How do you differentiate given #y = (sec)^2 x + (tan)^2 x#? Calculus Basic Differentiation Rules Chain Rule 1 Answer NJ Mar 2, 2018 #y' = 4sec^2xtanx# Explanation: #y = sec^2x + tan^2x# #y = secx secx + tanx tanx# #y' = (dy)/(dx) = (secxtanx)secx+secx(secxtanx)+ (sec^2x)tanx+tanx(sec^2x)# #y' = sec^2xtanx + sec^2xtanx+sec^2xtanx+sec^2xtanx# #y' = 4sec^2xtanx# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1104 views around the world You can reuse this answer Creative Commons License