What is the derivative of #y=sec^2(x)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer AJ Speller Sep 21, 2014 We need to use the chain rule #y=sec^2(x)# Let #u=sec(x)# #u'=sec(x)tan(x)# #y=u^2# #y'=2u*u'# #y'=2sec(x)* sec(x)tan(x)=2sec^2(x)tan(x)# 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) + tan^2(x)#? What is the derivative of #y=sec^3(x)#? What is the derivative of #y=sec(x) tan(x)#? What is the derivative of #y=x sec(kx)#? See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x) Impact of this question 8499 views around the world You can reuse this answer Creative Commons License