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Problem Determine d d x [ ( x + 3 ) sec x − x cos x 2 x ] \dfrac{d}{dx}\left[\dfrac{(x+3)\sec x - x\cos x}{2x}\right] d x d [ 2 x ( x + 3 ) sec x − x cos x ] sec x + ( x + 3 ) sec x tan x − cos x + x sin x 2 x \dfrac{\sec x + (x+3)\sec x\tan x - \cos x + x\sin x}{2x} 2 x sec x + ( x + 3 ) sec x tan x − cos x + x sin x ( sec x + ( x + 3 ) sec x tan x − cos x − x sin x ) ( 2 x ) − 2 ( ( x + 3 ) sec x − x cos x ) ( 2 x ) 2 \dfrac{\bigl(\sec x + (x+3)\sec x\tan x - \cos x - x\sin x\bigr)(2x) - 2\bigl((x+3)\sec x - x\cos x\bigr)}{(2x)^2} ( 2 x ) 2 ( sec x + ( x + 3 ) sec x tan x − cos x − x sin x ) ( 2 x ) − 2 ( ( x + 3 ) sec x − x cos x ) ( sec x + ( x + 3 ) sec x tan x − cos x + x sin x ) ( 2 x ) − 2 ( ( x + 3 ) sec x − x cos x ) ( 2 x ) 2 \dfrac{\bigl(\sec x + (x+3)\sec x\tan x - \cos x + x\sin x\bigr)(2x) - 2\bigl((x+3)\sec x - x\cos x\bigr)}{(2x)^2} ( 2 x ) 2 ( sec x + ( x + 3 ) sec x tan x − cos x + x sin x ) ( 2 x ) − 2 ( ( x + 3 ) sec x − x cos x ) sec x + ( x + 3 ) sec x tan x − cos x + x sin x 2 x − ( x + 3 ) sec x − x cos x x 2 \dfrac{\sec x + (x+3)\sec x\tan x - \cos x + x\sin x}{2x}\; -\; \dfrac{(x+3)\sec x - x\cos x}{x^{2}} 2 x sec x + ( x + 3 ) sec x tan x − cos x + x sin x − x 2 ( x + 3 ) sec x − x cos x Browse the problem library