How do you differentiate #sqrt(sqrtx)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Noah G Nov 26, 2016 We can rewrite #y= sqrt(sqrt(x))# as #y = (x^(1/2))^(1/2) = x^(1/4)# By the power rule: #dy/dx= 1/4x^(1/4 - 1)# #dy/dx= 1/4x^(-3/4)# #dy/dx = 1/(4x^(3/4))# Hopefully this helps! 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 915 views around the world You can reuse this answer Creative Commons License