Find \(\frac{∂z}{∂x} \enspace and \enspace \frac{∂z}{∂y}\) if z is defined as a function of x and y by the equation  \(x^3 + y^3 + z^3 + 6xyz = 1\).

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Solution:

x3 + y3 + z3 + 6xyz = 1

and suppose that z is defined as a function of x and y by (i).

Then differentiating (i) partially with respect to x

Note: This part is just for understanding

Since, x and z are variables, We have to apply product rule for last part.

\(\frac{∂z}{∂x} = \frac{∂ 6xyz}{∂x}\)

Since y is constant, We will take y outside,

\(\frac{∂z}{∂x} = 6y \frac{∂ xz}{∂x}\)

\(\frac{∂6xyz}{∂x} = 6y(z\frac{∂x}{∂x} + x\frac{∂z}{∂x})\)

\(\frac{∂6xyz}{∂x} = 6yz + 6xy\frac{∂z}{∂x})\)

\(3x^2 + \frac{∂z^3}{∂x} + 6yz + 6xy\frac{∂z}{∂x} = 0\)

\(3x^2 + 3z^2\frac{∂z}{∂x} + 6yz + 6xy\frac{∂z}{∂x} = 0\)

\(\frac{∂z}{∂x} = \frac{-3x^3 – 6yz}{3z^2 + 6xy}\)

\(\frac{∂z}{∂x} = \frac{-x^3 – 2yz}{z^2 + 2xy}\)

Then differentiating (i) partially with respect to y

\(3y^2 + \frac{∂z^3}{∂y} + 6xz + 6xy\frac{∂z}{∂y} = 0\)

\(3y^2 + 3z^2\frac{∂z}{∂y} + 6xz + 6xy\frac{∂z}{∂y} = 0\)

\(\frac{∂z}{∂y} = \frac{-3y^2 – 6xz}{3z^2 + 6xy}\)

\(\frac{∂z}{∂y} = \frac{-y^2 – 2xz}{z^2 + 2xy}\)

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