For xy+y2=8xy + y^2 = 8xy+y2=8, find y′y'y′ at (1,2)(1, 2)(1,2).A−25-\tfrac{2}{5}−52check_circleB25\tfrac{2}{5}52C−2-2−2D−15-\tfrac{1}{5}−51Explanationy+xy′+2yy′=0⇒y′(x+2y)=−y⇒y′(1,2)=−2/(1+4)=−2/5y + xy' + 2y y' = 0 \Rightarrow y'(x + 2y) = -y \Rightarrow y'(1, 2) = -2/(1+4) = -2/5y+xy′+2yy′=0⇒y′(x+2y)=−y⇒y′(1,2)=−2/(1+4)=−2/5.