I'm slightly confused by something, I imagine I'm thinking about it the wrong way so any help would be appreciated.
I'm faced with the equation $xy+ln ( \frac {x}{y})=2$
And am asked if the derivative $\frac{dx}{dy}$ exists at $(x,y)=(1,1)$
I've used the implicit differention formula $\frac {dx}{dy} =- \frac{\partial F}{\partial y}/\frac{\partial F}{\partial x}$ and have found it is defined and equal to $0$.
The bit which I'm confused about is, that if I plug (1,1) into the equation it is not satisfied, i.e $(1)(1) + ln (\frac{1}{1}) ≠2 $ and therefore on my graph, no line passes through that point, so how can the derivative be define at that point?
I'd be grateful for any guidance on this.
Nice work checking that the point actually lies on the curve defined by the equation!
To be honest, I expect that the person posing the problem did not check, but maybe I'm being unfair. Perhaps your answer is exactly what s/he was looking for. Anyhow, I guess my own general answer might be something like this:
For the family of curves defined by $$ xy+ln ( \frac {x}{y})=k, $$ $k \in \mathbb R$, there is a curve (namely the one for $k = 1$) that contains the point $(1, 1)$, and the derivative $dx/dy$ at that point on that curve, is $0$. But since the point $(1,1)$ does not lie on the $k = 2$ curve, the function implicitly defined by the $k = 2$ curve does not have a derivative at the point $(1,1)$.