Hàm số $y = \sqrt[5]{{{{\left( {{x^2} + 1} \right)}^2}}}$có đạo hàm là.

$D = \mathbb{R}\backslash \left\{ { - 1;4} \right\}$.     

$D = \mathbb{R}$.

$D = \left( { - \infty ; - 1} \right) \cup \left( {4; + \infty } \right)$.

$D = \left( { - \infty ; - 1} \right] \cup \left[ {4; + \infty } \right)$.

$y = {x^5}$

$y = {x^{ - 1}}$

$y = {x^{\sqrt 2 }}$

$y = {\left( {{x^{\sqrt 2 }}} \right)^2}$

$x > 0$

mọi $x$

$x < 1$

$x \ne 0$

$y' = \dfrac{4}{3}{x^{ - \frac{1}{3}}}$

$y' = \dfrac{4}{3}{x^{\frac{1}{3}}}$

$y' = \dfrac{3}{7}{x^{\frac{7}{3}}}$

$y' = \dfrac{3}{4}{x^{\frac{1}{3}}}$

$\dfrac{5}{3}{\left( {2x} \right)^{\dfrac{2}{3}}}$

$\dfrac{{10}}{3}{\left( {2x} \right)^{\dfrac{2}{3}}}$

$\dfrac{5}{6}{\left( {2x} \right)^{\dfrac{2}{3}}}$

$\dfrac{{10}}{3}{\left( {2x} \right)^{\dfrac{5}{3}}}$

$\left( {\sqrt[n]{x}} \right)' = \dfrac{1}{n}{x^{ - \dfrac{{n - 1}}{n}}}$

$\left( {\sqrt[n]{x}} \right)' = \dfrac{1}{n}{x^{\dfrac{{n - 1}}{n}}}$         

$\left( {\sqrt[n]{x}} \right)' = n{x^{ - \dfrac{{n - 1}}{n}}}$            

$\left( {\sqrt[n]{x}} \right)' = \dfrac{1}{n}{x^{ - \dfrac{{n + 1}}{n}}}$

$y = {x^{ - 4}}$.

$y = {x^4}$.

$y = {x^{ - \dfrac{3}{4}}}$.

$y = \sqrt[3]{x}$.