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

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Answer 1 · 2016-07-09T16:19:55

$(d g(x))/(d x)=sin x/cos^2 x$

$g(x)=sin^2 x+ cos ^2 x+ sec x" ; " (d g(x))/(d x)=?$

$"we can write as: "sin^2 x+cos ^2 x=1$

$g(x)=1+sec x$

$(d g(x))/(d x)=d/(d x)(1)+ d/(d x)(sec x)$

$"so ;"d/(d x)(1)=0" ; "(d)/(d x)(sec x)=sin x/cos^2 x$

$(d g(x))/(d x)=sin x/cos^2 x$

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