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

1 Answer
MrMangoLassi
Apr 5, 2018

= secx(sec^2x + tan^2x)=secx(sec2x+tan2x)

Explanation:

secx(tanx+cosx) = (tanx+cosx)/cosx = tanx/cosx + 1secx(tanx+cosx)=tanx+cosxcosx=tanxcosx+1
d/dx tanx/cosxddxtanxcosx 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)

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