Tính đạo hàm sau :a, y= (3x-1)^4 b, căn x trên 1-2x 30/09/2021 Bởi Madelyn Tính đạo hàm sau :a, y= (3x-1)^4 b, căn x trên 1-2x
Đáp án: $\begin{array}{l}a)y = {\left( {3x – 1} \right)^4}\\ \Rightarrow y’ = 4.\left( {3x – 1} \right)’.{\left( {3x – 1} \right)^3}\\ = 4.3.{\left( {3x – 1} \right)^3}\\ = 12.{\left( {3x – 1} \right)^3}\\b)y = \dfrac{{\sqrt x }}{{1 – 2x}}\\ \Rightarrow y’ = \dfrac{{\left( {\sqrt x } \right)’.\left( {1 – 2x} \right) – \sqrt x .\left( {1 – 2x} \right)’}}{{{{\left( {1 – 2x} \right)}^2}}}\\ = \dfrac{{\dfrac{1}{{2\sqrt x }}.\left( {1 – 2x} \right) – \sqrt x .\left( { – 2} \right)}}{{{{\left( {1 – 2x} \right)}^2}}}\\ = \dfrac{{\dfrac{{1 – 2x + 2\sqrt x .2\sqrt x }}{{2\sqrt x }}}}{{{{\left( {1 – 2x} \right)}^2}}}\\ = \dfrac{1}{{2\sqrt x {{\left( {1 – 2x} \right)}^2}}}\end{array}$ Bình luận
Đáp án:
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
a)y = {\left( {3x – 1} \right)^4}\\
\Rightarrow y’ = 4.\left( {3x – 1} \right)’.{\left( {3x – 1} \right)^3}\\
= 4.3.{\left( {3x – 1} \right)^3}\\
= 12.{\left( {3x – 1} \right)^3}\\
b)y = \dfrac{{\sqrt x }}{{1 – 2x}}\\
\Rightarrow y’ = \dfrac{{\left( {\sqrt x } \right)’.\left( {1 – 2x} \right) – \sqrt x .\left( {1 – 2x} \right)’}}{{{{\left( {1 – 2x} \right)}^2}}}\\
= \dfrac{{\dfrac{1}{{2\sqrt x }}.\left( {1 – 2x} \right) – \sqrt x .\left( { – 2} \right)}}{{{{\left( {1 – 2x} \right)}^2}}}\\
= \dfrac{{\dfrac{{1 – 2x + 2\sqrt x .2\sqrt x }}{{2\sqrt x }}}}{{{{\left( {1 – 2x} \right)}^2}}}\\
= \dfrac{1}{{2\sqrt x {{\left( {1 – 2x} \right)}^2}}}
\end{array}$