Tìm đạo hàm cấp hai của mỗi hàm số sau: a. \(y = \dfrac{{2x - 2}}{{x + 1}}\) b. \(y = \dfrac{{x +
Tìm đạo hàm cấp hai của mỗi hàm số sau:
a. \(y = \dfrac{{2x - 2}}{{x + 1}}\)
b. \(y = \dfrac{{x + 2}}{{{x^2} - x + 1}}\)
c. \(y = \dfrac{{{x^2} - 4x + 3}}{{x - 2}}\)
d. \(y = \dfrac{{x + 1}}{{{x^2} + 5x + 6}}\)
a) \(y = \dfrac{{2x - 2}}{{x + 1}}\)
\(\begin{array}{l}y' = \dfrac{4}{{{{\left( {x + 1} \right)}^2}}}\\y'' = \dfrac{{ - 4.2\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^4}}} = \dfrac{{ - 8}}{{{{\left( {x + 1} \right)}^3}}}\end{array}\)
b) \(y = \dfrac{{x + 2}}{{{x^2} - x + 1}}\)
\(\begin{array}{l}y' = \dfrac{{{x^2} - x + 1 - \left( {x + 2} \right)\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\\,\,\,\,\,\, = \dfrac{{{x^2} - x + 1 - 2{x^2} - 3x + 2}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\\,\,\,\,\,\, = \dfrac{{ - {x^2} - 4x + 3}}{{{{\left( {{x^2} - x + 1} \right)}^2}}}\\y'' = \dfrac{{\left( { - 2x - 4} \right){{\left( {{x^2} - x + 1} \right)}^2} - \left( { - {x^2} - 4x + 3} \right)2\left( {{x^2} - x + 1} \right)\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^4}}}\\\,\,\,\,\,\,\, = \dfrac{{\left( { - 2x - 4} \right)\left( {{x^2} - x + 1} \right) - 2\left( { - {x^2} - 4x + 3} \right)\left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^3}}}\\\,\,\,\,\,\,\, = \dfrac{{ - 2{x^3} + 2{x^2} - 2x - 4{x^2} + 4x - 4 + 4{x^3} + 16{x^2} - 12x - 2{x^2} - 8x + 6}}{{{{\left( {{x^2} - x + 1} \right)}^3}}}\\\,\,\,\,\,\,\, = \dfrac{{2{x^3} + 12{x^2} - 18x + 2}}{{{{\left( {{x^2} - x + 1} \right)}^3}}}\end{array}\)
c) \(y = \dfrac{{{x^2} - 4x + 3}}{{x - 2}}\)
\(\begin{array}{l}y' = \dfrac{{\left( {2x - 4} \right)\left( {x - 2} \right) - \left( {{x^2} - 4x + 3} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\\,\,\,\,\,\, = \dfrac{{2{x^2} - 8x + 8 - {x^2} + 4x - 3}}{{{{\left( {x - 2} \right)}^2}}}\\\,\,\,\,\,\, = \dfrac{{{x^2} - 4x + 5}}{{{{\left( {x - 2} \right)}^2}}} = 1 + \dfrac{1}{{{{\left( {x - 2} \right)}^2}}}\\y'' = \dfrac{{ - 2\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^4}}} = - \dfrac{2}{{{{\left( {x - 2} \right)}^3}}}\end{array}\)
d) \(y = \dfrac{{x + 1}}{{{x^2} + 5x + 6}} = \dfrac{{ - 1}}{{x + 2}} + \dfrac{2}{{x + 3}}\)
\(\begin{array}{l}
y' = \dfrac{1}{{{{\left( {x + 2} \right)}^2}}} - \dfrac{2}{{{{\left( {x + 3} \right)}^2}}}\\
y'' = \dfrac{{ - 2\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^4}}} + \dfrac{{2.2\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^4}}} = \dfrac{{ - 2}}{{{{\left( {x + 2} \right)}^3}}} + \dfrac{4}{{{{\left( {x + 3} \right)}^3}}}
\end{array}\)
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