Giải câu 1 bài Ôn tập chương 5: Đạo hàm

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

\(y' = \left ({{{x^3}} \over 3} - {{{x^2}} \over 2} + x - 5\right )'\)

\(=\frac{3x^2}{3}-\frac{2x}{2}+1 ={x^2} - x + 1\)

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

\(=2.\frac{1}{x}-4.\frac{1}{x^2}+5.\frac{1}{x^3}-\frac{6}{7}.\frac{1}{x^4}\)

\(=\frac{1}{x^4}\left ( 2x^3-4x^2+5x-\frac{6}{7} \right )\)

\(y'=\left ( \frac{1}{x^4} \right )'.\left ( 2x^3-4x^2+5x-\frac{6}{7} \right )+\left ( \frac{1}{x^4} \right ).\left ( 2x^3-4x^2+5x-\frac{6}{7} \right )'\)

\(=-\frac{4}{x^5}.\left ( 2x^3-4x^2+5x-\frac{6}{7} \right )+\left ( \frac{1}{x^4} \right ).\left ( 6x^2-8x+5\right )\)

\(=-\frac{8}{x^2}+\frac{16}{x^3}-\frac{20}{x^4}+\frac{24}{7x^5}+\frac{6}{x^2}-\frac{8}{x^3}+\frac{5}{x^4}\)

\(=-\frac{2}{x^2}+\frac{8}{x^3}-\frac{15}{x^4}+\frac{24}{7x^5}\)

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

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

\(= {{(3{x^2} - 6x + 7)'4x - (4x)'(3{x^2} - 6x + 7)} \over {16{x^2}}} \)

\(= {{(6x - 6)4x - 4(3{x^2} - 6x + 7)} \over {16{x^2}}} \)

\(= {{3{x^2} - 7} \over {4{x^2}}} \)

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

\(y' = \left[ {({2 \over x} + 3x)(\sqrt x  - 1)} \right]'\)

\(=\left ( - {2 \over {{x^2}}} + 3 \right )(\sqrt x - 1) + {1 \over {2\sqrt x }}.\left ( {2 \over x} + 3x \right )\)

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

\(= - {{4\sqrt x } \over {{2x^2}}} + {4 \over {{2x^2}}} + \frac{12x^2\sqrt{x}}{2x^2}- \frac{6x^2}{2x^2} + {{2\sqrt x } \over {{2x^2}}} + {{3x^2\sqrt x } \over {2x^2}} \)

\(= {{9{x^2}\sqrt x - 6{x^2} - 2\sqrt x + 4} \over {2{x^2}}} \) 

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

\(y' = \left ( {{1 + \sqrt x } \over {1 - \sqrt x }} \right )' = {{{1 \over {2\sqrt x }}(1 - \sqrt x ) + {1 \over {2\sqrt x }}(1 + \sqrt x )} \over {{{(1 - \sqrt x )}^2}}}\)

\(= \frac{1-\sqrt{x}+1+\sqrt{x}}{2\sqrt{x}(1-\sqrt{x})^2}\)

\(=\frac{1}{\sqrt{x}(1-\sqrt{x})^2} \)

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

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

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

\(=\frac{-2x^3+6x^2+7x^2-21x-(-2x^3+14x^2+10x+3x^2-21x-15)}{(x^2-3x)^2}\)

\(= {{ - 4{x^2} - 10x + 15} \over {{{({x^2} - 3x)}^2}}} \)