Question

Consider the surface defined by the equation x3 + y3 + z3 = 3xyz. Use implicit differ- ∂z entiation to find ∂x. (Note z is not actually a function of x and y here, but we can still find this partial derivative.)

Answer 1

Answer:    

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

Step-by-step explanation:

Given equation is $x^{3}+y^{3}+z^{3}=3xyz$ ………………(1)

we use derivative formula $\frac{d}{dx}(x^{n}) = n x^{n-1}$

$\frac{d}{dx}(x^{3)} = 3 x^{2}$

$\frac{d}{dx}(z^{3)} = 3 z^{2}$

And also apply 'u v' formula

$\frac{d}{dx}(uv}) = u\frac{d}{dx}(v)+v\frac{d}{dx}(u)$

Differentiating  equation (1) partially with respective to 'x' , we treated 'y' as constant.

$(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =3y(x\frac{∂z}{∂x}+z(1))$   ( here y treated as constant so the derivative of constant function is zero in addition but in multiplication the constant is keep as like 'y').

on simplification , we get

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

again simplification, we get

$3z^{2}\frac{∂z}{∂x}- 3yx\frac{∂z}{∂x}=3yz-3x^{2}$

taking common '$\frac{∂z}{∂x}$ on left on side , we get

$(3z^{2}- 3yx)\frac{∂z}{∂x}=3yz-3x^{2}$

dividing '$(3z^{2}- 3yx)$ on both sides, we get

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

Final answer:-

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