Tính đạo hàm của hs sau: a) y= x-1/5x-2 b) y= 2x+3/7+3x c) y= x^2+2x+3/3-4x d) y= x^2+7x+3/x^2-3x

Đáp án:

$\begin{array}{l}
a)y = \frac{{x - 1}}{{5x - 2}}\\
 \Rightarrow y' = \frac{{\left( {x - 1} \right)'.\left( {5x - 2} \right) - \left( {5x - 2} \right)'.\left( {x - 1} \right)}}{{{{\left( {5x - 2} \right)}^2}}}\\
 = \frac{{5x - 2 - 5x + 5}}{{{{\left( {5x - 2} \right)}^2}}}\\
 = \frac{3}{{{{\left( {5x - 2} \right)}^2}}}\\
b)y = \frac{{2x + 3}}{{7 + 3x}} = \frac{{2x + 3}}{{3x + 7}}\\
 \Rightarrow y' = \frac{{2.\left( {3x + 7} \right) - 3.\left( {3x + 3} \right)}}{{{{\left( {3x + 7} \right)}^2}}}\\
 = \frac{5}{{{{\left( {3x + 7} \right)}^2}}}\\
c)y = \frac{{{x^2} + 2x + 3}}{{3 - 4x}}\\
 \Rightarrow y' = \frac{{\left( {2x + 2} \right).\left( {3 - 4x} \right) - \left( { - 4} \right).\left( {{x^2} + 2x + 3} \right)}}{{{{\left( {3 - 4x} \right)}^2}}}\\
 = \frac{{ - 4{x^2} + 6x + 18}}{{{{\left( {3 - 4x} \right)}^2}}}\\
d)y = \frac{{{x^2} + 7x + 3}}{{{x^2} - 3x}}\\
 \Rightarrow y' = \frac{{\left( {2x + 7} \right)\left( {{x^2} - 3x} \right) - \left( {2x - 3} \right).\left( {{x^2} + 7x + 3} \right)}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\
 = \frac{{ - 8{x^2} - 6x + 9}}{{{{\left( {{x^2} - 3x} \right)}^2}}}
\end{array}$