$y=\frac{\sqrt{x^2+x+1}+sinx}{e^x+1}$ Tính đạo hàm

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$y=\frac{\sqrt{x^2+x+1}+sinx}{e^x+1}$
Tính đạo hàm

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Thiên Hương 5 years 2021-05-16T18:00:00+00:00 1 Answers 28 views 0

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  1. hongcuc2
    0
    2021-05-16T18:01:56+00:00
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    $\quad y =\dfrac{\sqrt{x^2 + x +1} + \sin x}{e^x +1}$

    $\to y’ =\dfrac{\left(\sqrt{x^2 + x +1} + \sin x\right)’.(e^x +1) – \left(\sqrt{x^2 + x +1} + \sin x\right).(e^x +1)’}{(e^x +1)^2}$

    $\to y’ = \dfrac{\left[\dfrac{(x^2 + x +1)’}{2\sqrt{x^2 + x + 1}} + (\sin x)’\right].(e^x +1)- \left(\sqrt{x^2 + x +1} + \sin x\right).(e^x)’}{(e^x +1)^2}$

    $\to y’ =\dfrac{\left[\dfrac{2x+1}{2\sqrt{x^2 + x + 1}} + \cos x\right](e^x +1) – e^x\left(\sqrt{x^2 + x +1} + \sin x\right)}{(e^x +1)^2}$

    $\to y’ = \dfrac{2x+1 + \cos x\sqrt{x^2 + x +1}}{e^x +1} – \dfrac{e^x\left(\sqrt{x^2 + x +1} + \sin x\right)}{(e^x +1)^2}$

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