Calculus teachers tend to say two things at the same time: “$\frac{dy}{dx}$ is not a fraction” and “you can cancel the $du$ as in a fraction”. Both are right, and that is precisely the strength and the danger of Leibniz’s notation. It looks like a fraction and, in most calculations, it behaves like one. Knowing where the resemblance holds and where it breaks is one of the subtlest skills in reading mathematics.

Where it works

The chain rule. If $y$ depends on $u$ and $u$ depends on $x$:

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}.$$

The $du$s “cancel”, as in a fraction. For $y=(x^2+1)^{10}$, with $u=x^2+1$: $\frac{dy}{du}=10u^9$ and $\frac{du}{dx}=2x$, so $\frac{dy}{dx}=20x(x^2+1)^9$.

The inverse function. If $y=e^x$, then $x=\ln y$, and

$$\frac{dx}{dy}=\frac{1}{dy/dx}=\frac1{e^x}=\frac1y.$$

Turning the fraction upside down gives the derivative of $\ln y$. A rule that in other notations needs a theorem comes out here by flipping the paper.

Separation of variables. For $\frac{dy}{dx}=ky$, the $dx$ is “moved to the other side”:

$$\frac{dy}{y}=k\,dx\quad\Longrightarrow\quad\ln y=kx+C\quad\Longrightarrow\quad y=Ce^{kx}.$$

A rigorous mathematician frowns at “moving it to the other side”, but the result is right, and there is a theorem guaranteeing it will always be right in this kind of calculation.

Where it breaks

The second derivative. The notation $\frac{d^2y}{dx^2}$ seems to say “$d^2y$ divided by $dx^2$”, and it would be tempting to conclude that $\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2$. For $y=x^2$: $\frac{d^2y}{dx^2}=2$, but $\left(\frac{dy}{dx}\right)^2=4x^2$. They have nothing to do with each other. Worse: the chain rule for the second derivative is not the product of the second derivatives:

$$\frac{d^2y}{dx^2}=\frac{d^2y}{du^2}\left(\frac{du}{dx}\right)^2+\frac{dy}{du}\cdot\frac{d^2u}{dx^2},$$

with an extra term that no “cancellation” announces.

Partial derivatives. For an ideal gas, $pV=nRT$. Each of the three quantities can be written as a function of the other two, and the partial derivatives are

$$\frac{\partial p}{\partial V}=-\frac{p}{V},\qquad\frac{\partial V}{\partial T}=\frac{nR}{p},\qquad\frac{\partial T}{\partial p}=\frac{V}{nR}.$$

If they were fractions, the product would cancel and give $1$. But

$$\frac{\partial p}{\partial V}\cdot\frac{\partial V}{\partial T}\cdot\frac{\partial T}{\partial p}=-\frac pV\cdot\frac{nR}p\cdot\frac{V}{nR}=-1.$$

The minus sign comes from something the notation hides: each $\partial$ keeps a different variable fixed, and that variable is not written. The $\partial p$ of the first factor is not the same object as the $\partial p$ of the third. Physicists know this and, in thermodynamics, write the fixed variable as a subscript: $\left(\frac{\partial p}{\partial V}\right)_T$.

So, is it or isn’t it?

For Leibniz, it was: $dy$ and $dx$ were infinitely small quantities, and $\frac{dy}{dx}$ was their ratio. In the nineteenth century, the rigour of Cauchy and Weierstrass replaced infinitesimals with limits, and $\frac{dy}{dx}$ became a single, indivisible sign. In the 1960s, the logician Abraham Robinson showed that it is possible, with care, to build genuinely infinitely small numbers, and in this non-standard analysis the derivative becomes a ratio again, up to an adjustment.

The short answer is: $\frac{dy}{dx}$ behaves like a fraction when there is one independent variable and one order of differentiation. Beyond that, you have to check.

Exercise: the inverse of sine

Use $\frac{dx}{dy}=\frac{1}{dy/dx}$ to differentiate $x=\arcsin y$, with $-1

If $y=\sin x$, then $\frac{dy}{dx}=\cos x=\sqrt{1-\sin^2x}=\sqrt{1-y^2}$ (with $\cos x>0$ on the right interval). So $\frac{dx}{dy}=\frac{1}{\sqrt{1-y^2}}$, which is the derivative of arcsine.

Exercise: the second derivative by the chain rule

Calculate $\frac{d^2y}{dx^2}$ for $y=u^3$ with $u=2x$, using the two-term formula, and check by differentiating $y=8x^3$ directly.

$\frac{d^2y}{du^2}=6u$, $\frac{du}{dx}=2$, $\frac{dy}{du}=3u^2$, $\frac{d^2u}{dx^2}=0$. The formula gives $6u\cdot4+0=24u=48x$. Directly: $y=8x^3$, $y'=24x^2$, $y''=48x$. It checks. Here the second term is zero because $u$ is linear; with $u=x^2$, it would not be.

Chapter index · ← From Δ to d: the difference that shrinks · The D that factorises: the derivative as an operator →