How do you differentiate #sqrt(sin^2(1/x) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Binayaka C. Jun 30, 2018 #f^'(x)=-1/(2 x^2*sqrt ((sin^2(1/x)))) * sin(2/x)# Explanation: #f(x)=sqrt(sin^2(1/x))= (sin^2(1/x))^(1/2)# #f^'(x)=1/2(sin^2(1/x))^(-1/2)* 2 sin(1/x)* cos(1/x)* -1/x^2# #f^'(x)=1/(2(sin^2(1/x))^(1/2))* sin(2/x)* -1/x^2# #f^'(x)=-1/(2x^2*sqrt((sin^2(1/x)))) * sin(2/x)# [Ans] 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 1303 views around the world You can reuse this answer Creative Commons License