What is the derivative of #sec^2 x - 1#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Shwetank Mauria Sep 28, 2016 #d/(dx)(sec^2x-1)=2sec^2xtanx# Explanation: We can do it two ways. One – as #d/(dx)secx=secxtanx# #d/(dx)(sec^2x-1)=2secx xx secxtanx# = #2sec^2xtanx# Two – #d/(dx)(sec^2x-1)# = #d/(dx)tan^2x# = #2tanx xx sec^2x# = #2sec^2xtanx# 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 10530 views around the world You can reuse this answer Creative Commons License