How do you differentiate #y=sqrtx+1+1/sqrtx#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Anjali G Nov 9, 2016 #y=sqrtx+1+1/sqrtx# #y=x^(1/2)+1+x^(-1/2)# #(dy)/(dx)=(1/2)x^(-1/2)-(1/2)x^(-3/2)# #(dy)/(dx)=[(1)/(2sqrtx)]-[(1)/(2(sqrtx)^3)]# #(dy)/(dx)=(1/(2sqrtx))(1-1/x)# 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 1368 views around the world You can reuse this answer Creative Commons License