Tìm đạo hàm \[y' = y'\left( x \right)\] của hàm số y = y(x) cho bởi phương trình tham số:

\[\left\{ {\begin{array}{*{20}{c}}{x = \ln \left( {1 + {t^2}} \right)}\\{y = 2t - 2\arctan t}\end{array}} \right.\]

\[{f^{\left( {2018} \right)}}\left( 2 \right) = - 2016!\]

\[{f^{\left( {2018} \right)}}\left( 2 \right) = - 2018!\]

\[{f^{\left( {2018} \right)}}\left( 2 \right) = - 2017!\]

\[{f^{\left( {2018} \right)}}\left( 2 \right) = - 2019!\]

a = 1; b = 0

a = 1; b = 1

a = 0; b = 1

a = - 1; b = 0

\[\frac{{2f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]

\[\frac{{{e^{f\left( {2x - 1} \right)}}f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]

\[\frac{{2.{e^{f\left( {2x - 1} \right)}}f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]

\[ - \frac{{2.{e^{f\left( {2x - 1} \right)}}f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]

\[\frac{{x + 1}}{2} + \frac{{{{\left( {x + 1} \right)}^3}}}{{16}} + \frac{3}{{256}}{\left( {x + 1} \right)^5}\]

\[\frac{{x + 1}}{2} + \frac{{{{\left( {x + 1} \right)}^3}}}{{16}} - \frac{3}{{256}}{\left( {x + 1} \right)^5}\]

\[\frac{{x + 1}}{2} + \frac{{{{\left( {x + 1} \right)}^3}}}{{16}}\]

\[\frac{1}{3}\left( {x + 1} \right)\left( {1 + \frac{x}{3} + \frac{{{x^2}}}{3}} \right)\]

1

0

2

+ ∞

- ∞

$\frac{-2}{3}$

0

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$-10d{x}^2$

2$d{x}^2$

2$e^{-1}d{x}^2$

-10$e^{-1}d{x}^2$

– 60

0

60

120

$\frac{5}{6}$

$\frac{7}{6}$

$\frac{3}{2}$

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$\frac{4}{15}$

$\frac{2}{15}$

$\frac{-4}{15}$

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