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

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

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

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

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

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