Tính đạo hàm 2x^2-4x+1/x-3

Đáp án:$y' = \dfrac{{2{x^2} - 12x + 11}}{{{{\left( {x - 3} \right)}^2}}}$

Giải thích các bước giải:

$\begin{array}{l}
y = \dfrac{{2{x^2} - 4x + 1}}{{x - 3}}\\
 \Leftrightarrow y' = \dfrac{{\left( {2{x^2} - 4x + 1} \right)'\left( {x - 3} \right) - \left( {2{x^2} - 4x + 1} \right).\left( {x - 3} \right)'}}{{{{\left( {x - 3} \right)}^2}}}\\
 = \dfrac{{\left( {4x - 4} \right).\left( {x - 3} \right) - \left( {2{x^2} - 4x + 1} \right).1}}{{{{\left( {x - 3} \right)}^2}}}\\
 = \dfrac{{4{x^2} - 12x - 4x + 12 - 2{x^2} + 4x - 1}}{{{{\left( {x - 3} \right)}^2}}}\\
 = \dfrac{{2{x^2} - 12x + 11}}{{{{\left( {x - 3} \right)}^2}}}
\end{array}$