How do you find the derivative of $y= 2secx + tanx$ ?

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Answer 1 · 2016-11-03T09:16:17

$(dy)/(dx)=secx(2tanx+secx)$

Since there is only a simple addition operation, $2secx$ and $tanx$ can be derived independently.

$d/(dx)2secx=2secxtanx->$ since $d/(dx)secx=secxtanx$

$d/(dx)tanx=sec^2x$

$:.(dy)/(dx)=2secxtanx+sec^2x=secx(2tanx+secx)$

How do you find the derivative of y= 2secx + tanxy= 2secx + tanx?