Implicit second order partial derivative

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Homework Statement



Given that the surface (x**5)(y**2)+(y**5)(z**3)+(z**3)(x**2)+4xyz=7 has the equation z=f(x,y) in a neighbourhood of the point (1,1,1) with f(x,y) differentiable, find the derivatives
(∂**2f)/(∂x**2) at (1,1)

Homework Equations





The Attempt at a Solution


I already solved for ∂f/∂x and ∂f/∂y for the first part of the problem implicitly, but how to I solve for the second order derivative? I've tried ∂f/∂x * ∂/dx to no avail. All of my other attempts were really just hunches which turned up nothing as well. Thanks for any help.
 

Answers and Replies

  • #2
SammyS
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Homework Statement



Given that the surface (x**5)(y**2)+(y**5)(z**3)+(z**3)(x**2)+4xyz=7 has the equation z=f(x,y) in a neighbourhood of the point (1,1,1) with f(x,y) differentiable, find the derivatives
(∂**2f)/(∂x**2) at (1,1)

Homework Equations



The Attempt at a Solution


I already solved for ∂f/∂x and ∂f/∂y for the first part of the problem implicitly, but how to I solve for the second order derivative? I've tried ∂f/∂x * ∂/dx to no avail. All of my other attempts were really just hunches which turned up nothing as well. Thanks for any help.
What did you get for ∂f/∂x and ∂f/∂y ?
 
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at point (1,1) I got -11/10 for both ∂f/∂x and ∂f/∂y
 
  • #4
SammyS
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at point (1,1) I got -11/10 for both ∂f/∂x and ∂f/∂y
What did you get for ∂f/∂x and ∂f/∂y in terms of x, y, and f in the neighborhood of (1, 1, 1), before you evaluated them at (1, 1, 1) ?
 
  • #5
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for ∂f/∂x i got -(5x**4*y**2+2z**3*x+4yz)/(3y**5*z**2+3z**2*x**2+4xy)
and
for ∂f/∂y i got -(2x**5*y+5y**4*z**3+4xz)/(3y**5*z**2+3z**2*x**2+4xy)
both of which were found using ∂f/∂x = -(∂/dx)/(∂/∂z)
 
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  • #6
SammyS
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I got the same for ∂f/∂x . I didn't do ∂f/∂y.

But I did it a different way.

At any rate take the partial w.r.t x of the following:
[itex]\displaystyle \left( \frac{\partial f}{\partial x}\right) \left(3y^5(f(x,y))^2+3(f(x,y))^2x^2+4xy\right)=-\left(5x^4y^2+2(f(x,y))^3x+4yf(x,y)\right)[/itex]​
Notice that I replaced z with f(x,y). After you've taken the derivative, change f(x,y) back to z, if you like.
 

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