Manipulating derivatives

Question

I am starting to learn some multivariable calculus and had a number of questions on some of the simplifications of differential expressions as I have not found a good resource that guides through.

1) Why is this invalid (I know it gives and incorrect answer): $\frac{d}{dx}\frac{dx}{dy}=\frac{d}{dy} \frac{dx}{dx}=\frac{d}{dy} (1)=0$ but this is valid: $\frac{d}{dx}\frac{dy}{dt}=\frac{d}{dx}\frac{dy}{dt}\frac{dt}{dt} =\frac{d}{dt}\frac{dy}{dt}\frac{dt}{dx}=\frac{d^2y}{dt^2}\frac{dt}{dx}\ $ As in in what makes one situation valid to manipulate these derivatives like fractions and the other not be.

2) How do I or can I simplify an expression like $ \frac{d}{dx}\frac{\partial f}{dx} $ in any obvious way?

Answer 1

This is to clarify my remarks on (2). It is too much to put in a comment.

New students invariably have a hard time picking up on why partial differentiation is considered a different thing than ordinary differentiation, even so much as to have its own unique symbolism. Let me give an example.

Let $z = x + 2$. The derivative is $\dfrac{dz}{dx} = 1$. That is the full answer. There is no other derivative.

But now, let $z = x + y$. To take the derivatives with respect to each variable, we treat the other variable as constant, and just use the normal derivative: $$\dfrac{\partial z}{\partial x} = 1,\quad \dfrac{\partial z}{\partial y} = 1$$

All well and good. But suppose I decide a change of variables will be useful. $x$ stays the same, but let $u = x + y$, so $y = u - x$. Then $z = u$ and $$\dfrac{\partial z}{\partial x} = 0,\quad \dfrac{\partial z}{\partial u} = 1$$

Do you see it? The meaning of $z$ does not change. The meaning of $x$ does not change. But when $y$ was taken as the other independent variable $$\dfrac{\partial z}{\partial x} = 1$$while when $u$ is taken as the other independent variable $$\dfrac{\partial z}{\partial x} = 0$$

The point is, when you see a normal derivative $\dfrac{dz}{dx}$, that is the full story. It depends on the reliance of $z$ on $x$ at the point in question, and nothing else.

But when there is more than one independent variable around, the derivative depends on more than just $z$ and $x$. It also depends on what other independent variables were chosen to go along with $x$. This is because holding those other variables constant means different things. When $y$ is held constant, $u = x + y$ changes with $x$. It is not constant. When $u$ is held constant, $y = u - x$ changes with $x$. These approach the point of differentiation from different directions.

The partial derivative reminds us of this. It is telling us that this derivative depends not just on the explicit variables $z$ and $x$, but also on the other variables that are not explicit in the notation.

This is why you do not mix partial differentiation with ordinary differentiation with respect to the same variable. The partial differentiation indicates that there are other independent variables required for its definition. But that does not change with the other differentiation. Those other independent variables are still out there. So calling the other derivative normal differentiation is inappropriate, ignoring factors on which the derivative depends.

Answer 2

It makes it clearer if you separate out the differentiation steps from the division steps and use parentheses to understand.

Oftentimes, $\frac{d}{dx}$ is used as an operator. That is, there is an implied $\frac{d}{dx}\left( ~~ \right)$. If it is being used in this way, you can't cancel, because it is an operation, not a multiplication. So, let's look at your first one:

$$ \frac{d}{dx} \frac{dx}{dy} $$

What the person who wrote this probably meant was: $$ \frac{d}{dx}\left(\frac{dx}{dy}\right) $$

That is, it is asking to take the derivative of the interior with respect to $x$. Personally, I think it is more clear if we separate out the differential from the full derivative, and write it like this: $$ \frac{d \left(\frac{dx}{dy}\right)}{dx} $$ That is actually what you are doing. Modern calculus notation seems to want to make it actively confusing what is going on, which leads to questions like this.

You actually can cancel differentials when they are used in multiplication. So, for instance, if you had: $$ \frac{dy}{dx}\frac{dx}{dy} $$ The differentials would cross-cancel and reduce to $1$. Again, the notation you are being given is making it unclear when you are multiplying vs. performing an operation. Usually the bare "d" is a clue to know that it is an operation.

Now, if you want to do cancellations with higher-order differentials, you have to do a still further change in notation. The typical $\frac{d^2y}{dx^2}$ does NOT allow for cancelling/reducing fractions. You have to use an expanded notation for the second derivative to get that ability. The second derivative in this expanded notation is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. In that notation, differentials are cancellable.

For more information about where that came from, see the paper "Extending the Algebraic Manipulability of Differentials". However, you may be able to derive it yourself, by recognizing that $\frac{dy}{dx}$ is a quotient, and, therefore, to take its derivative fully, you need to use the quotient rule!