Finding dz/dx and dz/dy using implicit differentiation

[mr_coffee]

hello everyone I'm stuck! anyone have any ideas?
I'm suppose to find dz/dx and dz/dy with implicit differentation. This is calc III!
http://img221.imageshack.us/img221/4000/lastscan4ou.jpg

 

[EnumaElish]

Is the question "x - z = arctan(yz). Find dz/dx. Find dz/dy." ?

Differentiate throughout: dx - dz = (ydz+zdy)arctan'(yz).

Then manipulate into dz = A + Bdx where A and B are functions of x, y and z. Divide by dx to get dz/dx = A/dx + B.

Similar for dz/dy.

 

[HallsofIvy]

One problem you have is that you have the wrong derivative for arctangent!

The derivative of arctan(x) is $$\frac{1}{1+ x^2}$$

if x-z= arctan(yz) then, writing z_x and z_y for the derivatives of z with respect to x and y respectively, we have
$$1-z_x= \frac{yz_x}{1+ y^2z^}$$
which you can solve for z_x and
$$-z_y= \frac{z+ yz_y}{1+ y^2z^2}$$
which you can solve for z_y.

 

[Fermat]

One thing I couldn't understand here was, what happened to dy/dx and dx/dy.

We have,

x - z = arctan(yz)

differentiating wrt x,

1 - dz/dx = d/dx{arctan(s)}, where s = yz
1 - dz/dx = d/ds{arctan(s)}.ds/dx
1 - dz/dx = 1/(1 + s²) * (y.dz/dx + z.dy/dx)
1 - dz/dx = (y.dz/dx + z.dy/dx) / (1 + y²z²)

Adopting HallsofIvy's notation,

1 - z_x = (yz_x + zy_x)/(1 + y²z²)

What have I missed out ?

 

[EnumaElish]

One thing I couldn't understand here was, what happened to dy/dx and dx/dy.

Could it be that the OP used total diff. "d" notation but HallsOfIvy interpreted it as partial diff. "$\partial$" notation?

 

[Fermat]

Could it be that the OP used total diff. "d" notation but HallsOfIvy interpreted it as partial diff. "$\partial$" notation?

That explains it.
Thanks.