How do you find the derivative of $[secx(tanx + cosx)]$ ?

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Answer 1 · 2018-04-05T16:52:26

$= secx(sec^2x + tan^2x)$

$secx(tanx+cosx) = (tanx+cosx)/cosx = tanx/cosx + 1$
$d/dx tanx/cosx$ using quotient rule

$u = tanx , u' = sec^2x$
$v = cosx , v' = -sinx$

$(u'v-v'u)/v^2$

$= (sec^2xcosx + sinxtanx)/cos^2x$
$= (cosx)/cos^4x + (sinxtanx)/cos^2x$
$= 1/cos^3x + tanx/cosxtanx$
$= sec^3x +tan^2x/cosx$

$= secx(sec^2x + tan^2x)$

How do you find the derivative of [secx(tanx + cosx)][secx(tanx + cosx)]?