My question is how should the total derivative (and functions in general) be interpreted in the multivariate case. I came up with two different ways and wish to know which is the agreed upon way.

Let's say we have the following function of f(x(y), y):

$$f(x(y), y)=x^2 + yx + y^2$$

x is a function of y following the relation:


Now, the there is no ambiguity for total derivative of y. It's simply the derivative of $(2y)^2+y(2y)+y^2$ with respect to y (alternatively the formula for total derivative can be used yielding the same result). The derivative equals $8y+4y+2y = 14y$

The total derivative of x on the other hand depends what we define the function sign to mean. If we define that x does not affect y, so that y is an independent variable (giving a not strict relationship for the above equation), then the total derivative reduces only to the direct effect x has on the function and is simply the partial derivative, giving:

$$\large \frac{\partial f(x,y)}{\partial x} = 2x+y$$

However, this is might be the wrong interpretation. Since x is a function of y, then by y must also be a function of x, if the 2nd equation presented is interpreted as strict relationship (rather than as dependent/independent variable). Thus we can solve for $y = \frac{x}{2}$ and then the total derivative with respect to x can be obtained from $$4\large x^2+x(0.5x) + (0.5x)^2 \rightarrow \large 2x+x+0.5x= 3.5x$$

Which of these is the correct interpretation? Both of the relationships of course can hold, so the question becomes whether $f(x(y), (y))$ is even a possible function or should we write it as $f(x(y), y(x))$. Then we should use completely different notation involving sub variables for the case where x depends on y but the relationship does not swing the other way around giving for example $f(x(t,z), y(t))$.

  • $\begingroup$ You have an implicit function $x^2+xy+y^2=0$ or a function of two variables $z=x^2+xy+y^2$ ? $\endgroup$ – Emilio Novati Dec 14 '15 at 20:42
  • $\begingroup$ @EmilioNovati Function of variables (edited to make it clearer). $\endgroup$ – Dole Dec 14 '15 at 20:45

It depends on whether $y$ is dependent on $x$ or not. It actually makes a big difference: if the growth of the function along the $x$-axis was independent from $y$ we would "lose" the increment given by the $y^2$ term. What does this mean?

If we consider $y$ as dependent on $x$, say $y(x)$, the partial derivative computation would lead to $$\displaystyle \frac {\partial} {\partial x} f(x,y(x))=\frac {\partial} {\partial x}(x^2+y(x)x+y(x)^2)=2x+y'(x)x+y(x)+2y'(x)y(x)$$

If $y(x)=\frac{x}{2}$, $y'(x)=\frac 12$, so

$$\displaystyle \frac {\partial} {\partial x}f(x,y(x))=2x+\frac {x}{2}+\frac{x}{2}+\frac x2 = \frac {7}{2}x$$

You are actually composing the two maps $y(x)=\frac x2$, $f(x,y)=x^2+yx+y^2$ into a third map $$\begin{align} y\circ f~:\mathbb{R}^2~~~ \overset{\mbox{f}}{\longrightarrow}~~~ \mathbb{R}~~~\overset{\mbox{y}}{\longrightarrow}~~~ \mathbb{R}~~~ \end{align} $$ $$\begin{align}~~~~~~~~~~~~~~~ (x,y)\longrightarrow f(x,y)\longrightarrow f(x,y(x)) \end{align}$$

You are stating that $\frac{\partial d}{\partial dx}(y \circ f)=\frac 32 x$: this is true indeed.

Instead, if $y$ is independent from $x$, as you have noticed that the partial derivative with respect to $x$ is

$$\displaystyle \frac {\partial} {\partial x} f(x,y)=\frac {\partial} {\partial x}(x^2+xy+y^2)=2x+y\overset{\mbox{{$\tiny{y=\frac x2}$}}}=\frac{5}{2}x$$

Substituting $y=\frac x2$ we obtain the law of the derivative of the function $f(x,y)$ at the section $y(x)=\frac x2$: we are composing the two functions and considering $y\circ f'$. This function is different from the first one:

$$\begin{align} y\circ f'~:\mathbb{R}^2~~~ \overset{\mbox{f}}{\longrightarrow}~~~ \mathbb{R}~~~\overset{\mbox{y}}{\longrightarrow}~~~ \mathbb{R}~~~ \end{align} $$ $$\begin{align}~~~~~~~~~~~~~~~ (x,y)\longrightarrow f'(x,y)\longrightarrow f'(x,y(x)) \end{align}$$

Indeed, the chain rule states that if $f$, $y$ are differentiable, then

$$(y\circ f)'(x)=y'(f(x))(f'(x))\neq y(f'(x))$$

in general: the two functions you are analyzing thus can be different and because of this, they can have different derivatives.

  • $\begingroup$ So does this mean that the notation $y(x)$ does not actually indicate that the function $(y)$ is only dependent on $x$. Because if that were the case, then as far as I can see $x$ can't be independent of $y$, making the independent scenario impossible using the notation. In other words $y(x)$ taking all the variables into account would have to mean $y(x, t)$, where $t$ are the others factors besides $x$ influencing $y$. $\endgroup$ – Dole Dec 14 '15 at 23:30

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