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

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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#

$= x \cdot sec k x tan k x \times k + sec k x$ $=x*seckxtankx×k + seckx$

$= sec k x (k x tan k x + 1)$ $=seckx(kxtankx+1)$

Answer 2 · 2018-03-23T07:00:12

$\Rightarrow sec (k x) (1 + k x \cdot tan k x)$ $=>sec(kx) (1 + kx*tankx)$

$d \frac{x sec (k x)}{d x}$ $d(xsec(kx))/dx$ = $d \frac{x}{d x} sec (k x)$ $d(x)/dxsec(kx)$ + $x d \frac{sec (k x)}{d x}$ $xd(sec(kx))/dx$

$d \frac{x sec (k x)}{d x}$ $d(xsec(kx))/dx$ = $sec (k x)$ $sec(kx)$ + $x \cdot k \cdot tan k x \cdot sec k x$ $x*k*tankx*seckx$

$\Rightarrow sec (k x) (1 + k x tan (k x))$ $=> sec (kx) ( 1 + kx tan (kx))$

Answer 3 · 2018-03-23T07:00:17

$\frac{1 + k x tan (k x)}{cos (k x)}$ $(1+kxtan(kx))/cos(kx)$

$y = \frac{x}{cos (k x)}$ $y=x/cos(kx)$

$y = \frac{u}{v}$ $y=u/v$
$y'=(v*u'-v'*u)/v^2$

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

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

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