Mọi người giúp em vs ạ

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Mọi người giúp em vs ạ
moi-nguoi-giup-em-vs-a

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RI SƠ 5 years 2021-04-20T05:35:59+00:00 1 Answers 18 views 0

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  1. haidang
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    2021-04-20T05:37:51+00:00
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    Đáp án:

    $\begin{array}{l}
    1)a)y = {x^4} + 3{x^2} + 2x\\
     \Leftrightarrow y’ = 4{x^3} + 6x + 2\\
    b)y = \dfrac{{x – 1}}{{3x + 1}}\\
     \Leftrightarrow y’ = \dfrac{{1.\left( {3x + 1} \right) – 3.\left( {x – 1} \right)}}{{{{\left( {3x + 1} \right)}^2}}}\\
     = \dfrac{{3x + 1 – 3x + 3}}{{{{\left( {3x + 1} \right)}^2}}}\\
     = \dfrac{4}{{{{\left( {3x + 1} \right)}^2}}}\\
    c)y = x.\tan x\\
     \Leftrightarrow y’ = \tan x + x.\dfrac{1}{{{{\cos }^2}x}}\\
    2)a)y = 2{x^3}\\
     \Leftrightarrow y’ = 6{x^2}\\
    PTTT:y = {y_o}’\left( {x – {x_0}} \right) + {y_0}\\
     + Khi:M\left( { – 1; – 1} \right)\\
     \Leftrightarrow {x_0} =  – 1;{y_0} =  – 1\\
     \Leftrightarrow PTTT:y = 6.{\left( { – 1} \right)^2}\left( {x + 1} \right) – 1\\
     \Leftrightarrow y = 6x + 5\\
    b)HSG:{y_0}’ = 6\\
     \Leftrightarrow 6x_0^2 = 6\\
     \Leftrightarrow {x_0} = 1/{x_0} =  – 1\\
     \Leftrightarrow \left[ \begin{array}{l}
    PTTT:y = 6\left( {x – 1} \right) + {2.1^3}\\
    PTTT:y = 6x + 5
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    PTTT:y = 6x – 4\\
    PTTT:y = 6x + 5
    \end{array} \right.
    \end{array}$

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