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Bài tập 58 trang 117 SGK Toán 12 NC

Bài tập 58 trang 117 SGK Toán 12 NC

Tìm đạo hàm của các hàm số sau:

a) \(y = {\left( {2x + 1} \right)^\pi }\)

b) \(y = 5\sqrt[3]{{{{\ln }^3}5x}}\)

c) \(y = \sqrt[3]{{\frac{{1 + {x^3}}}{{1 - {x^3}}}}}\)

d) \(y = {\left( {\frac{x}{b}} \right)^a}{\left( {\frac{a}{x}} \right)^b}\) với a > 0, b > 0

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Hướng dẫn giải chi tiết

 
 

a) \(y' = 2\pi {\left( {2x + 1} \right)^{\pi  - 1}}\)

b) \(y' = \frac{{{{\left( {{{\ln }^3}5x} \right)}^\prime }}}{{5\sqrt[5]{{{{\left( {{{\ln }^3}5x} \right)}^4}}}}} = \frac{{3{{\ln }^2}5x}}{{5x\sqrt[5]{{{{\ln }^{12}}5x}}}}\)

c) Đặt \(u = \frac{{1 + {x^3}}}{{1 - {x^3}}};y\prime  = \frac{{u'}}{{3\sqrt[3]{{{u^2}}}}}\)

\(u' = \frac{{3{x^2}\left( {1 - {x^3}} \right) - 3{x^2}\left( {1 + {x^3}} \right)}}{{{{\left( {1 - {x^3}} \right)}^2}}} = \frac{{6{x^2}}}{{\left( {1 - {x^3}} \right)}}\)

Do đó:

\(\begin{array}{l}
y' = \frac{{2{x^2}}}{{{{\left( {1 - {x^3}} \right)}^2}}}.\frac{1}{{\sqrt[3]{{{{\left( {\frac{{1 + {x^3}}}{{1 - {x^3}}}} \right)}^2}}}}}\\
 = \frac{{2{x^2}}}{{\sqrt[3]{{{{\left( {1 - {x^3}} \right)}^4}{{\left( {1 + {x^3}} \right)}^2}}}}}
\end{array}\)

d)

\(\begin{array}{l}
y' = \left[ {{{\left( {\frac{x}{b}} \right)}^a}} \right]{\rm{'}}{\left( {\frac{a}{x}} \right)^b} + {\left( {\frac{x}{b}} \right)^a}\left[ {{{\left( {\frac{a}{x}} \right)}^b}} \right]'\\
 = \frac{a}{b}{\left( {\frac{x}{a}} \right)^{a - 1}}{\left( {\frac{a}{x}} \right)^b} + {\left( {\frac{x}{b}} \right)^a}b{\left( {\frac{a}{x}} \right)^{b - 1}}.\left( { - \frac{a}{{{x^2}}}} \right)\\
 = {\left( {\frac{x}{b}} \right)^a}{\left( {\frac{a}{x}} \right)^b}\frac{{a - b}}{x}
\end{array}\)

-- Mod Toán 12 HỌC247

 
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