How can (dy/dx) = (1/(dx/dy)) be true when isolating different variables and differentiating them indicates this isn't the case?

Question

Let $y=x^3$. We know that... $$ \frac{dy}{dx}=3x^2 $$ The derivative of $y$ with respect to $x$ is $3x^2$. Everyone can agree on this. But what if we were to isolate the other variable? $$ y=x^3 $$ $$ \sqrt[3]{y}=\sqrt[3]{x^3}=x $$ $$ \sqrt[3]{y}=x $$ And now if we decide to take the derivative of $x$ with respect to $y$, $$ \frac{dx}{dy}=\frac{1}{3x^{2/3}} $$ $\frac{1}{\frac{dx}{dy}}=\frac{1}{\frac{1}{3x^{2/3}}}=3x^{2/3}$, which is not equal to $\frac{dy}{dx}=3x^2$. So...$\frac{dy}{dx}$ is not $\frac{1}{\frac{dx}{dy}}$? What am I missing here? Shouldn't this violate the Inverse Function Theorem?

P.S I know you're supposed to implicitly differentiate...but what I'm interested in is why the above procedures don't yield the same results as implicit differentiation (correctly) does.

Derivatives aren't really fractions. It's just a useful comparison when using the chain rule. — CyclotomicField, Oct 13 '22 at 16:26
@insipidintegrator oh my god. you are absolutely right. this is what always thinking of x as the input variable does to a person — voltamatron, Oct 13 '22 at 17:04
To just expand, the inverse function of $f(x)=x^3$ is $f^{-1}(x)= \sqrt[3]x$. — insipidintegrator, Oct 14 '22 at 03:42
I have revised it back to the original because editing the question to replace the x with y basically takes away the understandability (is that a word?) and relevance of the question. — insipidintegrator, Oct 14 '22 at 03:51

score 3 · Accepted Answer · answered Oct 13 '22 at 17:00

3

Recall that $y=x^3$, hence $$\begin{cases} \frac{1}{dx/dy}=3y^{\frac{2}{3}}\\ y=x^3\end{cases}\implies \frac{1}{dx/dy}=3x^2=\frac{dy}{dx}.$$

answered Oct 13 '22 at 17:00

ecrin

1,423

How can (dy/dx) = (1/(dx/dy)) be true when isolating different variables and differentiating them indicates this isn't the case?

1 Answers1