How do you find #dy/dx# given #x=1/(1+sqrty)#?
1 Answer
Feb 25, 2017
Explanation:
differentiate both sides
#color(blue)"implicitly with respect to x"#
#"Express "1/(1+y^(1/2))=(1+y^(1/2))^-1" and"# differentiate using the
#color(blue)"chain rule"#
#x=(1+y^(1/2))^-1#
#rArr1=-(1+y^(1/2))^-2xx 1/2y^(-1/2).dy/dx#
#dy/(dx)(-1/(2y^(1/2)(1+y^(1/2))^2))=1#
#rArrdy/dx=-1/(2sqrty(1+sqrty)^2)#
