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In calculus, is $ \dfrac{\mathrm{d}{x}}{\mathrm{d}{y}} = \dfrac{1}{\left( \dfrac{\mathrm{d}{y}}{\mathrm{d}{x}} \right)} $? I’m so confused about this matter. What would be a proof of it?

Edit: By the Chain Rule, $ \dfrac{\mathrm{d}{y}}{\mathrm{d}{x}} \cdot \dfrac{\mathrm{d}{x}}{\mathrm{d}{y}} = \dfrac{\mathrm{d}{y}}{\mathrm{d}{y}} = 1 $, so this confuses me.

Berrick Caleb Fillmore
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Lanz
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    This is imprecise. What you are asking is: what is $(f^{-1})'$ in terms of $f'$? – wj32 Feb 02 '13 at 04:39
  • There is nothing confusing in $\frac{dy}{dx}.\frac{dx}{dy}=1$. On the opposite, this is just a confirm that $\frac{dx}{dy}=1/\frac{dy}{dx}$. –  Feb 06 '15 at 11:06

2 Answers2

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The precise statement is as follows.

Let $ I $ be an open interval, and suppose that $ f: I \to \mathbb{R} $ is one-to-one and continuous on $ I $. If $ f $ is differentiable at $ a \in I $ and $ f'(a) \neq 0 $, then $ f^{-1}: f[I] \to I $ is differentiable at $ b = f(a) $ and $$ (f^{-1})'(b) = \frac{1}{f'(a)}. $$

Using Leibniz’s notation, the formula above can be expressed as $$ \frac{dx}{dy} \Bigg|_{y = b} = \frac{1}{\left( \dfrac{dy}{dx} \Bigg|_{x = a} \right)}. $$

Haskell Curry
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If the real variables $x$ and $y$ are dependent functionally on each other, say by the equations $y = f(x)$ and $x = g(y)$, then where things are differentiable, we do indeed have an equation

$$ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $$

and the most straightforward proof is as you indicated by the chain rule:

$$ \frac{dy}{dx} \frac{dx}{dy} = \frac{dy}{dy} = 1$$

and then solving the equation

$$ \frac{dy}{dx} \frac{dx}{dy} = 1$$

for $\frac{dy}{dx}$.

The argument can be rephrased in terms of functions rather than dependent variables: from the identity

$$ x = g(f(x)) $$

we differentiate to obtain

$$ 1 = g'(f(x)) f'(x) $$

and thus the analogous identity

$$ f'(x) = \frac{1}{g'(f(x))} $$

More generally, in the language of differential forms, the expression $\frac{dy}{dx}$ is defined to be the expression satisfying

$$ dy = \frac{dy}{dx} dx $$

if that is possible. When both $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are defined, then we have two equations

$$ dy = \frac{dy}{dx} dx \qquad \qquad dx = \frac{dx}{dy} dy $$

and can substitute to obtain

$$ dy = \frac{dy}{dx} \frac{dx}{dy} dy $$

and so we again have

$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $$

wherever $dy$ is nonvanishing. An example of a situation where both are defined is if the variables $x$ and $y$ are functionally related by a differentiable equation $f(x,y) = 0$. Taking the differential gives

$$ f_1(x,y) dx + f_2(x,y) dy = 0$$

where $f_1$ is the derivative of the two-variable function $f$ with respect to the first variable. When both $f_1$ and $f_2$ are nonzero, we can solve the equation in both ways to see that both $\frac{dx}{dy}$ and $\frac{dy}{dx}$ are defined.

  • Here is the chain rule explicitly applied to this problem. By definition, $x = f^{-1} ( f(x))$ where the inverse function exists. Differentiating both sides with respect to $x$, $1 = \frac{d f^{-1} (x)}{d f(x)} \frac{d f(x)}{dx}$ and given that $y = f(x), x = f^{-1} (y)$ where $x$ is a dummy variable, $1 = \frac{dx}{dy} \frac{dy}{dx} \implies \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$. – Toby Mak Jun 06 '21 at 06:00