Use implicit differentiation to find
∂z/∂x and ∂z/∂y.
x^3 + y^3  +z^3 + 6xyz=  1

Question

Use implicit differentiation to find 
∂z/∂x and ∂z/∂y.
x^3 + y^3  +z^3 + 6xyz=  1

anonymous · Answer

does any one remember

ganeshie8 · Answer

when you differentiate with.respect.to "x", y an z are treated as contants

anonymous · Answer

yes but its respect as function of z

ganeshie8 · Answer

$\large   x^3 + y^3  +z^3 + 6xyz=  1$

$\large  \frac{\partial }{\partial x} (x^3 + y^3  +z^3 + 6xyz=  1)$

anonymous · Answer

so do i take both der of x and also to z so do i chain the second one at z

ganeshie8 · Answer

lets see

anonymous · Answer

z is defined implicitly as a function of

anonymous · Answer

x

ganeshie8 · Answer

$\large  \frac{\partial }{\partial x} (x^3 + y^3  +z^3 + 6xyz=  1)$


$\large  3x^2 + 3z^2 \frac{\partial z}{\partial x} + 6y( x\frac{\partial z}{\partial x} + z)=  0$

anonymous · Answer

i kn 3x^2+3z^2dz/dx+6yz(dz/dx)+6xy(dz/dx)=0

ganeshie8 · Answer

you can solve $\frac{\partial z}{\partial x}$ from above

ganeshie8 · Answer

you may also first solve $z$ explicitly, and then find the partial and verify if you get the same

ganeshie8 · Answer

nvm, we cannot solve $z$ explicitly

anonymous · Answer

i was about to ask you to show me =P

anonymous · Answer

okay thx

ganeshie8 · Answer

lol

ganeshie8 · Answer

u wlc :)