How do you differentiate #f(x)=sqrt(1/csc(2/x ) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Sonnhard Jun 22, 2018 #f'(x)=-csc(2/x)/(x^2*sqrt(sin(2/x)))# Explanation: Note that #csc(x)=1/sin(x)# so we get by the chain rule #f'(x)=1/2*(1/csc(2/x))^(-1/2)*csc(x/2)*(-2/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 1513 views around the world You can reuse this answer Creative Commons License