Đáp án:
`1, y= 6sin x +2cos x -3tan x +4cotx`
`=> y' = 6cosx -2sinx - 3/(cos²x) - 4/(sin²x)`
`2, y= cos 4x`
`=> y' =-(4x)^'.sin 4x = -4sin4x`
`3, y= sin² x`
`=> y' = 2.sinx.(sinx)' = 2sinx.cosx= sin2x`
`4, y= x\sqrt{sin2x}`
`=> y' = \sqrt{sin2x} +(\sqrt{sin2x})^'.x`
`=> y' = \sqrt{sin2x} + \frac{2cos2x}{2\sqrt{sin2x}} .x`
`=> y' = \frac{ sin2x + cos2x.x}{\sqrt{sin2x}}`
