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

Mar 23, 2018

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

$= \sec k x \left(k x \tan k x + 1\right)$

Mar 23, 2018

$\implies \sec \left(k x\right) \left(1 + k x \cdot \tan k x\right)$

#### Explanation:

Using product rule,

$d \frac{x \sec \left(k x\right)}{\mathrm{dx}}$ = $d \frac{x}{\mathrm{dx}} \sec \left(k x\right)$ + $x d \frac{\sec \left(k x\right)}{\mathrm{dx}}$

$d \frac{x \sec \left(k x\right)}{\mathrm{dx}}$ = $\sec \left(k x\right)$ + $x \cdot k \cdot \tan k x \cdot \sec k x$

$\implies \sec \left(k x\right) \left(1 + k x \tan \left(k x\right)\right)$

Mar 23, 2018

$\frac{1 + k x \tan \left(k x\right)}{\cos} \left(k x\right)$

#### Explanation:

$y = \frac{x}{\cos} \left(k x\right)$

$y = \frac{u}{v}$
$y ' = \frac{v \cdot u ' - v ' \cdot u}{v} ^ 2$

$u = x$
$u ' = 1$
$v = \cos \left(k x\right)$
$v ' = - k \sin \left(k x\right)$

$y ' = \frac{\cos \left(k s\right) + k x \sin \left(k x\right)}{\cos {\left(k x\right)}^{2}} = \frac{1 + k x \tan \left(k x\right)}{\cos} \left(k x\right)$