Đề thi THPT QG - 2021 - mã 103

Trên khoảng \(\left( {0; + \infty } \right)\), đạo hàm của hàm số \(y = {x^{\frac{4}{3}}}\) 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 \(x\)

\(x < 1\)

\(x \ne 0\)

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

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

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

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

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

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

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

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

\(y = {x^{ - 4}}\).

\(y = {x^4}\).

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

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