# Proving the derivative for a composite function

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 ?

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

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