Can we take a derivative with respect to $y$?

Question

For example, suppose we have the function $f(x)= y$ . Can we take the derivative of the function with respect to $y$? Or maybe we can't because of the definition of derivative?

Answer 1

We can differentiate $y=f(x)$ with respect to y if we also know that $x$ is a function of $y$ in which case we have $$ 1=f'(x) \frac {dx}{dy}$$

For example for $y=\tan (x)$, we have $1=( 1+\tan ^2 x) \frac {dx}{dy}$ which implies $\frac {dx}{dy} =\frac {1}{1+y^2}$

That is $$\frac {d}{dy} \tan ^{-1} (y)= \frac {1}{1+y^2}$$

Answer 2

I like the other answer already given. This will be an additional answer from a different perspective (a more direct perspective as opposed to the function composition perspective)

$f$ is the function, not $f(x)$. $f(x)$ is an output. You defined $f$ as a function $f: x\in X \mapsto y\in Y$ where $X$ and $Y$ are some sets. The derivative of a function is defined with respect to the function's domain variable, which you labeled as $x$. So from our definition of $f$ and the definition of the derivative, $\frac{d}{dy}f$ doesn't make sense. It would only make sense if you were working with another function $g: y\in Y \mapsto g(y)$. There is no such function. You only have what is presently defined, which is $f$.

Also, $\frac{d}{dy} y$ is even worse because $y$ isn't even a function. It's an output - a number - a coordinate. In actuality, this operation does make sense (we actually do have a function of $y$ defined in this problem). First we look at $y = f(x)$. If you want to view $y$ as an output of another function, then $y$ has to be (in order to not change the relationship $y = f(x)$) the output of the identity function $id_Y$ over the set $Y$. That is $id_Y: y \mapsto y$ and therefore our $($output $y) = f(x)$ becomes $id_Y(y) = f(x)$

Now we have $$\frac{d}{dy}id_Y = \frac{d}{dy}f $$

$$ 1 = \frac{d}{dy}f$$ And we are back to where we started in the first paragraph - a derivative which does not make any sense - we have an ill-defined expression on the right hand side. By definition of the standard derivative, we usually have $\frac{d}{dx}f = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$. $f$ is only defined over the set $X$. It isn't defined over the set $Y$, $U$, or any other set (even though these sets may be equal to $X$, subsets of $X$, or contain subsets in common with subsets of $X$. $X$ is the domain). So we can take $1 = \frac{d}{dy}f$ as a definition. The rate of change of a function with respect to variations in it's range is $1$. Or we can defined a new operation: derivatives with respect to variations in the range as: given $f: x\in X \mapsto y\in Y$ where I now use the convention that the codomain $Y$ is the range of the function (this is important), then

$$\frac{d}{dy}f := \lim_{h \to 0} \frac{f(x) + h - f(x)}{h} = \lim_{h\to 0} \frac{h}{h} = 1 $$

And we get that

$$1 = 1 $$