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My question is regarding step 3. Why is the partial of z with respect to the partial of x equal to df/du rather than the partial of f with respect to the partial of u. I dont think im understand the basics right. What is the difference in this case, and what exactly are we looking for here. I need intuition for whats going on.

(The question is to prove that last statement you see in the image after the word "thus" is true.)

enter image description here

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I believe they use that d because after the substitution, f is in terms of only one variable u, but they use the partial for u because u is in terms of two variables x and y. – yunone Mar 29 '11 at 3:13
up vote 5 down vote accepted

We have two functions here: $$\begin{align*} &u\colon \mathbb{R}^2\to\mathbb{R}\\ &f\colon\mathbb{R}\to\mathbb{R}. \end{align*}$$ The function $u$ is given by $u(x,y) = x-y$, and $z=f\circ u$. So $z$ is a real-valued function of two variables.

But note that $f$ is a real function of real variable. So $f$ does not have partial derivatives, it has regular derivatives, because $f$ is a function of a single variable. The fact that we are evaluating it at a function of two variables does not change the fact that $f$ is a function of a single variable.

Forget for a second about all of these functions. Say you have a function $g\colon\mathbb{R}\to\mathbb{R}$; to fix ideas, say $g(x) = 3x^2-2$. What if $a$ is a constant, and we want to find the derivative of $g(x-a)$ with respect to $x$? We can just use the Chain Rule! Writing $v(x) = x-a$, we have $$\frac{d}{dx}g(x-a) = \frac{d}{dx}g(v(x)) = \frac{dg}{dv}\frac{dv}{dx} = g'(v(x)))v'(x) = g'(x-a),$$ and since $g'(x) = 6x$, then $\frac{d}{dx}g(v(x)) = g'(x-a) = 6(x-a)$. Here, $v$ is also a function of a single variable, so derivatives are appropriate.

Now suppose we consider instead $z=g\circ u\colon\mathbb{R}^2\to\mathbb{R}$ (same $u$ as above, $u=x-y$). If we want to find the partial derivative of $z$ with respect to $x$, we do this by fixing $y$; so it's just like trying to take the derivative of $g(x-a)$, except that instead of calling the constant $a$ we call it $y$. So we aren't taking partial derivatives of $g$, we are taking regular derivatives. So $$\frac{\partial z}{\partial x} = \frac{d}{dx}g(u(x,y)) = \frac{dg}{du}\frac{\partial u}{\partial x}.$$ We use partials for $u$ because $u$ is a function of two variables, but derivatives for $g$ because $g$ is not: it's a function of a single variable.

Back to your example: both $z$ and $u$ are functions of two variables, so we can talk about partials of both $z$ and $u$. But $f$ is a function of a single variable (into which we are plugging the output of $u$), so derivatives are appropriate, not partials.

What is the partial derivative of $z$ with respect to $x$? We consider the function as a function in which $y$ is fixed, and only $x$ changes.

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When you say that $f$ is a function of 1 variable, are you referring to $u$ as that variable? – maq Mar 29 '11 at 3:47
@mohabitar: Yes. $u\colon\mathbb{R}^2\to\mathbb{R}$, and $f\colon\mathbb{R}\to\mathbb{R}$. You are computing $z=f\circ u$, so you are evaluating $f$ at $u(x,y)$. But $f$ is still a function from $\mathbb{R}$ to $\mathbb{R}$, so one input, one output. – Arturo Magidin Mar 29 '11 at 3:53

partial derivative $\left(\frac{\partial u}{\partial x}\right)$ which says that you have 2 unknown parameters in $u$ i.e., $x$ and $y$ are unknown. But, when u write as $\left(\frac{du}{dx}\right)$, it says that you have only one unknown parameter i.e., $x$ and others are kept constant i.e., other constants are known i.e., $y$.

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