(y\right) = \ln\left(y\right)$ as $y$ is changing. If you were to draw a Cartesian coordinate plane with the horizontal axis being $y$ and the vertical axis being $\ln\left(y\right)$, then \eqref{eq1} says that at any given point $y = y_0$ where $\ln\left(y_0\right)$ is defined, the slope of the tangent line to the curve at that point would be $\frac{1}{y_0}$. In particular, this is true regardless of any other way you might represent $y$, such as $y = g\left(x\right)$. The only thing which affects the tangent line slope is how $\ln\left(y\right)$ is changing wrt $y$, regardless of any underlying function affecting how $y$ is changing, such as where $y = g\left(x\right)$, except to the extent, if dealing with only real numbers, where $y \le 0$ for some $x$ so $\ln\left(y\right)$ is not defined. Also, the inverse function $x = g^{-1}\left(y\right)$ needs to exist in a region around whatever point you're taking the derivative.

Note, however, that if you wanted to take the derivative wrt to $x$ instead, where $y = g\left(x\right)$ and $g\left(x\right)$ is differentiable, then you would get the different result that pwerth has stated in a comment. In particular, the chain rule gives that

$$\cfrac{d\left(g\left(x\right)\right)}{dx} = \cfrac{g'\left(x\right)}{g\left(x\right)} \tag{2}\label{eq2}$$

This is, of course, different because now you are finding how the value of the curve is changing wrt to a different variable, i.e., $x$, where $y$ is connected to it by $y = g\left(x\right)$. Thus, $y$ will, in general, be different than $x$ and, thus, there is no reason to expect the derivative wrt $y$ and the derivative wrt $x$ to be the same.

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