Question

How do you find `dy/dx` given `x=1/(1+sqrty)`?

Answer 1

`dy/dx=-1/(2sqrty(1+sqrty)^2)`

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)`