Given that the surface $x^{6}y^{5}+y^{4}z^{5}+z^{9}x^{7}+4xyz=7$ has the equation $z=f(x,y)$ in a neighborhood of the point (1, 1, 1) with f(x,y) differentiable, find:

$\displaystyle\frac{\partial^{2} f}{\partial x^2}(1,1)$

I have already found an intermediate expression:

$\displaystyle\frac{\partial f}{\partial x}=\frac{-6x^{5}y^{5}-7z^{9}x^{6}-4yz}{5y^{4}z^{4}+9z^{8}x^{7}+4xy}$

At the point (1, 1), we have that it's value is -17/18. However, I have tried finding the second partial derivative in two different ways, differentiating both sides of the expression and then differentiating the complete quotient, but i get different results and my answer is still wrong. Could someone guide me in the right direction or have an easier method to solve this exercise? Thank you.


If $$x^{6}y^{5}+y^{4}z^{5}+z^{9}x^{7}+4xyz=7,~~~z=f(x,y)$$ then $$6x^5y^5+\color{red}{\underline{5z^4z'y^5}}+\color{blue}{\underline{9z^8z'x^7}}+\color{green}{\underline{7z^9x^6}}+4yz+4xyz'=0$$ and so $$30x^5y^5+\color{red}{\underline{20z^3z'^2y^5+5z^4z''y^5}}+\color{blue}{\underline{72z^7z'^2x^7+9z^8z''x^7+63z^8z'x^6}}+\color{green}{\underline{63z^8z'x^6+42z^9x^6}}+4yz'+4yz'+4xyz''=0$$

  • $\begingroup$ Yay! color! (I like your profile update with the cat!) $\endgroup$ – Namaste Nov 19 '13 at 2:52
  • $\begingroup$ @amWhy: Thanks!. I learned doing that from you. :-) Yes, I got it from my FB page last night. :-) $\endgroup$ – mrs Nov 19 '13 at 2:55
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    $\begingroup$ I had the same result as you, but i completely forgot about the minus sign when passing the expression over to the other side (-1114/729)! Thank you very much for your patience! $\endgroup$ – arcbloom Nov 19 '13 at 4:54
  • $\begingroup$ @fogvajarash: I am glad. Welcome! :-) $\endgroup$ – mrs Nov 19 '13 at 5:47

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