$\frac {\operatorname d\!y}{\operatorname d\!x}$ for $\sqrt{xy}=1$

Question

To find $\displaystyle \dfrac {\operatorname d\!y}{\operatorname d\!x}$ for $\sqrt{xy}= 1$:

$$\dfrac{\sqrt y}{2\sqrt x}+\dfrac{y'\sqrt x}{2\sqrt y}=0\\ \dfrac{y'\sqrt x}{2\sqrt y}=\dfrac{-\sqrt y}{2\sqrt x}\\ y'=\dfrac{-y}x$$

Is this correct? and can you show me how to get the same answer solving using $\ln$?

Answer 1

$$\sqrt{xy}=1\implies \ln(\sqrt{xy})=\ln1$$

As $\displaystyle\ln y^m=m\ln y$ where both logarithm is defined unlike $\displaystyle\ln1=\ln(-1)^2=2\ln(-1)$ $$\implies \frac12\ln(xy)=0$$

As $\displaystyle\ln ab=\ln a+\ln b$ where both logarithm is defined unlike $\displaystyle\ln6=\ln (-1)(-6)=\ln(-1)+\ln(-6)$ $$\implies\ln |x|+\ln |y|=0$$

Differentiating we get $$\frac{d(\ln|x|)}{dx}+\frac{d(\ln|y|)}{dy}\frac{dy}{dx}=\frac{d(\ln 1)}{dx}$$

$$\implies\frac1x+\frac1y\frac{dy}{dx}=0$$

Answer 2

$\sqrt{xy}=1\implies xy=1 \implies xy'+y=0 \implies y'=-\dfrac yx$.
So, yeah, this is true.

Answer 3

$$\ln \sqrt {xy} = \ln 1$$ $$\frac{\ln{x}+\ln{y}}{2}=0$$ $$\ln y=-\ln x$$ $$\dfrac{\operatorname d}{\operatorname d\!x} \ln y= - \dfrac{\operatorname d}{\operatorname d\!x}\ln x $$ $$\dfrac{\operatorname d \ln y }{\operatorname d\!y} \dfrac{\operatorname dy}{\operatorname d\!x} =-\frac{1}{x}$$ $$\dfrac{\operatorname dy}{\operatorname d\!x} =-\frac{y}{x}$$