How do you find the derivative of #g(x)=sin^2x+cos^2x+secx#?

1 Answer
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#