How do you find the derivative of #g(x)=sin^2x+cos^2x+secx#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer ali ergin Jul 9, 2016 #(d g(x))/(d x)=sin x/cos^2 x# Explanation: #g(x)=sin^2 x+ cos ^2 x+ sec x" ; " (d g(x))/(d x)=?# #"we can write as: "sin^2 x+cos ^2 x=1# #g(x)=1+sec x# #(d g(x))/(d x)=d/(d x)(1)+ d/(d x)(sec x)# #"so ;"d/(d x)(1)=0" ; "(d)/(d x)(sec x)=sin x/cos^2 x# #(d g(x))/(d x)=sin x/cos^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 2793 views around the world You can reuse this answer Creative Commons License