How do you find #(dy)/(dx)# given #4x=-5y^2-x^2y+4#?
1 Answer
Dec 29, 2017
Explanation:
#"differentiate "color(blue)"implicitly with respect to x"#
#"differentiate "x^2y" using the "color(blue)"product rule"#
#rArr4=-10y.dy/dx-(x^2dy/dx+2xy)+0#
#rArr4=-10ydy/dx-x^2dy/dx-2xy#
#rArrdy/dx(-10y-x^2)=-2xy-4#
#rArrdy/dx=(-(2xy+4))/(-(10y+x^2))=(2xy+4)/(10y+x^2)#
