What is the derivative of #y=sec(2x) tan(2x)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer WADU HEK Apr 14, 2018 2sec(2x)#(sec^2(2x) + tan^2(2x))# Explanation: # y' = (sec(2x))(tan(2x))' + (tan(2x))(sec(2x))'# (Product Rule) # y' = (sec(2x))(sec^2(2x))(2) + (tan(2x))(sec(2x)tan(2x))(2) # (Chain rule and derivatives of trig) # y' = 2sec^3(2x) + 2sec(2x)tan^2(2x) # # y' = 2sec(2x)(sec^2(2x) + tan^2(2x)) # Answer link Related questions What is Derivatives of #y=sec(x)# ? What is the Derivative of #y=sec(x^2)#? What is the Derivative of #y=x sec(kx)#? What is the Derivative of #y=sec ^ 2(x)#? What is the derivative of #y=4 sec ^2(x)#? What is the derivative of #y=ln(sec(x)+tan(x))#? What is the derivative of #y=sec^2(x)#? What is the derivative of #y=sec^2(x) + tan^2(x)#? What is the derivative of #y=sec^3(x)#? What is the derivative of #y=sec(x) tan(x)#? See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x) Impact of this question 7886 views around the world You can reuse this answer Creative Commons License