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 ?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.