Đạo ham Giúp mk 3 câu này nha

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Đạo ham
Giúp mk 3 câu này nha
dao-ham-giup-mk-3-cau-nay-nha

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Thu Cúc 5 years 2021-05-16T02:50:28+00:00 2 Answers 27 views 0

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  1. dangkhoa
    0
    2021-05-16T02:51:34+00:00
    Reply

    Đáp án:

    \(\begin{array}{l}
    4)\quad y’ = \dfrac{2x+1}{2(x+1)\sqrt{x+1}}\\
    5)\quad y’ = – \dfrac{1}{(\sin x – \cos x)^2}\\
    6)\quad y’ = \dfrac{-2}{\cos^2(-2x+3)}
    \end{array}\) 

    Giải thích các bước giải:

    \(\begin{array}{l}
    4)\quad y = \dfrac{2x+3}{\sqrt{x+1}}\\
    \to y’ = \dfrac{(2x+3)’\sqrt{x+1} – (2x+3).\left(\sqrt{x+1}\right)’}{x+1}\\
    \to y’ = \dfrac{2\sqrt{x+1} – (2x+3)\cdot\dfrac{(x+1)’}{2\sqrt{x+1}}}{x+1}\\
    \to y’ = \dfrac{2\sqrt{x+1} – (2x+3)\cdot\dfrac{1}{2\sqrt{x+1}}}{x+1}\\
    \to y’ = \dfrac{\dfrac{4(x+1) – (2x+3)}{2\sqrt{x+1}}}{x+1}\\
    \to y’ = \dfrac{2x+1}{2(x+1)\sqrt{x+1}}\\
    5)\quad y = \dfrac{\sin x}{\sin x – \cos x}\\
    \to y’ = \dfrac{(\sin x)'(\sin x – \cos x) – \sin x(\sin x – \cos x)’}{(\sin x – \cos x)^2}\\
    \to y’ = \dfrac{\cos x(\sin x – \cos x) – \sin x(\cos x + \sin x)}{(\sin x – \cos x)^2}\\
    \to y’ = \dfrac{-\sin^2x – \cos^2x}{(\sin x – \cos x)^2}\\
    \to y’ = – \dfrac{1}{(\sin x – \cos x)^2}\\
    6)\quad y = \tan(-2x+3)\\
    \to y’ = \dfrac{(-2x+3)’}{\cos^2(-2x+3)}\\
    \to y’ = \dfrac{-2}{\cos^2(-2x+3)}
    \end{array}\) 

  2. Thunguyet
    0
    2021-05-16T02:51:54+00:00
    Reply

    4.

    $y’=\dfrac{2\sqrt{x+1}-(2x+3).(\sqrt{x+1})’ }{\sqrt{x+1}^2}$

    $=\dfrac{2\sqrt{x+1}-(2x+3).\dfrac{1}{2\sqrt{x+1}} }{x+1}$

    $=\dfrac{4(x+1)-(2x+3)}{(x+1).2\sqrt{x+1}}$

    $=\dfrac{2x+1}{2\sqrt{x+1}(x+1)}$

    5.

    $y’=1:\dfrac{\sin x-\cos x}{\sin x}$

    $=1:\Big(1-\cot x\Big)$

    $=\dfrac{1}{1-\cot x}$

    $y’=-\dfrac{(1-\cot x)’}{(1-\cot x)^2}$

    $=-\dfrac{1+\sin^2x}{(1-\cot x)^2}$

    6.

    $y’=\dfrac{(-2x+3)’}{\cos^2(-2x+3)}$

    $=\dfrac{-2}{2\cos^2(-2x+3)}$

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