Finding $ \frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y} $ if $ F(cx - az, cy-bz) = 0 $

Question

If it is given that $ F(cx - az, cy-bz) = 0 $, then find $ \dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$.

How do I go about doing this? I don't really understand which variables are independent and which aren't.

I'm new to partial derivatives. I know chain rule though. Some hints?

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As per Blah's answer, here's what I've done:

$$ F_x = F_u (c - az_x) + F_v(-bz_x) = 0 $$ $$ F_y = F_u (- az_y) + F_v(c -bz_y) = 0 $$ (The "$=0$" is there because $F$ is constant (zero) for all values of $x$ and $y$)

So, by doing some algebra, I get $z_x = \dfrac{cF_u} { bF_v + aF_u }$ and something similar for $z_y$.

But the above contains $F_u$ and $F_v$. How do I get rid of those?

Answer 1

How do I go about doing this?

Introduce new functions (in addition zu $z=z(x,y)$) $$ u(x,y)=cx-az(x,y) \qquad v(x,y)=cy-bz(x,y) $$ and compute $$ u_x=c-az_x \qquad u_y=-az_y \qquad v_x=-bz_x \qquad v_y=c-bz_y $$ Now use the chain rule to write $z_x,z_y$ in in terms of the partial derivatives of $F$

Answer 2

The equation $$F(cx - az, cy-bz) = 0$$ defines implicitly a function $(x,y)\mapsto z(x,y)$ under reasonable hypothesis (see http://en.wikipedia.org/wiki/Implicit_function_theorem). Applying the chain rule to $$\pmatrix{x\cr y}\longmapsto \pmatrix{cx-az(x,y)\cr cy-bz(x,y)}\longmapsto F(cx - az(x,y), cy-bz(x,y))$$ and using that $F(cx - az(x,y), cy-bz(x,y))=0$ forall $(x,y)$ you will obtain a system of two equations with two unknowns $\partial z/\partial x$ and $\partial z/\partial y$.