Respuesta :
Answer:
[tex]\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}[/tex]
Step-by-step explanation:
Given equation is [tex]x^{3}+y^{3}+z^{3}=3xyz[/tex] ………………(1)
we use derivative formula [tex]\frac{d}{dx}(x^{n}) = n x^{n-1}[/tex]
[tex]\frac{d}{dx}(x^{3)} = 3 x^{2}[/tex]
[tex]\frac{d}{dx}(z^{3)} = 3 z^{2}[/tex]
And also apply 'u v' formula
[tex]\frac{d}{dx}(uv}) = u\frac{d}{dx}(v)+v\frac{d}{dx}(u)[/tex]
Differentiating equation (1) partially with respective to 'x' , we treated 'y' as constant.
[tex](3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =3y(x\frac{∂z}{∂x}+z(1))[/tex] ( 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
[tex](3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =(3yx\frac{∂z}{∂x}+3yz)[/tex]
again simplification, we get
[tex]3z^{2}\frac{∂z}{∂x}- 3yx\frac{∂z}{∂x}=3yz-3x^{2}[/tex]
taking common '[tex]\frac{∂z}{∂x}[/tex] on left on side , we get
[tex](3z^{2}- 3yx)\frac{∂z}{∂x}=3yz-3x^{2}[/tex]
dividing '[tex](3z^{2}- 3yx)[/tex] on both sides, we get
[tex]\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}[/tex]
Final answer:-
[tex]\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}[/tex]
