What is the Derivative of #y=x sec(kx)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Wataru Sep 21, 2014 By Product Rule, #y'=1cdot sec(kx)+x cdot[sec(kx)]'# by Chain Rule, #=sec(kx)+x cdot [sec(kx)tan(kx)cdot k]# by factoring out #sec(kx)#, #=sec(kx)[1+kxtan(kx)]# 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=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)#? 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 9828 views around the world You can reuse this answer Creative Commons License