What is $ \frac{d \ E(ln(y)|X)}{d \ y}$ in OLS? Assume that the true model (DGP) is
$ ln(y) \ = \ \beta_0 \ + \ \beta_1 ln(x_1) \ + \ \cdots \ + \ \beta_k x_k \ + \ \varepsilon \hspace{3em} \text{where } \ \begin{bmatrix} x_1\\
\vdots\\
x_k\\
\varepsilon \end{bmatrix} \sim IID  \ \text{ and }  \ \varepsilon \sim N(0, \sigma^2) $
The estimated regression model is
$ ln(y) \ = \ \widehat{\beta_0} \ + \ \widehat{\beta_1} ln(x_1) \ + \ \cdots \ + \ \widehat{\beta_k} x_k \ \hspace{3em} \forall \ i \in 1 \ldots n$
So what is the result of $\frac{d\ E(ln(y)|X)}{d\ y}$?
Since $\frac{d\ ln(y)}{d\ y} = \frac{1}{y}$, should it be something like $\frac{1}{ E(ln(y)|X)}$? And how can I prove it?
Thank you for your help!

My motivation:
I was told that given the dependent variable is logged, the coefficient of the logged independent variable can be explained as the percentage change: "a one percent change in $ln(x_1)$ is associated with a $\beta_1$ percent change in $ln(y)$." And the math was:
$\begin{equation*}
\beta_1 \hspace{1em} = \hspace{1em} \displaystyle \frac{\partial \ ln(y) }{ \partial \ ln(x_1)} \hspace{1em} = \hspace{1em} \frac{ \displaystyle \frac{\partial \ ln(y) }{\partial \ y} \ \partial \ y }{ \hspace{.7em}  \displaystyle \frac{\partial \ ln(x_1) }{\partial \ x_1 } \ \partial \ x_1 \hspace{.7em}  } \hspace{1em} = \hspace{1em} \frac{\displaystyle \frac{1 }{ y } \ \partial \ y}{  \hspace{.7em}  \displaystyle \frac{1 }{ x_1 } \ \partial \ x_1  \hspace{.7em}  } \hspace{1em} = \hspace{1em} \frac{  \displaystyle  \frac{\partial \ y}{y} }{ \hspace{.7em}  \displaystyle \frac{  \partial \ x_1}{x_1} \hspace{.7em}  } \hspace{1em} = \hspace{1em} \frac{  \displaystyle  \frac{\Delta \ y}{y} }{ \hspace{.7em}  \displaystyle \frac{  \Delta \ x_1}{x_1} \hspace{.7em}  } 
\end{equation*}$
I was trying to make sense of this percent-percent explanation of $\beta_1$. However, I thought that the marginal effect of the independent variable $ln(x_1)$ should be defined by $\displaystyle \frac{\partial \ E(ln(y)|X)}{\partial \ ln(x_1)}$ rather than $\displaystyle \frac{\partial \ ln(y)}{\partial \ ln(x_1)}$ . In this case, the denominator $ \displaystyle \frac{\partial \ ln(x_1) }{\partial \ x_1 } \ \partial \ x_1$ is still the same. But how about the numerator?
By definition, $E(ln(y)|X) \equiv \displaystyle \int \ y f(y|X) \ dy \ $ and I have no idea how to show that the numerator can be expressed into something like $ \displaystyle \frac{\Delta \ y}{y}$...

The true regression function $u(\mathbf{x}) \equiv \mathbb{E}(\ln y | \mathbf{X} = \mathbf{x})$ is a function of the regressor vector $\mathbf{x}$, not the response variable $y$.  (Remember that the expectation operation is an integration --- it "integrates out" the variable of integration.)  Thus, if you take its partial derivative with respect to $y$ you get the trivial result:$^\dagger$
$$\frac{\partial}{\partial y} \mathbb{E}(\ln y | \mathbf{x}) = \frac{\partial u}{\partial y}(\mathbf{x}) = 0.$$

$^\dagger$ ...assuming you even consider the partial derivative operation to be well-defined with respect to a variable that is not part of the function, which is another matter.
