How do you take the derivative of # tan^2(4x)#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer Michael Jul 2, 2015 #(d[tan^2(4x)])/(dx)=8sec^2(4x)tan(4x)# Explanation: Use the chain rule: #(d[tan^2(4x)])/(dx)=2sec^2(4x).tan(4x)xxd((4x))/dx# #=2sec^2(4x).tan(4x)xx4# #=8sec^2(4x)tan(4x)# Answer link Related questions What is the derivative of #y=cos(x)# ? What is the derivative of #y=tan(x)# ? How do you find the 108th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x)# from first principle? How do you find the derivative of #y=cos(x^2)# ? How do you find the derivative of #y=e^x cos(x)# ? How do you find the derivative of #y=x^cos(x)#? How do you find the second derivative of #y=cos(x^2)# ? How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x) Impact of this question 11249 views around the world You can reuse this answer Creative Commons License