# Multivariable function definition

1. Nov 13, 2014

### Nikitin

Hi. I am a bit confused on the definition of multivariable functions. Say you have $f(x) = x^2 + x$ and $g(x,a) = x^2 + a$ where $a=x$. Is $g(x,a)$ then a mathematically legal multivariable function? Because if you take $\frac{\partial f(x)}{\partial x}=2x +1$ you'll get a different result from $\frac{\partial g(x,a)}{\partial x} = 2x$ even though $g(x,a)$ and $f(x)$ are the exact same functions.

Or am I missing something?

2. Nov 13, 2014

### ShayanJ

$\frac{\partial g(x,a)}{\partial x}=\frac{\partial g(x,a)}{\partial x}+\frac{\partial g(x,a)}{\partial a}\frac{da}{dx}=2x+1$

3. Nov 13, 2014

### Nikitin

Isn't that the total derivative? I thought the partial derivative was supposed to keep all other variables constant (in this case, a).

4. Nov 13, 2014

### ShayanJ

We use partial derivative only when the function being differentiated is a function of several variables. But a depends only on x.
I should correct myself though!
$\frac{d g(x,a)}{dx}=\frac{\partial g(x,a)}{\partial x}+\frac{\partial g(x,a)}{\partial a}\frac{da}{dx}=2x+1$
The $\frac{d}{dx}$ notation, if used on a function of several variables, means the change of the function caused by the change of x through any kind of x dependence of the function, even the indirect ones. But $\frac{\partial}{\partial x}$ is only for explicit dependence.
So in contrast to the above, we have:
$\frac{\partial g(x,a)}{\partial x}=2x$
For functions of one variable, the two derivatives are equal.

Last edited: Nov 13, 2014
5. Nov 13, 2014

### Nikitin

OK, I see. so g(x,a) is NOT a function of several variables, right?

But what if you have a function $h(x,y,z)$ such that $x=x(t),y=y(t),z=z(t)$. In that case, $h(x,y,z)=h(t)$ is NOT a multivariable function without a partial derivative with regards to x, y or z?

6. Nov 13, 2014

### PeroK

You need to be more careful about what you are doing. First, let's define:

$g(x, y) = x^2 + y$

That's a multi-variable function with $\frac{\partial{g}}{\partial{x}} = 2x$ and $\frac{\partial{g}}{\partial{y}} = 1$

Now, if you evaluate the first partial derivative along the line $y=x$ you get:

$\frac{\partial{g}}{\partial{x}}(x, x) = 2x$

If, however, you define $g(x, y) = x^2 + x$ then $\frac{\partial{g}}{\partial{x}} = 2x + 1$ and $\frac{\partial{g}}{\partial{y}} = 0$

But, this is now a different muti-variable function, with different partial derivatives.

7. Nov 13, 2014

### Stephen Tashi

Using symbols is an attempt to abbreviate thought, but it isn't systematic enough to replace thinking. it is a cultural tradition in writing mathematics to "abuse notation". It's common to see the same symbols used to represent distinct things. It also common to see symbols used in a way that is meaningless when they are interpreted literally.

The general form of what you have asked about is this scenario: Let $g(x,a)$ be a real valued function of two real variables. Let $r(w)$ be a real valued function of one real varable. Define a real valued function of one real variable by $s(y) = g(y,r(y))$. Using that sort of symbolism, there is no ambiguity about the number of variables that functions $g,$ and $s$ have.

In your question, you have followed the above scenario (using $r(w) = w$ ). The ambiguity in the way that your are using symbols leads to confusion. You don't have symbolism that clearly identifies the difference between $g(x,a)$ and $s(y)$. The way you use symbols in your question follows cultural traditions, so I'm not saying you have to change your style. However, you do have to keep in mind that writing in that style obscures some technicalities.