## University Calculus: Early Transcendentals (3rd Edition)

$-\dfrac{\sin x}{1+\cos ^2 x}$
Here, $y=-f'(\sec x)$ and $y'=\dfrac{d(\sec x)}{dx} =-\dfrac{(\sec x\tan x)}{\sec^2 x+1}$ or, $=-\dfrac{(1/\cos x)(\sin x/\cos x)}{1/\cos^2 x+1}$ or, $=-\dfrac{\sin x}{1+\cos ^2 x}$