y=2$Sin^{2}$x-x+15 giúp mình ạ

Question

y=2$Sin^{2}$x-x+15
giúp mình ạ

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Farah 3 years 2021-04-26T12:05:58+00:00 2 Answers 17 views 0

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  1. vankhanh
    0
    2021-04-26T12:07:03+00:00
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    \(\begin{array}{l}
    \quad y = 2\sin^2x – x + 15\\
    \to y’ = 2(\sin^2x)’ – (x’) + (15)’\\
    \to y’ = 2.2\sin x.(\sin x)’ – 1 + 0\\
    \to y’ = 4\sin x.\cos x – 1\\
    \to y’ = 2\sin2x – 1
    \end{array}\)

  2. King
    0
    2021-04-26T12:07:21+00:00
    Reply

    Đáp án: `y’=2sin2x -1`

     

    Giải thích các bước giải:

     `y= 2sin²x -x+15`

    `=> y’= 2.2sinx.cosx -1+0`

    `=> y’=2sin2x -1`

    ADCT:

    `sin^{n}x = n.sin^{n-1}x.cosx`

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )