A calculus problem by Hobart Pao

Solution.

$z^2 = x + \sqrt{y+\sqrt{x+\sqrt{y+...}}}$ $z^2 - x = \sqrt{y+\sqrt{x+\sqrt{y+...}}} = \sqrt{y+ z}$ $z^4 + x^2 -2z^2 x = y +z$ $4z^3\dfrac{\partial z}{\partial x} + 2x - 2 \left( 2zx \dfrac{\partial z}{\partial x} + z^2 \right) = \dfrac{\partial z}{\partial x}$ $(4z^3 - 4zx) \dfrac{\partial z}{\partial x} - 2z^2 + 2x = \dfrac{\partial z}{\partial x}$ $\dfrac{\partial z}{\partial x} = \dfrac{-2z^2 + 2x }{1-4z^3 + 4zx}$

At $(x, y, z)=(2,2,2)$ , we get $\dfrac{-2 \cdot 2^2 + 2 \cdot 2}{1-4 \cdot 8 + 4 \cdot 2^2} = \dfrac{4}{15}; 4+15=\boxed{19}$