Tìm đạo hàm của mỗi hàm số sau :
\(\begin{array}{l}
a)y = {\left( {{x^7} + x} \right)^2}\\
b)y = ({x^2} + 1)(5 - 3{x^2})\\
c)y = \frac{{2x}}{{{x^2} - 1}}\\
d)y = \frac{{5x - 3}}{{{x^2} + x + 1}}\\
e)y = \frac{{{x^2} + 2x + 2}}{{x + 1}}\\
f)y = x(2x - 1)(3x + 2)
\end{array}\)
a)
\(\begin{array}{l}
y = {\left( {{x^7} + x} \right)^2} = {x^{14}} + 2{x^8} + {x^2}\\
\Rightarrow y\prime = 14{x^{13}} + 16{x^7} + 2x
\end{array}\)
b)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
y\prime = ({x^2} + 1)\prime (5 - 3{x^2})\\
+ ({x^2} + 1)(5 - 3{x^2})\prime
\end{array}\\
\begin{array}{l}
= 2x(5 - 3{x^2}) - 6x({x^2} + 1)\\
= 4x - 12{x^3}
\end{array}
\end{array}\)
c)
\(y\prime = \frac{{2({x^2} - 1) - 2x(2x)}}{{{{({x^2} - 1)}^2}}} = \frac{{ - 2({x^2} + 1)}}{{{{({x^2} - 1)}^2}}}\)
d)
\(\begin{array}{l}
y\prime = \frac{{5({x^2} + x + 1) - (5x - 3)(2x + 1)}}{{{{({x^2} + x + 1)}^2}}}\\
= \frac{{ - 5{x^2} + 6x + 8}}{{{{(x2 + x + 1)}^2}}}
\end{array}\)
e)
\(\begin{array}{l}
y\prime = \frac{{(2x + 2)(x + 1) - ({x^2} + 2x + 2)}}{{{{(x + 1)}^2}}}\\
= \frac{{{x^2} + 2x}}{{{{(x + 1)}^2}}}
\end{array}\)
f)
\(\begin{array}{l}
y = (2{x^2} - x)(3x + 2)\\
\Rightarrow y\prime = (4x - 1)(3x + 2) + {(2{x^2} - x)^3}\\
= 18{x^2} + 2x - 2
\end{array}\)
-- Mod Toán 11
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