How do you find #(dy)/(dx)# given #xcosy+y=x^3#?
1 Answer
Oct 26, 2016
Explanation:
differentiate
#color(blue)"implicitly with respect to x"#
#• "Use the "color(blue)"product rule"# on the term xcosy
#rArrx(-siny).dy/dx+cosy.1+1.dy/dx=3x^2#
#rArrdy/dx(-xsiny+1)=3x^2-cosy#
#rArrdy/dx=(3x^2-cosy)/(1-xsiny)#
