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

1 Answer
MrMangoLassi
Apr 5, 2018

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

Explanation:

#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)#

Related questions
See all questions in Derivative Rules for y=cos(x) and y=tan(x)
Impact of this question
2754 views around the world
You can reuse this answer
Creative Commons License