a.y=(1x-2x^2)^5 b.y=(x^3-2x^2+1)^11 c.y=(x^2-2x)^5

0 bình luận về “a.y=(1x-2x^2)^5 b.y=(x^3-2x^2+1)^11 c.y=(x^2-2x)^5”

1. `y = (x – 2x^2)^{5}`

   `-> y’ = 5.(x-2x^2)^{4}.(x-2x^2)’ = -20x(x-2x^2)^4`

   `y = (x^3 – 2x^2 + 1)^{11}`

   `-> y’ = 11.(x^3 – 2x^2 + 1)^{10}.(x^3 – 2x^2 + 1)’ = 11.(3x^2 – 4x)(x^3 – 2x^2 + 1)^{10}`

   `y = (x^2 – 2x)^{5}`

   `-> y’ = 5.(x^2 – 2x)^{4}.(x^2 – 2x)’ = 5(2x – 2).(x^2 – 2x)^{4}`

    

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2. Đáp án:

   \(c)y’ = \left( {10x – 10} \right){\left( {{x^2} – 2x} \right)^4}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)y = {(1x – 2{x^2})^5}\\
   y’ = 5\left( {1 – 4x} \right){\left( {1x – 2{x^2}} \right)^4}\\
    = \left( {5 – 20x} \right){\left( {1x – 2{x^2}} \right)^4}\\
   b)y = {({x^3} – 2{x^2} + 1)^{11}}\\
   y’ = 11\left( {3{x^2} – 4x} \right){\left( {{x^3} – 2{x^2} + 1} \right)^{10}}\\
    = \left( {33{x^2} – 44x} \right){\left( {{x^3} – 2{x^2} + 1} \right)^{10}}\\
   c)y’ = 5\left( {2x – 2} \right){\left( {{x^2} – 2x} \right)^4}\\
    = \left( {10x – 10} \right){\left( {{x^2} – 2x} \right)^4}
   \end{array}\)

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