What is the derivative of #Sec^2 pi(x) #? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Monzur R. Jun 7, 2017 #dy/dx = 2pisec^2pixtanpix# Explanation: Use the chain rule to differentiate #y=sec^2pix# Let #u=secpix# and #y=u^2# #dy/(du) = 2u=2secpix# #(du)/dx= pisecpixtanpix# Chain rule: #dy/dx=dy/(du) xx (du)/dx# #dy/dx = 2secpix xx pisecpixtanpix = 2pisec^2pixtanpix# 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 9004 views around the world You can reuse this answer Creative Commons License