Question

It’s implicit differentiation problem into finding the tangent line for the given point

Answer 1

the assumption being that whilst "x" is just a simple variable, "y" is really a function in x-terms, so

`sin(xy)=y\implies \stackrel{\textit{\Large chain~rule}}{cos(xy)\stackrel{product~rule}{\left( y+x\cfrac{dy}{dx} \right)}}=\cfrac{dy}{dx}\implies ycos(xy)+cos(xy)x\cfrac{dy}{dx}=\cfrac{dy}{dx} \\\\\\ ycos(xy)=\cfrac{dy}{dx}-cos(xy)x\cfrac{dy}{dx}\implies ycos(xy)=\cfrac{dy}{dx}( ~~ 1-cos(xy)x ~~ ) \\\\\\ \left. \cfrac{ycos(xy)}{1-cos(xy)x}=\cfrac{dy}{dx} \right|_{(\pi )/(2),1}\implies \cfrac{1cos\left( (\pi )/(2)\cdot 1 \right)}{1-cos\left( (\pi )/(2)\cdot 1 \right)(\pi )/(2)}\implies 0`

`~\dotfill\\\\ \begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{0}(x-\stackrel{x_1}{(\pi )/(2)})\implies {\Large \begin{array}{llll} y=1 \end{array}}`