# Difficulty understanding partial derivatives [closed]

If I have a function: $$z=f(x, y)$$

And I want to find: $$(\frac{ \partial y}{ \partial x}) _{z}$$

How do I do this? My approach is that if z is kept constant, $$dz$$ would be $$0$$. Therefore, using the relation:

$$dz=( \frac{ \partial z}{ \partial x})dx + ( \frac{ \partial z}{ \partial y})dy$$

I can set dz = $$0$$. I the divided both sides by dy, getting:

$$\frac{ \partial z}{ \partial x} \frac{dx}{dy} = - \frac{ \partial z}{ \partial y}$$

But I don't know where to go from here. I also don't understand what $$\frac{dx}{dy}$$ means in this context, is this the same as $$(\frac{ \partial y}{ \partial x}) _{z}$$?

Thanks

We cannot blindly manipulate symbols as if the notation takes care of everything. The starting point is that $$z=$$const. defines implicitly a function $$x\mapsto y(x)\,:$$ $$z=f(x,y(x))\,.$$ For this and what follows to work let's assume that $$\partial_yf\not=0\,.$$

The total derivative of $$z$$ is zero (as you noted): $$0=\partial_xf+\partial_yf\;y'$$ Then: $$\Big(\frac{\partial y}{\partial x}\Big)_z=y'=-\frac{\partial_x f}{\partial_ yf}\,.$$

$$dy/dx$$ in this context is the same as $$(\partial y/\partial x)_z$$. For a fixed $$z$$, you have a curve $$y=y(x)$$ in the x-y plane and $$dy/dx$$ gives the tangent of this curve in this plane.