How do you find the derivative of #y=sqrt(x)/(1+sqrt(x))#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Konstantinos Michailidis Jul 7, 2016 Write this as follows #y=(sqrtx+1-1)/(1+sqrtx)=1-1/(1+sqrtx)# Hence #dy/dx=(d(1+sqrtx)/dx)/[1+sqrtx]^2=1/[2*sqrtx*(1+sqrtx)^2]# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 2080 views around the world You can reuse this answer Creative Commons License