seems pretty simple but the $\displaystyle tan(\phi)$ on the left side threw me off

$\displaystyle tan(\phi) = \frac {X_L}{R}$ ; find $\displaystyle \frac {\partial \phi}{\partial R}$

Answer key states in the first step that:

$\displaystyle \frac {\partial \phi}{\partial R}= \frac {1}{(\frac {X_L}{R^2}) + 1} * \frac {-X_L}{R^2}$

not sure how they got $\displaystyle \frac {1}{(\frac {X_L}{R^2}) + 1} $