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?
Your error is that $$\dfrac{\mathrm d x}{\mathrm d y}=\dfrac{1}{\sqrt{1-\color{crimson}y^2}}$$
which simplifies to
$$\frac{1}{\sqrt{1-sin^2(x)}}$$
$$=\frac{1}{\sqrt{cos^2(x)}}$$ $$=\frac{1}{cos(x)}$$ $$=\frac{1}{\frac{dy}{dx}}$$
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 $\frac{dx}{dy}$.
