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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

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2 Answers 2

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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: $$ \Big(\frac{\partial y}{\partial x}\Big)_z=y'=-\frac{\partial_x f}{\partial_ yf}\,. $$

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$dy/dx$ 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.

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