How would you calculate this partial derivative?

Question

Given the function $z^{2}x-yz+2xy=4$ how would you find $z_{x}$? I've tried doing it by rearranging to obtain a function in the form $z=f(x,y)$, but this doesn't seem to be possible.

Answer 1

As suggested by @John, the quadratic equation in terms of z will be

xz^2 - yz + (2xy - 4) = 0

Now, z can be written as (formula for roots of quadratic equation):

z = (y +/- (y^2 - 4x(2xy - 4))^(1/2))/(2x)

Simplifying further,

z = (y +/- (y^2 - 8yx^2 + 16x))^(1/2))/(2x)

Now, dz/dx can be found by differentiating this expression w.r.t. x

I get following result:

dz/dx = -y/(2x^2) +/- (1-xy)/(x(y^2-8yx^2+16x)^3/2) -/+ ((y^2-8yx^2+16x)^1/2)/(2x^2)