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

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 $$\frac{dx}{dy}$$.