Answer
$\dfrac{\partial (A)}{\partial b}=\dfrac{c \cos A -b}{bc \sin A}$
Work Step by Step
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, $\dfrac{\partial (A)}{\partial b}=\dfrac{c \cos A -b}{bc \sin A}$
