Question

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

Answer 1

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

OR,

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

Explanation:

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.!