How do you differentiate #sqrt(sin^3(1/x) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer James May 7, 2018 the answer #dy/dx=[-3sin^2(1/x)*cos(1/x)]/[2*x^2*sqrt(sin^3(1/x))]# Explanation: Firstly suppose: #u=1/x# #(du)/dx=-1/x^2# #r=sin^3(u)# #(dr)/(du)=3sin^2(u)*cos(u)# #y=sqrtr# #dy/(dr)=1/(2sqrtr)# now #dy/(dx)=dy/(dr)*(dr)/(du)(du)/dx# #dy/dx=1/(2sqrtr)*3sin^2(u)*cos(u)[-1/x^2]# #dy/dx=[-3sin^2(1/x)*cos(1/x)]/[2*x^2*sqrt(sin^3(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 1265 views around the world You can reuse this answer Creative Commons License