Given: $ \begin{eqnarray} R & = & \ln(u^2 + v^2 + w^2)\\ u & = & x + 6y\\ v & = & 2x - y\\ w & = & 4xy \end{eqnarray} $
I am trying to determine $\frac{\partial R}{\partial x}$ when $x = y = 2$.
Apparently I am incorrect: (Sorry for the lack of appropriate formatting)
$\frac{\partial R}{\partial x} = \frac{\partial R}{\partial u} \frac{\partial u} {\partial x}+ \frac{\partial R}{\partial v} \frac{\partial v}{\partial x} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial x}$
$= (1 / (u^2 + v^2 + w^2)) ((2u) (1) + (2v) (2) + (2w) (4y))$
$= (1 / (196 + 4 + 256)) (28 + 8 + 256)$
$= 73/114$