How do you differentiate #4/cosx + 1/ tanx#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer maira · Monzur R. Mar 3, 2018 #d/dx(4secx + cotx)= 4secxtanx - csc^2x# Explanation: #y=4/cosx+1/tanx# #dy/dx= d/dx(4secx + cotx) = 4secxtanx - csc^2x# 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 6879 views around the world You can reuse this answer Creative Commons License