How do you differentiate #y = cos sec^2(x)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer Anees Apr 13, 2015 #(dy)/(dx)=-2sec^2xtanx(sinsec^2(x))# #y=(cossec^2(x))# Differentiating both side with respect to 'x' #(dy)/(dx)=d/(dx)(cossec^2(x))# #(dy)/(dx)=-sinsec^2(x)d/(dx)(sec^2(x))# #(dy)/(dx)=-sinsec^2(x)(2secx)(d/(dx)(secx))# #(dy)/(dx)=-sinsec^2(x)(2secx)(secxtanx)# #(dy)/(dx)=-sinsec^2(x)(2sec^2xtanx)# #(dy)/(dx)=-2sec^2xtanx(sinsec^2(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 3102 views around the world You can reuse this answer Creative Commons License