Counterexample to $ \frac{\mathrm{d}{y}}{\mathrm{d}{x}} = \frac{1}{\left( \frac{\mathrm{d}{x}}{\mathrm{d}{y}} \right)} $

Question

I saw this question: Is $ \frac{\mathrm{d}{x}}{\mathrm{d}{y}} = \frac{1}{\left( \frac{\mathrm{d}{y}}{\mathrm{d}{x}} \right)} $?

If this is true then see the following example:

$y = sin(x)$, then $\frac{dy}{dx} = cos(x)$

But also $x = sin^{-1}(y)$, so $\frac{dx}{dy} = \frac{1}{\sqrt{1-x^2}}$

Therefore, clearly $\frac{dy}{dx} \neq \frac{1}{\frac{dx}{dy}}$

How is this possible?

Answer 1

Your error is that $$\dfrac{\mathrm d x}{\mathrm d y}=\dfrac{1}{\sqrt{1-\color{crimson}y^2}}$$

which simplifies to

$$\frac{1}{\sqrt{1-sin^2(x)}}$$

$$=\frac{1}{\sqrt{cos^2(x)}}$$ $$=\frac{1}{cos(x)}$$ $$=\frac{1}{\frac{dy}{dx}}$$

Answer 2

First of all, note that $\frac{dx}{dy}=\frac{1}{\sqrt{1-y^2}}=\frac{1}{\sqrt{1-\sin^2x}}=\frac{1}{\left|\cos x\right|}$. Now compare $\frac{dy}{dx}$ and $\frac{dx}{dy}$.