Partial derivatives: implicit differentiation?

Assuming that z is a function of x and y:

1) Differentiating both sides with respect to x:

2x + 0 + 6z ∂z/∂x = 0 ==> ∂z/∂x = -x/(3z).

Differentiating both sides with respect to y:

0 + 4y + 6z ∂z/∂y = 0 ==> ∂z/∂x = -2y/(3z).

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2) Differentiating both sides with respect to x:

e^z ∂z/∂x = yz + xy ∂z/∂x

==> ∂z/∂x = yz/(e^z - xy).

Differentiating both sides with respect to y:

e^z ∂z/∂y = xz + xy ∂z/∂y

==> ∂z/∂y = xz/(e^z - xy).

I hope this helps!

Let's take a normal case, an equation in two variables(on your comfort), 2x^2 + 8x.Y^three = e^y + x^three differentiate with admire to x, then 4x + 8y^three + 8x .3y^2 .(dy/dx) = e^y . (dy/dx) + 3x^2 (24xy^2 - e^y) (dy/dx) = 3x^2 - 4x - 8y^three and, dy/dx = (8y^three - 3x^2 + 4x) / (e^y - 24xy^2) the above is an example of implicit differentiation. As you might notice it's a pure final result of the chain rule of differentiation and offers the by-product in terms of both variables. As for partial differentiation, z = 2x^2 + 8x.Y^three would have a partial by-product by means of differentiation with admire to x of, &z / &x = 4x + 8y^three ,(where & is for the partial differential notation) As you will find, the result does now not depend on the chain rule and as an alternative finds the derivatives wrt x, treating another time period as in the event that they had been steady. There are countless functions, for instance, partial derivatives are crucial to see whether the Cauchy-Riemann conditions are convinced for a function in problematic analysis. Hope this helps!