0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

which simplifies to

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

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

$\endgroup$
0
$\begingroup$

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 $\frac{dx}{dy}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.