Can Leibniz Notation Be Treated As a Quotient?

Question

Why is saying $\frac{dy}{dx}\frac{dy}{dx}=y\frac{d^2y}{(dx)^2 }$ not valid? Does Leibniz notation (and thinking of it as an infinitesimal quotient) not work for higher-order derivatives?

Answer 1

Consider $y=3x$ then $\frac{dy}{dx} =3$. So the left hand side of your equation would be 9. However, $\frac{d^{2}y}{dx^{2}} =0$, so the right hand side of your equation would be 0 for this example, but $0\neq 9$. Thus, the function $y=3x$ provides a counterexample.

Leibniz notation does work for higher derivatives, the nth derivative of y is denoted by $\frac{d^{n}y}{dx^{n}}$. However, treating the derivative as a quotient is called abuse of notation and is considered more as a convenient tool to help students remember certain rules like the chain rule. There is actually a way of treating infinitesimals rigorously is treated in what is known as non-standard analysis, but this isn't really the most common approach.

Answer 2

Your intuition is correct that the derivative OPERATOR can be multiplied together in such a manner i.e. $\frac{d}{dx}\frac{d}{dx} = \frac{d^2}{dx^2}$. However, the expression $\frac{dy}{dx} = \frac{d}{dx}(y)$ is an evaluated derivative on the function $y$ we write the $y$ in the numerator with $d$ as a notational convenience. When you evaluate a derivative, it no longer behaves like the derivative operator and is now simply a function. Another way to look at it is if you consider $\frac{d}{dx}$ as a "function" then the product $\frac{d}{dx}\frac{d}{dx}$ is simply the composition of two derivative functions, conversely $\frac{dy}{dx} \frac{dy}{dx}$ is the product of two evaluated functions. So we can restructure your question of follows : does $f(x)*f(x) = f(f(x))$ ? the answer is obviously no. However, $(f\circ f)(x) = f(f(x))$ is true by the definition of composition of functions.

Answer 3

Since $$f''(a)=\lim_{h\to0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$ under suitable conditions, I suppose one could conceptualize the $d^2 f$ in $d^2 f/dx^2$ as a "second-order" infinitesimal difference: i.e., as $f(x+2dx)-2f(x+dx)+f(x)$. Mind you, I've never liked thinking in terms of "infinitesimals" :-)