If $u=x.ln(xy)$ and $x^2+y^2+3xy -1=0$ then what will be $\frac{dy}{dx}$

Question

We have $$u=x.ln(xy)$$ and $$x^2+y^2+3xy -1=0$$ then how can we calculate $\frac{dy}{dx}$ I can calculate it using only second equation , but how can I use first equation? I understand that $u=f(x,y)$ and $y=f(x)$ but what next ? my book says $$\frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u }{\partial y}.\frac{dy}{dx}$$ what is this? how did we get it ? please explain in detail

Answer 1

I believe the formula is

$\frac{d}{dx}u=\frac{\partial u}{\partial x}+\frac{\partial u }{\partial y}.\frac{dy}{dx}$

This happens because $u=u(x,y)$ and $y$ is not independent with respect to $x$.

An example to clarify it is this: $$u=x^3$$ Say $y=x^2$, now $u=xy$ but obviously $\frac{d}{dx}u \neq \frac{\partial}{\partial x}u$ in the usual sense because the variables are not independent. In fact using the formula above $$\frac{d}{dx}u = y + x(2x) = 3x^2$$