函数$y=^{x^2,x\ge 1}_{2x,x<1}$在x=1处可导
举一反三
- 函数\(y = \ln \left( {1 + {x^2}} \right)\)的导数为( ). A: \( { { 2x} \over {1 + {x^2}}}\) B: \( - { { 2x} \over {1 + {x^2}}}\) C: \( { { 2x} \over {1 - {x^2}}}\) D: \( - { { 2x} \over {1 - {x^2}}}\)
- 函数\(y = {x^2} - \ln x\)的导数为（ ）. A: \(2x - {1 \over x}\) B: \(2x + {1 \over x}\) C: \( - 2x + {1 \over x}\) D: \( - 2x - {1 \over x}\)
- 如下程序的输出是什么？ #include <stdio.h> void Swap (int x, int y);int main() { int x = 1; int y = 2; printf ("x=%d,y=%d\n", x, y); Swap (x, y); printf ("x=%d,y=%d", x, y); } void Swap (int x, int y) { int temp; temp = x; x = y; y = temp; printf ("x=%d,y=%d\n", x, y); }? x=1,y=2x=2,y=1x=2,y=1|x=1,y=2x=1,y=2x=2,y=1|x=1,y=2x=2,y=1x=1,y=2|x=1,y=2x=1,y=2x=1,y=2
- 如下程序的输出是什么？ #include [stdio.h] void Swap (int x, int y); int main() { int x = 1; int y = 2; printf ("x=%d,y=%d\n", x, y); Swap (x, y); printf ("x=%d,y=%d", x, y); } void Swap (int x, int y) { int temp; temp = x; x = y; y = temp; printf ("x=%d,y=%d\n", x, y); } A: x=1,y=2x=2,y=1x=2,y=1 B: x=1,y=2x=1,y=2x=2,y=1 C: x=1,y=2x=2,y=1x=1,y=2 D: x=1,y=2x=1,y=2x=1,y=2
- 已知\( y = \ln (1 + {x^2}) \)，则\( y' \)为( ). A: \( { { 2x} \over {1 + {x^2}}} \) B: \( {x \over {1 + {x^2}}} \) C: \( {1 \over {1 + {x^2}}} \) D: \( { { {x^2}} \over {1 + {x^2}}} \)