How do you find #(dy)/(dx)# given #x^3-xy+y^3=1#?
1 Answer
Jul 22, 2017
Explanation:
#"differentiate "color(blue)"implicitly with respect to x"#
#"differentiate the term " xy" using the "color(blue)"product rule"#
#rArr3x^2-x.dy/dx-y.1+3y^2dy/dx=0#
#rArrdy/dx(3y^2-x)=y-3x^2#
#rArrdy/dx=(y-3x^2)/(3y^2-x)#
