A: \({1 \over {12}}(5\sqrt 2 + 6\sqrt 5 {\rm{ - }}1)\)
B: \({1 \over {12}}(6\sqrt 5 + 5\sqrt 2 {\rm{ - }}1)\)
C: \({1 \over {12}}(5\sqrt 5 + 6\sqrt 2 {\rm{ - }}1)\)
D: \({1 \over {12}}(5\sqrt 5 + 6\sqrt 2 + 1)\)
举一反三
- 计算\(\int_L {\sqrt y } ds\),其中\(L\)是抛物线\(y=x^2\)上点\((0,0)\)与\((1,1)\)之间的一段弧。 A: \({1 \over {12}}(6\sqrt 5 - 1)\) B: \({1 \over {12}}(5\sqrt 6 - 1)\) C: \({1 \over {12}}(5\sqrt 5 - 1)\) D: \({1 \over {12}}(5\sqrt 5 + 1)\)
- 函数\(y = \sqrt {1{\rm{ - }}x} \)的导数为( ). A: \({\rm{ - }}{1 \over {2\sqrt {1{\rm{ - }}x} }}\) B: \({1 \over {2\sqrt {1{\rm{ - }}x} }}\) C: \({1 \over {\sqrt {1{\rm{ - }}x} }}\) D: \( - {1 \over {\sqrt {1{\rm{ - }}x} }}\)
- 求函数$y = {{1 + \root 3 \of {{x^2}} - \sqrt {2x} } \over {\sqrt x }}$的导数$y' = $( ) A: $ {1 \over 2}{x^{ - {3 \over 2}}} + {1 \over 6}{x^{ - {5 \over 6}}}$ B: $ - {1 \over 2}{x^{ - {3 \over 2}}} + {1 \over 6}{x^{ - {5 \over 6}}}$ C: ${1 \over 2}{x^{ - {3 \over 2}}} - {1 \over 6}{x^{ - {5 \over 6}}}$ D: ${1 \over 3}{x^{ - {3 \over 2}}} - {1 \over 6}{x^{ - {5 \over 6}}}$
- 求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$
- 函数\(y = \arcsin x\)的导数为( ). A: \( - {1 \over {\sqrt {1 + {x^2}} }}\) B: \({1 \over {\sqrt {1 + {x^2}} }}\) C: \({1 \over {\sqrt {1 - {x^2}} }}\) D: \( - {1 \over {\sqrt {1 - {x^2}} }}\)
内容
- 0
\(已知L是抛物线y=x^2上点O(0,0)与点A(1,1)之间的一段弧,则\int_{L}\sqrt{y}ds=(\,)\) A: \[\frac{1}{12}(5\sqrt{5}-1)\] B: \[\frac{1}{12}(3\sqrt{3}-1)\] C: \[\frac{1}{13}(5\sqrt{5}-1)\] D: \[\frac{1}{13}(3\sqrt{3}-1)\]
- 1
函数\(y = {\left( {\arcsin x} \right)^2}\)的导数为( ). A: \(2\arcsin x{1 \over {\sqrt {1 - {x^2}} }}\) B: \( - 2\arcsin x{1 \over {\sqrt {1 - {x^2}} }}\) C: \(2\arcsin x{1 \over {\sqrt {1 + {x^2}} }}\) D: \( - 2\arcsin x{1 \over {\sqrt {1 + {x^2}} }}\)
- 2
\( \int_0^1 {dx} \int_ { { x^2}}^x { { {\left( { { x^2} + {y^2}} \right)}^{ - {1 \over 2}}}dy} \) =( ) A: \( \sqrt 2 + 1 \) B: \( \sqrt 2 - 1 \) C: \( \sqrt 2 \) D: \( \pi \)
- 3
\( d(\arcsin x) \)=( ). A: \( {1 \over {\sqrt {1 - {x^2}} }}dx \) B: \( - {1 \over {\sqrt {1 - {x^2}} }}dx \) C: 0 D: 1
- 4
函数\( y = x + \sqrt {1 - x} \)的极大值为( ) A: 0 B: 1 C: \( {5 \over 4} \) D: \( {4 \over 5} \)