Why aren't $z_1=f(xy)$ and $z_2=f(x/y)$ functions of 2 variables?
Hagen von Eitzen answered that $z_1, z_2$
depend on only one variable - there's no comma between the parentheses.
the function $f(xy)=e^{xy}sin(xy)+(xy)^3$ may look like a multivariable function in x and y, but it can be written more simply as a univariate function, $f(t)=e^tsint+t^3$.
Yes, I agree that defining $t_1=xy$ transforms $f(xy)$ into $f(t_1)$, and $t_2= \dfrac xy$ transforms $f(\dfrac xy)$ into $f(t_2)$. Yes, $f(t_1)$ and $f(t_2)$ are single-variable. I agree that $z_1, z_2$ can be re-defined to be functions of ONE variable.
But I'm still befuddled why $z_1, z_2$ can't be construed as TWO variables. I feel that redefining with $t$ misrepresents the original 2 independent variables x, y. These definitions don't change the original fact that f did depend on 2 independent variables!
1 answer
You are misrepresenting what Hagen von Eitzen wrote. He did not write that $z_1$ and $z_2$ depend on only one variable. He wrote that $f$ is a function of only one variable. That's a massive difference.
The question this refers to was whether $z_1=f'(xy)$ or $z_2=f'(x/y)$ are ambiguous, and the correct answer was that it is not because $f$ is a function of only one variable. Note that $f'(xy)$ is not the same as $(f(xy))'$, which would be ambiguous assuming $x$ and $y$ are both independent variables, unless we previously established a convention that e.g. primes always refer to a differentiation on $x$.
The notation $f'(xy)$ means that we take the function $f$, which takes one argument (let's name it $t$), take the derivative of that function with respect of its argument (no ambiguity here, there's only one argument after all), and then insert $xy$ as argument into the derivative we just calculated.
Note that the difference is important even in cases where only one variable is involved. For example, consider $z_3=f'(3x)$. This is not the derivative of $f(3x)$, the latter would be written as $z_4=(f(3x))'$ and by the chain rule would evaluate to $z_4=3f'(3x)$. Clearly $3f'(3x)\ne f'(3x)$ unless $f'(t)=0$ for all $t$.
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