Bài 58 sách giải tích 12 nâng cao trang 117

Tìm đạo hàm của các hàm số sau:

LG a

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

Lời giải chi tiết:

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

LG b

\(y = \root 5 \of {{{\ln }^3}5x} \)

Phương pháp giải:

Áp dụng: \(\left( {\root n \of u } \right)' = {u' \over {n\root n \of {{u^{n - 1}}} }}\)

Lời giải chi tiết:

\(y' = {{\left( {{{\ln }^3}5x} \right)'} \over {5\root 5 \of {{{\left( {{{\ln }^3}5x} \right)}^4}} }} \)

\( = \frac{{3{{\ln }^2}5x.\left( {\ln 5x} \right)'}}{{5\sqrt[5]{{{{\left( {{{\ln }^3}5x} \right)}^4}}}}} \) \(= \frac{{3{{\ln }^2}5x.\frac{{\left( {5x} \right)'}}{{5x}}}}{{5\sqrt[5]{{{{\ln }^{12}}5x}}}} \) \(= \frac{{3{{\ln }^2}5x.\frac{5}{{5x}}}}{{5\sqrt[5]{{{{\ln }^{10}}5x.{{\ln }^2}5x}}}} \) \(= \frac{{3{{\ln }^2}5x.\frac{1}{x}}}{{5{{\ln }^2}5x\sqrt[5]{{{{\ln }^2}5x}}}} \) \(= \frac{3}{{5x\sqrt[5]{{{{\ln }^2}5x}}}}\)

Cách khác:

\(\begin{array}{l}
y = \sqrt[5]{{{{\ln }^3}5x}} = {\left( {\ln 5x} \right)^{\frac{3}{5}}}\\
y' = \left[ {{{\left( {\ln 5x} \right)}^{\frac{3}{5}}}} \right]'\\
= \frac{3}{5}{\left( {\ln 5x} \right)^{\frac{3}{5} - 1}}\left( {\ln 5x} \right)'\\
= \frac{3}{5}{\left( {\ln 5x} \right)^{ - \frac{2}{5}}}.\frac{5}{{5x}}\\
= \frac{3}{5}.\frac{1}{{{{\left( {\ln 5x} \right)}^{\frac{2}{5}}}}}.\frac{1}{x}\\
= \frac{3}{{5x\sqrt[5]{{{{\ln }^2}5x}}}}
\end{array}\)

LG c

\(y = \root 3 \of {{{1 + {x^3}} \over {1 - {x^3}}}} \) 

Lời giải chi tiết:

Đặt \(u = {{1 + {x^3}} \over {1 - {x^3}}} \Rightarrow y = \sqrt[3]{u}\Rightarrow y' = {{u'} \over {3\root 3 \of {{u^2}} }}\)

\(\begin{array}{l}
u' = \frac{{\left( {1 + {x^3}} \right)'\left( {1 - {x^3}} \right) - \left( {1 + {x^3}} \right)\left( {1 - {x^3}} \right)'}}{{{{\left( {1 - {x^3}} \right)}^2}}}\\
= \frac{{3{x^2}\left( {1 - {x^3}} \right) - \left( {1 + {x^3}} \right)\left( { - 3{x^2}} \right)}}{{{{\left( {1 - {x^3}} \right)}^2}}}\\
= \frac{{3{x^2} - 3{x^5} + 3{x^2} + 3{x^5}}}{{{{\left( {1 - {x^3}} \right)}^2}}}\\
= \frac{{6{x^2}}}{{{{\left( {1 - {x^3}} \right)}^2}}}\\
\Rightarrow y' = \frac{{\frac{{6{x^2}}}{{{{\left( {1 - {x^3}} \right)}^2}}}}}{{3\sqrt[3]{{{{\left( {\frac{{1 + {x^3}}}{{1 - {x^3}}}} \right)}^2}}}}}
\end{array}\)

\(= {{2{x^2}} \over {{{\left( {1 - {x^3}} \right)}^2}}}.{1 \over {\root 3 \of {{{\left( {{{1 + {x^3}} \over {1 - {x^3}}}} \right)}^2}} }} \)

\( = \frac{{2{x^2}}}{{\sqrt[3]{{{{\left( {1 - {x^3}} \right)}^6}.\frac{{{{\left( {1 + {x^3}} \right)}^2}}}{{{{\left( {1 - {x^3}} \right)}^2}}}}}}}\)

\(= {{2{x^2}} \over {\root 3 \of {{{\left( {1 - {x^3}} \right)}^4}{{\left( {1 + {x^3}} \right)}^2}} }}\)

LG d

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

Lời giải chi tiết:

\(y = {\left( {\frac{x}{b}} \right)^a}.{\left( {\frac{a}{x}} \right)^b} = \frac{{{x^a}}}{{{b^a}}}.\frac{{{a^b}}}{{{x^b}}} = \frac{{{a^b}}}{{{b^a}}}.{x^{a - b}}\)

\(y' = \left( {\frac{{{a^b}}}{{{b^a}}}{x^{a - b}}} \right)' = \frac{{{a^b}}}{{{b^a}}}.\left( {a - b} \right)\left( {{x^{a - b - 1}}} \right)\)