How do you differentiate #f(t)=root3(1+tant)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Alan N. Feb 23, 2017 #f'(t) = sec^2t/(3(1+tant)^(2/3))# Explanation: #f(t) = root3(1+tant)# #= (1+tant)^(1/3)# #f'(t) = 1/3(1+tant)^(-2/3) * d/dt(1+tant)# [Power rule and Chain rule] #= 1/3(1+tant)^(-2/3) * (0+ sec^2t)# [Standard differential] #=sec^2t/(3(1+tant)^(2/3))# 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 2961 views around the world You can reuse this answer Creative Commons License