# dx/dy in three different ways

Up a level : Differentiation, derivatives
Previous page : Implicit differentiation
Next page : The circle and implicit derivatives

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 is 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

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

The last step can be done since, if the derivative exists, both Δx and Δy will go toward 0 simultaneously.

This would now give us, for our function, y=x2 that

$\frac{{dx}}{{dy}} = \frac{1}{{2x}}$

but since

$x = \sqrt y$

we get

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

Using implicit differentiation

We may also take the derivative of y=x2 with respect to x to get

$1 = 2x \cdot x' = 2x\frac{{dx}}{{dy}}$

or

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

Another example

Let

$y = {x^3}$

• The derivative of the inverse function

$x = {y^{1/3}}$

So

$\frac{{dx}}{{dy}} = \frac{1}{3}{y^{\frac{1}{3} - 1}} = \frac{1}{3}{y^{ - \frac{2}{3}}} = \frac{1}{{3{y^{2/3}}}}$

• The reciprocal

The derivative dy/dx is y´=3x2, so

$\frac{{dx}}{{dy}} = \frac{1}{{3{x^2}}}$

But

$x = {y^{1/3}}$

so

${x^2} = {y^{2/3}}$

and thus

$\frac{{dx}}{{dy}} = \frac{1}{{3{y^{2/3}}}}$

• The implicit differential

The derivative with respect to y of

$y = {x^3}$

Gives us

$1 = 3{x^2}\frac{{dx}}{{dy}}$

So

$1 = 3{x^2}\frac{{dx}}{{dy}} = \frac{1}{{3{x^2}}}$

And then we do as above.

Up a level : Differentiation, derivatives
Previous page : Implicit differentiation
Next page : The circle and implicit derivativesLast modified: Jan 21, 2018 @ 22:34