Question

(a) The function z is defined implicitly as a function in x and y by the equation x4+y4+z4+xyz=1. Find zx and zy

(b) The function z is defined implicitly as a function in x and y by the equation ez=xyz Find zx and zy.

Answer 1

(a) For the implicit function x⁴ + y⁴ + z⁴ + xyz = 1), the partial derivatives are
`\(z_x = -(yx^2)/(z^2)\)`and
`\(z_y = -(xy^2)/(z^2)\).`

(b) For
`\(e^z = xyz\)`, the partial derivatives are
`\(z_x = ye^z\)` and
`\(z_y = xe^z\).`

Step-by-step explanation:

(a) To find
`\(z_x\)`and
`\(z_y\)` for the equation x⁴ + y⁴ + z⁴ + xyz = 1), we differentiate implicitly with respect to x and y . For
`\(z_x\)`, we differentiate the equation with respect to x treating y as a constant, and for c
`\(z_x\)` , we differentiate with respect to y , treating x as a constant. The results are
`\(z_x = -(yx^2)/(z^2)\) and \(z_y = -(xy^2)/(z^2)\).`

(b) For
`(e^z = xyz\)`, we differentiate implicitly to find
`\(z_x\)` and
`\(z_y\)`. The derivative of
`\(e^z\)`with respect to x (treating y as a constant) gives
`\(z_x = ye^z\)`, and the derivative with respect to y (treating x as a constant) gives
`\(z_y = xe^z\).`

In summary, the partial derivatives for the given implicit functions are
`(z_x = -(yx^2)/(z^2)\), \(z_y = -(xy^2)/(z^2)\)` for (a), and
`\(z_x = ye^z\), \(z_y = xe^z\) for (b).`These derivatives provide the rate of change of \(z\) with respect to x and (y), respectively.