# Is $\frac{\mathrm{d}{x}}{\mathrm{d}{y}} = \frac{1}{\left( \frac{\mathrm{d}{y}}{\mathrm{d}{x}} \right)}$?

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.

• This is imprecise. What you are asking is: what is $(f^{-1})'$ in terms of $f'$?
– wj32
Feb 2, 2013 at 4:39
• There is nothing confusing in $\frac{dy}{dx}.\frac{dx}{dy}=1$. On the opposite, this is just a confirm that $\frac{dx}{dy}=1/\frac{dy}{dx}$.
– user65203
Feb 6, 2015 at 11:06

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 $$\frac{dx}{dy} \Bigg|_{y = b} = \frac{1}{\left( \dfrac{dy}{dx} \Bigg|_{x = a} \right)}.$$

• An important point is evaluating the derivatives at the correct points as shown here. Feb 2, 2013 at 4:48

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, $1 = \frac{dx}{dy} \frac{dy}{dx} \implies \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$. Jun 6, 2021 at 6:00