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

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Answer 1 · 2014-09-21T16:25:19

$y'=1cdot sec(kx)+x cdot[sec(kx)]'$

$=sec(kx)+x cdot [sec(kx)tan(kx)cdot k]$

by factoring out $sec(kx)$ ,

$=sec(kx)[1+kxtan(kx)]$

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