High School Calculus. If $y=x^2$ then why is $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$ My Math teacher told me that  $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$. I asked for a proof and he gave me the result by using the chain rule. I don't understand why this is the case for every function because the function $y=x^2$ doesn't have an inverse function.  If this is the case then why is $\frac{dx}{dy}$ not $\frac{-1}{2x}$? Maybe it only works for injective functions or we assume only one of the values to be true.
 A: There is such a thing as a function being locally one-to-one, and you frequently, at least tacitly, rely on that idea in many of the implicit differentiation problems that get assigned in first-semester calculus.
To say that $f$ is locally one-to-one at a point $a$ in the domain of $f$ means that there is some open interval, say $(a-\varepsilon,a+\varepsilon)$ within the domain of $f,$ for which the restriction of $f$ to that interval is one-to-one. The formula you give applies to such restrictions. Since derivatives are an inherently local idea, this is not problematic.
Notice that in implicit differentiation problems they give you something like $x^2 + 3y^2 = 1,$ and expect you to deduce that $2x+ 6y\dfrac{dy}{dx} =0,$ even though the equation does that implicitly defines $y$ as a function of $x$ does not have a unique solution for $x.$ You're just applying differentiation in a small neighborhood of each point on the curve.
A: $\newcommand\R{\mathbb R}$Let's avoid Leibniz notation. In Lagrange notation, the inverse function theorem for a differentiable function $f\colon\R\to\R$ states that when $f(a)=b$ then
$$
g'(b) = \frac 1{f'(a)},
$$
where $g$ is a local inverse of $f$ at $a$. This means, that $g$ is a function $g\colon U\to V$, where $U$ is an open set containing $f(a)$, $V$ is an open set containing $a$ and we have
\begin{align*}
f(g(y)) &= y \quad\text{for all $y\in U$}, \\
g(f(x)) &= x \quad\text{for all $x\in V$}.
\end{align*}
Let's apply this to $f(x) = x^2$ with derivative $f'(x)=2x$. For $x=2$ we have $f(x)=4$ and we can pick the local inverse $g_1\colon \R_{>0}\to\R_{>0}, y\mapsto \sqrt y$. We see that $g_1'(y) = \frac 1{2\sqrt{y}}$ and can verify
$$
g_1'(4) = \frac{1}{4} = \frac{1}{f'(2)}.
$$
For $x=-2$, we can not pick the same $g_1$ as a local inverse, since $-2$ is not even contained in $V=\R_{>0}$ and there is no hope in making $V$ bigger since we would get
$$
g_1(f(-2)) = g_1(4) = 2 \neq -2.
$$
But we can instead pick $g_2\colon \R_{>0}\to\R_{<0}, y\mapsto -\sqrt{y}$ to get a correct local inverse and again verify the inverse function theorem:
$$
g_2'(4) = -\frac{1}{4} = \frac{1}{f'(-2)}.
$$
A: If $y=x^2$ then $$\frac{dy}{dx}=2x$$
and $\frac{dy}{dx}$ is given by using implicity derivative w.r.t $y$:
$$1=2x\frac{dx}{dy}\to \frac{dx}{dy}=\frac{1}{2x}$$
A: The branch choice doesn't matter, since if $x=-y^{1/2}$ then $\frac{dx}{dy}=-\frac{1}{2}y^{-1/2}=\frac{1}{2x}$. A general proof that $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$ is easy if you write each side as a limit by definition.
A: The formula ${dx\over dy}={1\over dy/dx}$ as a general principle does not refer to a global situation, but to a window $W$ in the $(x,y)$-plane, centered at some point $(x_0,y_0)$, within which the "variables" $x$ and $y$ are dependent on each other in such a way that you can write $y=\phi(x)$, whereby  $\phi(x_0)=y_0$, as well as $x=\psi(y)$, whereby $\psi(y_0)=x_0$. Assume that the points $(x,y)\in W$ satisfying $y=\phi(x)$, resp., $x=\psi(y)$ are lying on a curve $\gamma$ through the point $(x_0,y_0)$ whose tangent $t_0$ at $(x_0,y_0)$ is neither vertical nor horizontal. Inspecting a figure showing the window $W$ you can convince yourself that the slope of $t_0$  when viewed with respect to the $x$-axis, i.e., as tangent to $y=\phi(x)$, is $\phi'(x_0)$, and  when viewed with respect to the $y$-axis, i.e., as tangent to $x=\psi(y)$ is $\psi'(y_0)$. It follows that $\psi'(y_0)={1\over\phi'(x_0)}$.
