How do you find #dy/dx# given #x=1-2y^2#?
1 Answer
Jun 1, 2017
Explanation:
#color(blue)"differentiate implicitly with respect to x"#
#"differentiate "-2y^2" using the "color(blue)"chain rule"#
#rArr1=0-4y.dy/dx#
#rArr4ydy/dx=-1#
#rArrdy/dx=-1/(4y)#
