Bài tập 23 trang 205 SGK Toán 11 NC
Bài tập 23 trang 205 SGK Toán 11 NC
Tính đạo hàm của mỗi hàm số sau
\(\begin{array}{l}
a)y = \frac{{2x + 3}}{{{x^2} - 5x + 5}}\\
b)y = \frac{1}{{{{({x^2} - x + 1)}^5}}}\\
c)y = {x^2} + x\sqrt x + 1\\
d)y = (x + 1){(x + 2)^2}{(x + 3)^3}\\
e)y = \sqrt {\frac{{{x^2} + 1}}{x}}
\end{array}\)
a)
\(\begin{array}{l}
y\prime = \frac{{2({x^2} - 5x + 5) - (2x + 3)(2x - 5)}}{{{{({x^2} - 5x + 5)}^2}}}\\
= \frac{{ - 2{x^2} - 6x + 25}}{{{{({x^2} - 5x + 5)}^2}}}
\end{array}\)
b)
\(\begin{array}{l}
y\prime = \frac{{ - 5({x^2} - x + 1)4(2x - 1)}}{{{{({x^2} - x + 1)}^{10}}}}\\
= \frac{{ - 5(2x - 1)}}{{{{({x^2} - x + 1)}^6}}}
\end{array}\)
c) \(y\prime = 2x + \sqrt x + x.\frac{1}{{2\sqrt x }} = 2x + \frac{3}{2}\sqrt x \)
d)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
y\prime = {(x + 2)^2}{(x + 3)^2}\\
+ (x + 1).2(x + 2){(x + 3)^3}\\
+ (x + 1){(x + 2)^2}3{(x + 3)^2}
\end{array}\\
{ = 2(x + 2){{(x + 3)}^2}(3{x^2} + 11x + 9)}
\end{array}\)
e) \(y\prime = \frac{{1 - \frac{1}{{{x^2}}}}}{{2\sqrt {x + \frac{1}{x}} }} = \frac{{{x^2} - 1}}{{2{x^2}\sqrt {\frac{{{x^2} + 1}}{x}} }}\)
-- Mod Toán 11
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