Question

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

Answer 1

`(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`