dx/dy in three different ways

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Say you have a function

$y = {x^2},\quad x \geqslant 0$

The derivative of this is then

$y' = \frac{{dy}}{{dx}} = 2x$

Now, what is dx/dy? The derivative dy/dx gives us how fast y is changing as we change x, and dx/dy gives us how fast x is changing as we change y. We can find this in several ways.

Using the inverse function

The most natural way might be to make x the subject and then take the derivative. We thus switch the roles of x and y. We get

$x = \sqrt y$

and then

$\frac{{dx}}{{dy}} = \frac{1}{{2\sqrt y }}$

Here we have to remember that x is now a function of our variable y.

Using the reciprocal

If dy/dx=m, could it be that dx/dy=1/m? Could it be that simple?

Normally we see dy/dx as one symbol the derivative of y with respect to x. But could the parts dy and dx take on roles as separate entities? Let us say we have some function, and that the derivative at a particular point is m, and if we draw a tangent line at that point, then the tangent line would have the slope m at every point.

We may then choose to let dy and dx represent any two values such that dy/dx=m. If so then we may start to manipulate them as any variables and thus do the following

$\frac{{dy}}{{dx}} = m,\quad dy = mdx,\quad 1 = m\frac{{dx}}{{dy}},\quad \frac{1}{m} = \frac{{dx}}{{dy}}$

We can also show this by

$\begin{gathered} \frac{1}{{\;\frac{{dy}}{{dx}}\;}} = \frac{1}{{\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\frac{{\Delta y}}{{\Delta x}}}} \hfill \\ \quad \quad = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta x}}{{\Delta y}} = \mathop {\lim }\limits_{\Delta y \to 0} \frac{{\Delta x}}{{\Delta y}} = \frac{{dx}}{{dy}} \hfill \\ \end{gathered}$