Multiple Choice 1 point If y=sinx+tanx then (d^2y)/(dx^2)= -cosx+sec^2xtanx cosx+sec^2x -sinx+sec^2x -sinx+2sec^2xtanx

To find the second derivative of \( y = \sin x + \tan x \), we first need to find the first derivative and then differentiate it again.<br /><br />First, let's find the first derivative of \( y \):<br /><br />\[ y = \sin x + \tan x \]<br /><br />The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \tan x \) is \( \sec^2 x \). Therefore,<br /><br />\[ \frac{dy}{dx} = \cos x + \sec^2 x \]<br /><br />Next, we need to find the second derivative, which is the derivative of \( \frac{dy}{dx} \):<br /><br />\[ \frac{d^2y}{dx^2} = \frac{d}{dx} (\cos x + \sec^2 x) \]<br /><br />The derivative of \( \cos x \) is \( -\sin x \), and the derivative of \( \sec^2 x \) is \( 2 \sec^2 x \tan x \). Therefore,<br /><br />\[ \frac{d^2y}{dx^2} = -\sin x + 2 \sec^2 x \tan x \]<br /><br />So, the correct answer is:<br /><br />\[ -\sin x + 2 \sec^2 x \tan x \]<br /><br />Therefore, the correct choice is:<br /><br />\[ -\sin x + 2 \sec^2 x \tan x \]