What is the derivative of #y= sec^3 x+ tan^2 x sec x#?

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Answer 1 · 2017-12-19T10:50:34

# dy/dx=(6sec^2x-1)secxtanx,#

OR,

# (6tan^2x+5)secxtanx.#

Note that, #y=sec^3x+tan^2xsecx,#

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

#=secx{sec^2x+(sec^2x-1)},#

#=secx(2sec^2x-1).#

#:. y=2sec^3x-secx.#

#:. dy/dx=d/dx{2sec^3x-secx},#

#=2d/dx{(secx)^3}-d/dx{secx},#

#=2*3(secx)^2d/dx{secx}-d/dx{secx},#

#=6sec^2xd/dx{secx}-d/dx{secx},#

#=(6sec^2x-1)d/dx{secx},#

# rArr dy/dx=(6sec^2x-1)secxtanx,#

or, what is the same as,

#{6(tan^2x+1)-1}secxtanx=(6tan^2x+5)secxtanx.#

Enjoy Maths.!