# How does the idea of a differential $\text{d}x$ work if derivatives are not fractions?

I was going through some old work and had a question about differentials.

I know that derivatives $$\frac{\text{d}y(x)}{\text{d}x}$$ are not fractions, but rather an operator $$\frac{\text{d}}{\text{d}{x}}$$ on the function $$y(x)$$. (read this).

But when we have this definition: $$\text{d}y=\frac{\text{d}y}{\text{d}x}\text{d}x$$, is this meant to be something that we have defined simply because it works, or can we express this without "rearranging the fraction".

If for example, we tried to do the same with multivariable functions, this "cancelling" logic does not necessarily work.

I've used this substitution many times, but never fully understood what the differential on its own was representing, so any explanation about that would be appreciated.

• If I were to try to explain, I would just end up regurgitating teh first few chapters of this text. Much of the conceptual confusion you encounter in calculus is because you aren't taught about vectors and co-vectors first. Jul 19, 2021 at 13:07
• I think dy/dx is a fraction. It also works just fine with multivariable equations. The difficulty is higher-order differentials, but see my paper "Extending the Algebraic Manipulability of Differentials" to see how to make those work as well. Jul 19, 2021 at 13:58

One way of viewing them is as one-forms. See Help understanding expression for Derivative of a function for the multivariable situation. The idea is that (in single variable) given a differentiable function $$f:\Bbb{R}\to\Bbb{R}$$, and a point $$a\in\Bbb{R}$$, the quantity $$f'(a)\in\Bbb{R}$$ is what we geometrically think of as the slope at the point $$(a,f(a))$$ of the graph of $$f$$ (in fact logically speaking, this ought to be a definition for the term "slope at a point").

Now, what I'm suggesting to you is rather than thinking of the single number $$f'(a)$$, we consider the linear transformation $$L_{f,a}:\Bbb{R}\to\Bbb{R}$$ defined as \begin{align} L_{f,a}(h):=f'(a)\cdot h \end{align} What is the significance of this linear transformation? By rewriting the definition of the derivative $$f'(a)=\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}$$, we get \begin{align} f(a+h)-f(a)&=L_{f,a}(h)+R_a(h) \end{align} where $$R_a(h)$$ is the "remainder" term which is "small" in the sense that $$\lim\limits_{h\to 0}\frac{R_a(h)}{h}=0$$. Now, traditional notation demands that the LHS be denoted as $$\Delta f_a(h)$$ and $$L_{f,a}$$ be denoted as $$Df_a(h)$$ or $$df_a(h)$$. Therefore, we get the very memorable equation \begin{align} \Delta f_a(h)&=df_a(h)+R_a(h) \end{align} "actual change in function at a point equals a linear term plus a small error term". Note that this is nothing but a simple algebraic rewriting of the definition of a derivative, but it is very powerful because the same idea can be used in higher dimensions: we're shifting our primary perspective from slopes to linear approximation simply because linear algebra is a very well-studied and powerful tool for systematically organizing all this information (in one dimension, linear algebra is almost trivial which is why we don't emphasize this perspective).

So, now for a differentiable function $$f:\Bbb{R}\to\Bbb{R}$$, rather than considering the derivative $$f':\Bbb{R}\to\Bbb{R}$$, we instead consider the object $$df$$, which for every point $$a\in\Bbb{R}$$ gives a linear transformation $$df_a:\Bbb{R}\to\Bbb{R}$$, where the interpretation is that for a "displacement vector from the point $$a$$" $$h\in\Bbb{R}$$, $$df_a(h)$$ is the linear approximation of the actual error $$\Delta f_a(h)$$.

That's all there is to the definition of $$df$$; it's just a simple re-interpretation of the function $$f'$$. Next, what does $$dx$$ mean? Well, now we understand that $$d$$ acts on differentiable functions, so what function is $$x$$? Well, it is tradition to use $$x:\Bbb{R}\to\Bbb{R}$$ to mean the identity function, i.e for any point $$a\in\Bbb{R}$$, we set $$x(a):=\text{id}_{\Bbb{R}}(a):=a$$. Now, it is easily verified that $$dx_a=\text{id}_{\Bbb{R}}$$ (all this is saying is that $$x'(a)=1$$ for all $$a$$). Therefore, \begin{align} df_a(h)&=f'(a)\cdot h=f'(a)\cdot dx_a(h)= (f'dx)_a(h) \end{align} This is why if we don't write the displacement $$h$$ anywhere, nor the point of evaluation of derivative $$a$$, we end up with $$df=f'\,dx$$, where now both sides have a proper definition.

Note:

Throughout this answer, since we're only dealing with functions defined on vector spaces such as $$\Bbb{R}$$ (or in my other linked answer, $$\Bbb{R}^n$$), I have avoided a careful distinction between the vector space and its tangent space at a point. But hopefully, with this introduction, future encounters with $$df_a$$ being defined at the tangent space at $$a$$ wouldn't seem so random.

• And to add one more thing: Once you arrive at $df = f' dx$ it is quite natural to allow rewriting this as $f' = \frac{df}{dx}$ (though the fraction-free form is surprisingly useful to have around). Jul 19, 2021 at 13:40
• @EikeSchulte perhaps it may be natural but I am of the opinion that just because one can doesn't mean one should. As is standard in linear algebra, defining quotients is always a tricky business and it's better to avoid it if possible. Here I would prefer to avoid such a notational rewriting because interpreting it properly would require us to reinstate all the arguments: $f'(a)=\frac{df_a(1)}{dx_a(1)}$, and suddenly all the simplicity is lost. And going into several variables, there's no straight-forward way in which $\frac{\partial f}{\partial x^i}$ is a quotient of $df$ and $dx^i$. Jul 19, 2021 at 13:47
• (of course we can say $\frac{\partial f}{\partial x^i}(a)=\frac{df_a(e_i)}{dx^i_a(e_i)}$, where $e_i=(0,\cdots, 1,\cdots 0)$ is the $i^{th}$ standard basis vector... but again writing out all of this in gory detail just to provide a proper interpretation as a quotient simply to preserve some "fraction-ness" seems too far of a stretch for me) Jul 19, 2021 at 13:49
• I kind of agree with that but mentioning the fraction connects to the original question. Maybe it’s best if I just leave my comment up, so the connection is there without turning into a recommendation. (And true: fractions for partial derivatives are (basically) nonsense.) Jul 19, 2021 at 13:49

Well we don't actually treat $$\frac{dy}{dx}$$ as fractions. Every time you see it being manipulated as a fraction just remind yourself that whatever you're doing is actually a proved theorem that you don't yet know. Leibniz notation is used when defining derivatives and that notation is beautifully laid out in such a way that they can be treated like fractions for most cases.

For example, you know $$\frac{dy/dt}{dx/dt} = \frac{dy}{dt}$$ Here you didn't just cancel out $$dt$$. If you, try this as an exercise, just use the limits definition of a derivative and solve it out, you'll see that this is indeed true in the way derivatives are defined and they need not be fractions for this.

Here's the limit definition of derivatives. $$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{y(x + \Delta x) - y(x)}{\Delta x}$$

Now if the two variables $$x$$ and $$y$$ are defined parametrically and you want to find the rate at which the function $$y$$ is changing with respect to function $$x$$, we have to be aware of the fact that both of those functions are dependent on a variable $$t$$. Now if you want to find the rate of change, you have to find how those two functions change with $$t$$. That's why the notation is changed to $$\frac{dy/dt}{dx/dt}$$. Now the limit definition of it becomes,

$$\frac{dy/dt}{dx/dt} = \frac{\lim_{\Delta t\to 0} (\frac{y(t + \Delta t) - y(t)}{\Delta t})}{\lim_{\Delta t \to 0} (\frac{x(t + \Delta t) - x(t)}{\Delta t})}$$

You'd see that that can be written as,

$$\frac{dy/dt}{dx/dt} = \lim_{\Delta t \to 0} \frac{(\frac{y(t + \Delta t) - y(t)}{\Delta t})}{(\frac{x(t + \Delta t) - x(t)}{\Delta t})}$$

Now the $$\Delta t$$ terms are cancels in the respective denominators and you're left with,

$$\frac{dy/dt}{dx/dt} = \lim_{\Delta t \to 0} \frac{y(t + \Delta t) - y(t)}{x(t + \Delta t) - x(t)}$$

Now you treat $$dy$$ as $$y(t + \Delta t) - y(t)$$ and $$dx$$ as $$x(t + \Delta t) - x(t)$$ as small change in $$y$$ and $$x$$ respectively. That's why the notation $$dy/dx$$. Also as $$\Delta t \to 0$$ is defined as $$dt$$. This is completely valid. Try to prove it for other theorems but be careful that for $$dx/dy = \frac{1}{dy/dx}$$ to be true the function should be invertible, too.

Having an intuition that they work like fractions helps but you need to be aware that $$\frac{d}{dx}$$ is an operator.