How do you differentiate #f(x)=csc(lnx) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Shwetank Mauria Jun 28, 2016 #(df)/(dx)=-1/xcsc(lnx)cot(lnx)# Explanation: As #f(x)=csc(lnx)=csc(g(x))#, where #g(x)=lnx# As according to chain rule #(df)/(dx)=(df)/(dg)xx(dg)/(dx)# #(df)/(dx)=-csc(lnx)cot(lnx)xx1/x=-1/xcsc(lnx)cot(lnx)# 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 2047 views around the world You can reuse this answer Creative Commons License