How do you find the derivative of #f(x)=sin (1/x^2)# using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer maganbhai P. Apr 1, 2018 #f^'(x)=-2/x^3cos(1/x^2)# Explanation: Here, #f(x)=sin(1/x^2)=sin(x^-2)# Using Chain Rule,we get #f^'(x)=cos(x^-2)d/(dx)(x^-2)# #=cos(1/x^2)(-2x^(-2-1))# #=-2cos(1/x^2)(x^-3)# #=-2/x^3cos(1/x^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 1207 views around the world You can reuse this answer Creative Commons License