Thomas' Calculus 13th Edition

$v_x= \dfrac{\ln v}{\ln u \ln v -1}$
We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to $x$. $1=v_x \ln u +(v) (u^{-1} ) u_x$ and $0=u_x \ln v +(u) (v^{-1} ) v_x$ So, $u_x= \dfrac{-(u/v) v_x}{\ln v}$ Now, $1=v_x \ln u +(v) (u^{-1} ) (u_x= \dfrac{-(u/v) v_x}{\ln v})$ $\implies v_x= \dfrac{\ln v}{\ln u \ln v -1}$