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

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Answer 1 · 2018-03-23T06:55:17

As below.

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 · 2018-03-23T07:00:12

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

#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 · 2018-03-23T07:00:17

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

#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)#