Consider the equation xz^2 − 4 yz − 6 lnz = −1 as defining z implicitly as a function of x and y. The values of \dfrac∂z∂x and \dfrac∂z∂y at (3,1,1) are and .
(This problem used to have "log" instead of "ln", but the answer was the same, because in webwork "log" means the natural logarithm.

Question

Consider the equation xz^2 − 4 yz − 6 lnz = −1 as defining z implicitly as a function of x and y. The values of \dfrac∂z∂x and \dfrac∂z∂y at (3,1,1) are  and .
(This problem used to have "log" instead of "ln", but the answer was the same, because in webwork "log" means the natural logarithm.

turingtest · Answer

the values of what at what are... what?

jamesj · Answer

You want to find partial dz/dx and partial dz/dy ?

anonymous · Answer

the ones i need ! and yes to james

anonymous · Answer

∂z/∂x=1/16
∂z/∂y=-1

anonymous · Answer

ksram i get to -1 as well but on the other one when i tried the answer its wrong !

turingtest · Answer

you say "the partial derivatives (dz/dx) and (dz/dy) at (3,1,1) are__and___ so are you not missing something?

anonymous · Answer

the spaces are the answers i need ! which one of them as ksram said is -1

anonymous · Answer

found it its 1/4

anonymous · Answer

is 1/16 correct?

turingtest · Answer

no it's 1/4