How do you differentiate #f(x)=e^tan(lnx)# using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Sasha P. Oct 29, 2015 #f'(x) = (e^(tan(lnx))sec^2(lnx))/x# Explanation: #f'(x) = (e^(tan(lnx)))' = e^(tan(lnx)) * (tan(lnx))' = e^(tan(lnx)) * sec^2(lnx) * (lnx)' = e^(tan(lnx)) * sec^2(lnx) * 1/x# #f'(x) = (e^(tan(lnx))sec^2(lnx))/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 1752 views around the world You can reuse this answer Creative Commons License