In higher-order derivatives, such as $\frac{d^2y}{dx^2}$, why is the exponent on $d$ in the numerator and $x$ in the denominator?

Question

Why are the exponents in higher order derivatives the following: $$ \frac{d^2y}{dx^2} $$

Why is the exponent on the $d$ in the numerator and $x$ in the denominator?

Answer 1

$\frac {d}{dx}$ is the differnential operator. $\frac {d}{dx} y = \frac {dy}{dx}$ is the differential operator applied to $y$ giving the derivative of y.

$\frac {d}{dx} \frac {dy}{dx} = \frac {d^2y}{dx^2}$ is the differential operator applied to the derivative of y, yielding the second derivative of $y.$

Answer 2

You are right, it's not consistent. In one case you are interpreting a $dx$-like symbol - in particular, $dx$, as though it's a single "quantity" to be squared, while in the other, you are interpreting the $dx$-like symbol $dy$ as though it were composed of two parts, "$d$" and $y$, and "squaring" the first, but not the second.

The operator explanation is right - it should really be written $\left[\frac{d}{dx}\right]^2 y$, where the power is on the operator, and signifies repeated composition, just as when you have a function $f(x)$, that a power on the function symbol, $f^n(x)$ should always mean $n$-fold composition of $f$ with itself, but in many circumstances ends up ambiguous, such as $\sin^2(x) = [\sin(x)]^2$ and not $\sin(\sin(x))$ as it logically should - which is in turn inconsistent with that $\sin^{-1}(x)$ typically is seen as a compositional power and not $\frac{1}{\sin(x)}$. And then we "simplify" this "fraction" to $\frac{d^2}{dx^2}$ by "distributing the power".

However, there is another way you can kind of make sense of this and that is that if you want to imagine the two as "quantities", the $d^2 y$ is a different "kind" of quantity than $dx^2$. In particular, look at the limit definition:

$$\frac{d^2 y}{dx^2} = \lim_{\Delta x \rightarrow 0} \frac{y(x + 2 \Delta x) - 2 y(x + \Delta x) + y(x)}{[\Delta x]^2}$$

versus

$$\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{y(x + \Delta x) - y(x)}{\Delta x}$$

In particular, the denominator is a simple power of $\Delta x$, but the numerator becomes a rather more complex difference expression in the second case versus the first case. Thus the power of $\Delta x$ goes nicely to the power of $dx$, but the numerator seems like it should get a different treatment. You can actually think of the numerator as two applications of the difference operator

$$\Delta_h[f] = f(x + h) - f(x)$$

so that

$$\frac{d^2 y}{dx^2} = \lim_{\Delta x \rightarrow 0} \frac{\Delta_{\Delta x}^2[y]}{[\Delta x]^2}$$

Now doesn't the right-hand side look more suggestively like the left?

There you go.