How do you find #(dy)/(dx)# given #y^2+siny=sinx#?
1 Answer
Oct 20, 2016
Explanation:
differentiate
#color(blue)"implicitly with respect to x"#
#2y.dy/dx+cosy.dy/dx=cosx#
#rArrdy/dx(2y+cosy)=cosx#
#rArrdy/dx=(cosx)/(2y+cosy)#