## Thomas' Calculus 13th Edition

$\dfrac{\partial (A)}{\partial b}=\dfrac{c \cos A -b}{bc \sin A}$
Recall the law of cosines $a^2=b^2+c^2-2bc \cos A$ Differentiate the given function with respect to $a$ . $2a =-2bc (-\sin A \dfrac{\partial (A)}{\partial a}$ This implies that $\dfrac{\partial (A)}{\partial a}=\dfrac{a}{bc \sin A}$ Differentiate the given function with respect to $a$ and $b$. $0 =2b-2c \cos A-2bc (-\sin A \dfrac{\partial (A)}{\partial b}$ This implies that $\dfrac{\partial (A)}{\partial b}=\dfrac{2c \cos A -2b}{2bc \sin A}$ So, $\dfrac{\partial (A)}{\partial b}=\dfrac{c \cos A -b}{bc \sin A}$