作业帮非常感激.1、求导数：函数,不定积分
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/12 06:21:12
非常感激.1、求导数：函数,不定积分
放大之后把图片另存为再打开就可以看清楚了,
放大之后把图片另存为再打开就可以看清楚了,
(1)：y = x²sin2x
y' = sin2x • 2x + x² • 2cos2x
= 2xsin2x + 2x²cos2x
= 2x(sin2x + xcos2x)
(2)：∫ xlnx dx = ∫ lnx d(x²/2)
= (x²/2)lnx - (1/2)∫ x² d(lnx)
= (1/2)x²lnx - (1/2)∫ x² • 1/x dx
= (1/2)x²lnx - (1/2)∫ x dx
= (1/2)x²lnx - (1/2)(x²/2) + C
= (x²/4)(2lnx - 1) + C
(3)：∫ (2 - 3x)¹² dx
令u = 2 - 3x,du = (2)'dx - (3x)'dx = -3dx
原式= ∫ u¹² • du/(-3)
= (-1/3) • u¹³/13 + C
= (-1/39)(2 - 3x)¹³ + C
(4)：
f(x) = 3x³ - 9x + 7
f'(x) = 9x² - 9
f''(x) = 18x
f'(x) = 0 => x = ±1
f''(-1) < 0,取得极大值,f''(1) > 0,取得极小值
极大值f(-1) = 13,极小值f(1) = 1
在端点,f(-3) = -47,f(3) = 61
故在x∈[-3,3]中,最大值为61,最小值为-47
再问： 非常感谢你的帮忙，请再帮忙做几道题，实太太感谢你了！
再答： (1)：y = cosx，y' = - sinx，y'' = - cosx，y''' = sinx (2)：∫ (2^x + x²) = ∫ 2^x dx + ∫ x² dx = (2^x)/ln2 + x³/3 + C (3)：y = ax^(b + c) y' = a(b + c)x^(b + c - 1) (4)： lim(x→2) (x² - x + k)/(x - 2) = 3，分子能够因式分解，并且能约去x - 2项 所以(x² - x + k) = (x - 2)(x + m) x² - x + k = x² + (m - 2)x - 2m { m - 2 = -1 => m = 1 { k = - 2m => k = (-2)(1) = -2 所以k = -2 (5)：∫ dx/(3 + 2x) = (1/2)∫ d(2x)/(3 + 2x) = (1/2)∫ d(3 + 2x)/(3 + 2x) = (1/2)ln|3 + 2x| + C (6)：d(？) = (2^x)ln2 dx d(？)/dx = (2^x)ln2 ？= (ln2)∫ 2^x dx = (ln2) • (2^x)/ln2 + C = 2^x + C 故d(2^x + C) = (2^x)ln2 dx，C为任意的常数
再问： 实在太感谢你了！还有到你几道题能再麻烦你吗?希望能得到你的帮助,谢谢!
再答： (1)：∫ lnx/x³ dx = ∫ lnx d[-1/(2x²)] = - lnx/2x² + (1/2)∫ x² d(lnx) = -lnx/2x² + (1/2)∫ x dx = -lnx/2x² + x²/4 + C (2)：y = 1/x，y' = -1/x²，y(4) = 1/4 y' = -1/16 y = (-1/16)(x - 4) + 1/4 y = (-1/16)x + 1/2 (3)：y = x³ - 3x² - 9x + 10 y' = 3x² - 6x - 9 = 3(x² - 2x - 3) y'' = 6x - 6 = 6(x - 1) y' = 0 => x = 3 或 x = -1 y''(-1) < 0，取得极大值，y''(3) > 0，取得极小值 极大值f(-1) = 15，极小值f(3) = -17 在端点：f(-2) = 8，f(6) = 64 所以在[-2，6]的最大值是64，最小值是-17 (4)：y = 2 + x - x²，抛物线开口向下 y' = -2x + 1 y' = 0 => x = 1/2 所以极大值f(1/2) = 9/4
y' = sin2x • 2x + x² • 2cos2x
= 2xsin2x + 2x²cos2x
= 2x(sin2x + xcos2x)
(2)：∫ xlnx dx = ∫ lnx d(x²/2)
= (x²/2)lnx - (1/2)∫ x² d(lnx)
= (1/2)x²lnx - (1/2)∫ x² • 1/x dx
= (1/2)x²lnx - (1/2)∫ x dx
= (1/2)x²lnx - (1/2)(x²/2) + C
= (x²/4)(2lnx - 1) + C
(3)：∫ (2 - 3x)¹² dx
令u = 2 - 3x,du = (2)'dx - (3x)'dx = -3dx
原式= ∫ u¹² • du/(-3)
= (-1/3) • u¹³/13 + C
= (-1/39)(2 - 3x)¹³ + C
(4)：
f(x) = 3x³ - 9x + 7
f'(x) = 9x² - 9
f''(x) = 18x
f'(x) = 0 => x = ±1
f''(-1) < 0,取得极大值,f''(1) > 0,取得极小值
极大值f(-1) = 13,极小值f(1) = 1
在端点,f(-3) = -47,f(3) = 61
故在x∈[-3,3]中,最大值为61,最小值为-47
再问： 非常感谢你的帮忙，请再帮忙做几道题，实太太感谢你了！
再答： (1)：y = cosx，y' = - sinx，y'' = - cosx，y''' = sinx (2)：∫ (2^x + x²) = ∫ 2^x dx + ∫ x² dx = (2^x)/ln2 + x³/3 + C (3)：y = ax^(b + c) y' = a(b + c)x^(b + c - 1) (4)： lim(x→2) (x² - x + k)/(x - 2) = 3，分子能够因式分解，并且能约去x - 2项 所以(x² - x + k) = (x - 2)(x + m) x² - x + k = x² + (m - 2)x - 2m { m - 2 = -1 => m = 1 { k = - 2m => k = (-2)(1) = -2 所以k = -2 (5)：∫ dx/(3 + 2x) = (1/2)∫ d(2x)/(3 + 2x) = (1/2)∫ d(3 + 2x)/(3 + 2x) = (1/2)ln|3 + 2x| + C (6)：d(？) = (2^x)ln2 dx d(？)/dx = (2^x)ln2 ？= (ln2)∫ 2^x dx = (ln2) • (2^x)/ln2 + C = 2^x + C 故d(2^x + C) = (2^x)ln2 dx，C为任意的常数
再问： 实在太感谢你了！还有到你几道题能再麻烦你吗?希望能得到你的帮助,谢谢!
再答： (1)：∫ lnx/x³ dx = ∫ lnx d[-1/(2x²)] = - lnx/2x² + (1/2)∫ x² d(lnx) = -lnx/2x² + (1/2)∫ x dx = -lnx/2x² + x²/4 + C (2)：y = 1/x，y' = -1/x²，y(4) = 1/4 y' = -1/16 y = (-1/16)(x - 4) + 1/4 y = (-1/16)x + 1/2 (3)：y = x³ - 3x² - 9x + 10 y' = 3x² - 6x - 9 = 3(x² - 2x - 3) y'' = 6x - 6 = 6(x - 1) y' = 0 => x = 3 或 x = -1 y''(-1) < 0，取得极大值，y''(3) > 0，取得极小值 极大值f(-1) = 15，极小值f(3) = -17 在端点：f(-2) = 8，f(6) = 64 所以在[-2，6]的最大值是64，最小值是-17 (4)：y = 2 + x - x²，抛物线开口向下 y' = -2x + 1 y' = 0 => x = 1/2 所以极大值f(1/2) = 9/4