What's wrong with substituting $g(y) = 3$ into $f(z) = z^{5x + 1}$ to calculate $\frac{d\left(3^{(5x+1)}\right)}{dx}$?

Question

I don't understand this answer, but I've rewritten it to commence with $g(y) = 3$, to avoid ambiguity.

The root of the difficulty is that $x$ appears free in $f(z)$, but we are trying to "capture" it with $g(y)$, which is illegal. When we substitute $g(y)$ into $f(g(y))$, we have a variable clash:

$$ f(g(\color{red} y)) = 3^{5\color{blue}x + 1} $$

The red $y$ is a different variable from the blue $x$.

The original expression [I emboldened] had $x$ bound to the $\mathrm d x$, so by unbinding $\color{lightgreen}{it}$, we have changed the meaning of the expression:

$$ \frac{\mathrm d}{\mathrm d \color{blue} x} f(g(\color{red}y)) \ne \frac{\mathrm d}{\mathrm d \color{red}y} f(g(\color{red}y)) $$

1. What exactly does it mean to "capture" $x$ with $g(y)$?

2. Why's this illegal?

3. What's the original expression?

4. How was "$x$ bound to the $\mathrm d x$"?

5. What is "$\color{lightgreen}{it}$"?

6. How did we unbind "$\color{lightgreen}{it}$"?

Answer 1

First of all, look at the definition of $f(z) = z^{5x+1}$. So, $f$ is not only a function of $z$, but it is a function of two variables: $z$ and $x$, and they're not related to each other. By definition, $x$ can take any value which does not depend on $z$. So, the derivative of $f$ can be defined with respect to both $z$ and $x$, keeping the other variable constant. See partial derivatives for more details. For the same reason, $(5x+1)z^{5x}$ is the derivative of $f$ with respect to $z$, not with respect to $x$. Thus, \begin{equation} \frac{d}{dx}f(z,x)\neq (5x+1)z^{5x} = \frac{d}{dz}f(z,x) \end{equation} Writing $\frac{d}{dx}f(x) = (5x+1)z^{5x}$ is referred to as illegal here.