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

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Answer 1 · 2016-11-03T09:16:17

$\frac{d y}{d x} = sec x (2 tan x + sec x)$ $(dy)/(dx)=secx(2tanx+secx)$

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

$\frac{d}{d x} 2 sec x = 2 sec x tan x \to$ $d/(dx)2secx=2secxtanx->$ since $\frac{d}{d x} sec x = sec x tan x$ $d/(dx)secx=secxtanx$

$\frac{d}{d x} tan x = {sec}^{2} x$ $d/(dx)tanx=sec^2x$

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

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