Đáp án:
\(y'\left( 4 \right) = \dfrac{{17}}{{100}}\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = \dfrac{{\left( {2x - 3} \right).\dfrac{1}{{2\sqrt {{x^2} - 3x} }}\left( {x + 1} \right) - \sqrt {{x^2} - 3x} }}{{{{\left( {x + 1} \right)}^2}}}\\
= \dfrac{{\left( {2x - 3} \right)\left( {x + 1} \right) - 2\left( {{x^2} - 3x} \right)}}{{2{{\left( {x + 1} \right)}^2}\sqrt {{x^2} - 3x} }}\\
= \dfrac{{2{x^2} - x - 3 - 2{x^2} + 6x}}{{2{{\left( {x + 1} \right)}^2}\sqrt {{x^2} - 3x} }}\\
= \dfrac{{5x - 3}}{{2{{\left( {x + 1} \right)}^2}\sqrt {{x^2} - 3x} }}\\
y'\left( 4 \right) = \dfrac{{5.4 - 3}}{{2{{\left( {4 + 1} \right)}^2}\sqrt {{4^2} - 3.4} }} = \dfrac{{17}}{{100}}\\
y'\left( 5 \right) = \dfrac{{5.5 - 3}}{{2{{\left( {5 + 1} \right)}^2}\sqrt {{5^2} - 3.5} }} = 0,09662515073
\end{array}\)
