Tìm đạo hàm y’ của hàm số
a, y = (x^2 + 2x – 3) / (x + 2)
b, y = 1/ [căn(x^2 + 1)]
c, y = (-x^2 + 2x – 3) / (x – 2)
Tìm đạo hàm y’ của hàm số
a, y = (x^2 + 2x – 3) / (x + 2)
b, y = 1/ [căn(x^2 + 1)]
c, y = (-x^2 + 2x – 3) / (x – 2)
Đáp án:
$\begin{array}{l}
a)y = \dfrac{{{x^2} + 2x – 3}}{{x + 2}}\\
y’ = \dfrac{{\left( {2x + 2} \right).\left( {x + 2} \right) – {x^2} – 2x + 3}}{{{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{2{x^2} + 6x + 4 – {x^2} – 2x + 3}}{{{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{{x^2} + 4x + 7}}{{{{\left( {x + 2} \right)}^2}}}\\
b)y = \dfrac{1}{{\sqrt {{x^2} + 1} }}\\
y’ = \dfrac{{0.\sqrt {{x^2} + 1} – \left( {\sqrt {{x^2} + 1} } \right)’}}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2}}}\\
= \dfrac{{ – 2x.\dfrac{1}{{2\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = \dfrac{{ – x}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}\\
c)y = \dfrac{{ – {x^2} + 2x – 3}}{{x – 2}}\\
\Leftrightarrow y’ = \dfrac{{\left( { – 2x + 2} \right)\left( {x – 2} \right) + {x^2} – 2x + 3}}{{{{\left( {x – 2} \right)}^2}}}\\
= \dfrac{{ – 2{x^2} + 4x + 2x – 4 + {x^2} – 2x + 3}}{{{{\left( {x – 2} \right)}^2}}}\\
= \dfrac{{ – {x^2} + 4x – 1}}{{{{\left( {x – 2} \right)}^2}}}
\end{array}$