Question

What is the derivative of `y=x sec(kx)`?

Answer 1

As below.

Explanation:

Differentiating y w.r.t. x,

#dy/dx=x.{d/dxsec(kx)×d/dxkx}+sec(kx)×(d/dx)x#

`=x*seckxtankx×k + seckx`

`=seckx(kxtankx+1)`

Answer 2

`=>sec(kx) (1 + kx*tankx)`

Explanation:

Using product rule,

`d(xsec(kx))/dx` = `d(x)/dxsec(kx)` + `xd(sec(kx))/dx`

`d(xsec(kx))/dx` = `sec(kx)` + `x*k*tankx*seckx`

`=> sec (kx) ( 1 + kx tan (kx))`

Answer 3

`(1+kxtan(kx))/cos(kx)`

Explanation:

`y=x/cos(kx)`

`y=u/v`
`y'=(v*u'-v'*u)/v^2`

`u=x`
`u'=1`
`v=cos(kx)`
`v'=-ksin(kx)`

`y'=(cos(ks)+kxsin(kx))/(cos(kx)^2) =(1+kxtan(kx))/cos(kx)`