Cal 3 Surfaces: Finding Derivatives at (1,1,1)

[lisa92]

Given that the surface x^8y^7+y^3z^6+z^8x63+6xyz=9 has the equation z=f(x,y) in a neighbourhod of the point (1,1,1) with f(x,y) differentiable, find the derivatives

fx(1,1)=

fy(1,1)=

Attempt: I tried plugging (1,1,1) in fx but it wasnt rigth. We haven't seen this in class and I don't know what to do.

Thanks

 

[LCKurtz]

If a funtion z = f(x,y) is defined implicitly by F(x,y,z)= 0 as in your problem, you can use the formulas:

$$f_x = -\frac {F_x}{F_z}\hbox{ and }f_y=-\frac{F_y}{F_z}$$

 

[I like Serena]

Hi lisa92!

I'd like to show you how to derive the formulas of LCKurtz.

With F(x,y,z)=0 you have:

$d(F(x,y,z))=F_x(x,y,z)dx + F_y(x,y,z)dy + F_z(x,y,z)dz=0$


From this you can find:

$dz={-F_x(x,y,z)dx - F_y(x,y,z)dy \over F_z(x,y,z)}$


Taking partial derivatives (for which the other coordinate is considered constant):

${\partial z \over \partial x}={-F_x(x,y,z) \over F_z(x,y,z)}$

${\partial z \over \partial y}={-F_y(x,y,z) \over F_z(x,y,z)}$


This can also be written as the formulas LCKurtz gave:

$f_x={-F_x \over F_z}$

$f_y={-F_y \over F_z}$