Implicit differentiation problem

mr_coffee
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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
 
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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.
 
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 zx and zy 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 zx and
-z_y= \frac{z+ yz_y}{1+ y^2z^2}
which you can solve for zy.
 
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 - zx = (yzx + zyx)/(1 + y²z²)

What have I missed out ?
 
Fermat said:
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?
 
Last edited:
EnumaElish said:
Could it be that the OP used total diff. "d" notation but HallsOfIvy interpreted it as partial diff. "\partial" notation?
That explains it.
Thanks.
 
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