How do you find the derivative of #sec^2(x)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Bill K. Jul 27, 2015 #2sec^{2}(x)tan(x)# Explanation: Use the Chain Rule: #d/dx(sec^{2}(x))=2sec(x)*d/dx(sec(x))# #=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)#? 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 1540 views around the world You can reuse this answer Creative Commons License