# 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)#