Bài 1 trang 176 SGK Đại số và Giải tích 11

Tính đạo hàm của các hàm số sau

a) \(y = {{{x^3}} \over 3} - {{{x^2}} \over 2} + x - 5\)        

b) \(y = {2 \over x} - {4 \over {{x^2}}} + {5 \over {{x^3}}} - {6 \over {7{x^4}}}\)

c) \(y = {{3{x^2} - 6x + 7} \over {4x}}\)

d) \(y = ({2 \over x} + 3x)(\sqrt x  - 1)\)

e) \(y = {{1 + \sqrt x } \over {1 - \sqrt x }}\)

f) \(y = {{ - {x^2} + 7x + 5} \over {{x^2} - 3x}}\)

Lời giải

\(\begin{array}{l}
a)y' = {x^2} - x + 1\\
b)y' = - \dfrac{2}{{{x^2}}} + \dfrac{{4.2x}}{{{x^4}}} - \dfrac{{5.3{x^2}}}{{{x^6}}} + \dfrac{{6.4{x^3}}}{{7{x^8}}}\\
= - \dfrac{2}{{{x^2}}} + \dfrac{8}{{{x^3}}} - \dfrac{{15}}{{{x^4}}} + \dfrac{{24}}{{7{x^5}}}\\
c)y' = \dfrac{{\left( {6x - 6} \right).4x - 4\left( {3{x^2} - 6x + 7} \right)}}{{16{x^2}}}\\
 = \dfrac{{24{x^2} - 24x - 12{x^2} + 24x - 28}}{{16{x^2}}}\\
= \dfrac{{12{x^2} - 28}}{{16{x^2}}} = \dfrac{{3{x^2} - 7}}{{4{x^2}}}\\
d)y' = \left( { - \dfrac{2}{{{x^2}}} + 3} \right)\left( {\sqrt x - 1} \right) + \left( {\dfrac{2}{x} + 3x} \right).\dfrac{1}{{2\sqrt x }}\\
= \dfrac{{ - 2}}{{x\sqrt x }} + \dfrac{2}{{{x^2}}} + 3\sqrt x - 3 + \dfrac{1}{{x\sqrt x }} + \dfrac{3}{2}\sqrt x \\
\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 1}}{{x\sqrt x }} + \dfrac{2}{{{x^2}}} + \dfrac{{9\sqrt x }}{2} - 3\\
e)\,\,y' = \dfrac{{\dfrac{1}{{2\sqrt x }}\left( {1 - \sqrt x } \right) + \dfrac{1}{{2\sqrt x }}\left( {1 + \sqrt x } \right)}}{{{{\left( {1 - \sqrt x } \right)}^2}}}\\
\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{{\sqrt x {{\left( {1 - \sqrt x } \right)}^2}}}\\
f)\,\,y' = \dfrac{{\left( { - 2x + 7} \right)\left( {{x^2} - 3x} \right) - \left( {2x - 3} \right)\left( { - {x^2} + 7x + 5} \right)}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 2{x^3} + 13{x^2} - 21x + 2{x^3} - 17{x^2} + 11x + 15}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 4{x^2} - 10x + 15}}{{{{\left( {{x^2} - 3x} \right)}^2}}}
\end{array}\)