Tính đạo hàm của hàm số $y'=sin^{2}(\frac{\pi}{2}-2x)+\frac{\pi}

Question

Cơ bản

Tính đạo hàm của hàm số $y'=sin^{2}(\frac{\pi}{2}-2x)+\frac{\pi}{2}x-\frac{\pi}{4})$

4 câu trả lời

Answer 1

A.

$y'=-2sin(\pi-4x)$

Answer 2

$y'=2sin(\frac{\pi}{2}-x)cos(\frac{\pi}{2}-x)+\frac{\pi}{2}x)$

Answer 3

$y'=2sin(\frac{\pi}{2}-x)cos(\frac{\pi}{2}-x)+\frac{\pi}{2})$

Answer 4

$y'=-2sin(\pi-4x)+\frac{\pi}{2})$