11 年级数学 第 13 章：极限和导数的 NCERT 解决方案

练习 13.1

计算练习1到22中的以下极限。

1. lim_{x → 3} x + 3

解决方案

代入 x = 3，

lim_{x → 3} x + 3 = 3 + 3 = 6

2. lim_{x → π} (x - 22/7)

解决方案

代入 x = π，

lim_{x → π} (x - 22/7) = π - 22/7

3. lim_{r → 1} πr²

解决方案

代入 r = 1，

lim_{r → 1} πr² = π(1)² = π

4. lim_{x → 4} (4x + 3)/(x - 2)

解决方案

代入 x = 4，

lim_{x → 4} (4x + 3)/(x - 2) = (4(4) + 3)/(4 - 2)

= (16 + 3)/2 = 19/2

5. lim_{x → -1} (x¹⁰ + x⁵ + 1)/(x - 1)

解决方案

代入 x = -1，

lim_{x → -1} (x¹⁰ + x⁵ + 1)/(x - 1) = ((-1)¹⁰ + (-1)⁵ + 1)/(-1 - 1)

= (1 - 1 + 1)/(-2)

= -1/2

6. lim_{x → 0} ((x + 1)⁵ - 1)/x

解决方案

代入 x = 0，

lim_{x → 0} ((x + 1)⁵ - 1)/x = ((0 + 1)⁵ - 1)/0

分母为0。因此，给定极限未定义。

设 x + 1 = y ⇒ x = y - 1

因此，

lim_{y → 1} (y⁵ - 1)/(y - 1)

= lim_{y → 1} (y⁵ - 1⁵)/(y - 1)

我们知道 lim_{x → a} (xⁿ - aⁿ)/(x - a) = na^{n - 1}。因此，

lim_{y → 1} (y⁵ - 1⁵)/(y - 1)

= 5(1)^{5 - 1} = 5(1)⁴ = 5

7. lim_{x → 2} (3x² - x - 10)/(x² - 4)

解决方案

代入 x = 2，

lim_{x → 2} (3x² - x - 10)/(x² - 4) = (3(2)² - 2 - 10)/(2² - 4)

= (3(4) - 12)/(4 - 4)

= (12 - 12)/(4 - 4) = 0/0

极限为0/0。因此，

将 lim_{x → 2} (3x² - x - 10)/(x² - 4) 的分子和分母因式分解

= lim_{x → 2} (3x² - 6x + 5x - 10)/(x² - 2²)

= lim_{x → 2} (3x(x - 2) + 5(x - 2))/(x + 2)(x - 2)

= lim_{x → 2} (x - 2)(3x + 5)/(x + 2)(x - 2)

= lim_{x → 2} (3x + 5)/(x + 2)

现在，代入 x = 2，

lim_{x → 2} (3x + 5)/(x + 2) = (3(2) + 5)/(2 + 2)

= (6 + 5)/4 = 11/4

8. lim_{x → 3} (x⁴ - 81)/(2x² - 5x - 3)

解决方案

代入 x = 3，

lim_{x → 3} (x⁴ - 81)/(2x² - 5x - 3) = (3⁴ - 81)/(2(3)² - 5(3) - 3)

= (81 - 81)/(2(9) - 15 - 3)

= (81 - 81)/(18 - 18) = 0/0

极限为0/0。因此，

将 lim_{x → 3} (x⁴ - 81)/(2x² - 5x - 3) 的分子和分母因式分解

= lim_{x → 3} (x⁴ - 3⁴)/(2x² - 6x + x - 3)

= lim_{x → 3} ((x²)² - 9²)/(2x(x - 3) + 1(x - 3))

= lim_{x → 3} (x² - 9)(x² + 9)/(x - 3)(2x + 1)

= lim_{x → 3} (x² - 3²)(x² + 9)/(x - 3)(2x + 1)

= lim_{x → 3} (x - 3)(x + 3)(x² + 9)/(x - 3)(2x + 1)

= lim_{x → 3} (x + 3)(x² + 9)/(2x + 1)

现在，代入 x = 3，

lim_{x → 3} (x + 3)(x² + 9)/(2x + 1) = (3 + 3)(3² + 9)/(2(3) + 1)

= (6)(9 + 9)/(6 + 1)

= (6)(18)/(7)

= 108/7

9. lim_{x → 0} (ax + b)/(cx + 1)

解决方案

代入 x = 0，

lim_{x → 0} (ax + b)/(cx + 1) = (0 + b)/(0 + 1)

= b/1 = b

10. lim_{z → 1} (z^1/3 - 1)/(z^1/6 - 1)

解决方案

代入 z = 1，

lim_{z → 1} (z^1/3 - 1)/(z^1/6 - 1) = (1^1/3 - 1)/(1^1/6 - 1)

= (1 - 1)/(1 - 1) = 0/0

极限为0/0。因此，

将 lim_{z → 1} (z^1/3 - 1)/(z^1/6 - 1) 的分子和分母因式分解

= lim_{z → 1} ((z^1/6)² - 1²)/(z^1/6 - 1)

= lim_{z → 1} (z^1/6 - 1)(z^1/6 + 1)/(z^1/6 - 1)

= lim_{z → 1} z^1/6 + 1

现在，代入 z = 1，

lim_{z → 1} z^1/6 + 1 = 1^1/6 + 1 = 1 + 1

= 2

11. lim_{x → 1} (ax² + bx + c)/(cx² + bx + a)，a + b + c ≠ 0

解决方案

代入 x = 1，

lim_{x → 1} (ax² + bx + c)/(cx² + bx + a) = (a(1)² + b(1) + c)/(c(1)² + b(1) + a)

= (a + b + c)/(a + b + c) = 1

12. lim_{x → -2} (1/x + 1/2)/(x + 2)

解决方案

代入 x = -2，

lim_{x → -2} (1/x + 1/2)/(x + 2) = (1/(-2) + 1/2)/(-2 + 2)

= (1/2 - 1/2)/(2 - 2)

= 0/0

极限为0/0。因此，

将 lim_{x → -2} (1/x + 1/2)/(x + 2) 的分子和分母因式分解

= lim_{x → -2} ((2 + x)/2x)/(x+2)

= lim_{x → -2} ((2 + x)/2x)/(x + 2)

= lim_{x → -2} 1/2x

现在，代入 x = -2，

lim_{x → -2} 1/2x = 1/2(-2) = -1/4

13. lim_{x → 0} sin ax/bx

解决方案

代入 x = 0，

lim_{x → 0} sin ax/bx = sin a(0)/b(0)

= sin 0/b(0)

= 0/0

极限为0/0。因此，

将 lim_{x → 0} sin ax/bx 的分子和分母因式分解

表达式乘以并除以 a

lim_{x → 0} sin ax/bx × a/a

= lim_{x → 0} sin ax/ax × a/b

= a/b × lim_{x → 0} sin ax/ax

我们知道

lim_{x → 0} sin x/x = 1

因此，

a/b × lim_{x → 0} sin ax/ax = 1 × a/b

= a/b

14. lim_{x → 0} sin ax/sin bx，a, b ≠ 0

解决方案

代入 x = 0，

lim_{x → 0} sin ax/sin bx = sin a(0)/sin b(0)

= sin 0/sin 0

= 0/0

极限为0/0。因此，

将 lim_{x → 0} sin ax/sin bx 的分子和分母因式分解

分子乘以并除以 ax，分母乘以并除以 bx

lim_{x → 0} (sin ax/ax × ax)/(sin bx/bx × bx)

= a/b × {lim_{ax → 0} (sin ax/ax)}/{lim_{bx → 0} (sin bx/bx)}

我们知道

lim_{x → 0} sin x/x = 1

因此，

a/b × {lim_{ax → 0} (sin ax/ax)}/{lim_{bx → 0} (sin bx/bx)} = a/b × (1)/(1)

= a/b

15. lim_{x → π} sin (π - x)/π(π - x)

解决方案

代入 x = π，

lim_{x → π} sin (π - x)/π(π - x) = sin (π - π)/π(π - π)

= sin 0/π(0)

= 0/0

极限为0/0。因此，

将 lim_{x → π} sin (π - x)/π(π - x) 的分子和分母因式分解

= lim_{π - x → 0} sin (π - x)/π(π - x)

= 1/π × lim_{π - x → 0} sin (π - x)/(π - x)

我们知道

lim_{x → 0} sin x/x = 1

因此，

1/π × lim_{π - x → 0} sin (π - x)/(π - x) = 1/π × 1

= 1/π

16. lim_{x → 0} cos x/(π - x)

解决方案

代入 x = 0，

lim_{x → 0} cos x/(π - x) = cos 0/(π - 0)

= 1/π

17. lim_{x → 0} (cos 2x - 1)/(cos x - 1)

解决方案

代入 x = 0，

lim_{x → 0} (cos 2x - 1)/(cos x - 1) = (cos 2(0) - 1)/(cos 0 - 1)

= (cos 0 - 1)/(cos 0 - 1)

= (1 - 1)/(1 - 1)

= 0/0

极限为0/0。因此，

将 lim_{x → 0} (cos 2x - 1)/(cos x - 1) 的分子和分母因式分解

= lim_{x → 0} (1 - 2 sin² x -1)/(1 - 2 sin² x/2 -1)

= lim_{x → 0} (-2 sin² x)/(-2 sin² x/2)

= lim_{x → 0} sin² x/sin² x/2

分子乘以并除以 x²，分母乘以并除以 x²/4 = (x/2)²

lim_{x → 0} (sin² x/x² × x²)/[(sin² x/2)/(x/2)² × x²/4]

= 4 (lim_{x → 0} sin² x/x²)/(lim_{x → 0} sin² x/2/(x/2)²)

= 4 (lim_{x → 0} sin x/x)²/(lim_{x → 0} sin x/2/(x/2))²

我们知道

lim_{x → 0} sin x/x = 1

因此，

4 (lim_{x → 0} sin x/x)²/(lim^{x → 0} sin x/2/(x/2))² = 4 (1)/(1)

= 4

18. lim_{x → 0} (ax + x cos x)/(b sin x)

解决方案

代入 x = 0，

lim_{x → 0} (ax + x cos x)/(b sin x) = (a(0) + 0 cos 0)

= (0 + 0) = 0

极限为0/0。因此，

将 lim_{x → 0} (ax + x cos x)/(b sin x) 的分子和分母因式分解

= lim_{x → 0} x(a + cos x)/(b sin x)

= 1/b × lim_{x → 0} x(a + cos x)/sin x

= 1/b × lim_{x → 0} x/sin x × lim_{x → 0} (a + cos x)

= 1/b × 1/(lim_{x → 0} sin x/x) × lim_{x → 0} (a + cos x)

我们知道

lim_{x → 0} sin x/x = 1

因此，

1/b × 1/(lim_{x → 0} sin x/x) × lim_{x → 0} (a + cos x)

= 1/b × 1 × lim_{x → 0} (a + cos x)

现在，代入 x = 0，

1/b × 1 × lim_{x → 0} (a + cos x) = 1/b × lim_{x → 0} (a + cos 0)

= 1/b × (a + cos 0)

= (a + 1)/b

19. lim_{x → 0} x sec x

解决方案

代入 x = 0

lim_{x → 0} x sec x = 0 (sec 0) = 0

20. lim_{x → 0} (sin ax + bx)/(ax + sin bx)，a, b, a + b ≠ 0

解决方案

代入 x = 0，

= (sin a(0) + b(0))/(a(0) + sin b(0))

= (sin 0 + 0)/(0 + sin 0)

= sin 0/sin 0 = 0/0

极限为0/0。因此，

将 lim_{x → 0} (sin ax + bx)/(ax + sin bx) 的分子和分母因式分解

= lim_{x → 0} (sin (ax/ax)ax + bx)/(ax + sin (bx/bx)bx)

= (lim_{ax → 0} sin (ax/ax) × lim_{x → 0} ax + lim_{x → 0} bx)/(lim_{x → 0} ax + lim_{bx → 0} sin (bx/bx) × lim_{x → 0} bx)

我们知道

lim_{x → 0} sin x/x = 1

因此，

= (1 × lim_{x → 0} ax + lim_{x → 0} bx)/(lim_{x → 0} ax + 1 × lim_{x → 0} bx)

= (lim_{x → 0} ax + lim_{x → 0} bx)/(lim_{x → 0} ax + lim_{x → 0} bx)

= 1

21. lim_{x → 0} (cosec x - cot x)

解决方案

当 x = 0 时，cosec x 和 cot x 都未定义。因此，

将 lim_{x → 0} (cosec x - cot x) 的分子和分母因式分解

= lim_{x → 0} (1/sin x - cos x/sin x)

= lim_{x → 0} (1 - cos x)/sin x

= lim_{x → 0} (2 sin² x/2)/2 sin x/2 cos x/2

= lim_{x → 0} (sin x/2)/cos x/2

= lim_{x → 0} tan x/2

代入 x = 0，

lim_{x → 0} tan x/2 = tan 0/2 = tan 0 = 0

22. lim_{x → π/2} tan 2x/(x - π/2)

解决方案

代入 x = π/2，

lim_{x → π/2} tan 2x/(x - π/2) = tan 2(π/2)/(π/2 - π/2)

= tan π/(π/2 - π/2) = 0/0

极限为0/0。因此，

将 lim_{x → π/2} tan 2x/(x - π/2) 的分子和分母因式分解

设 x - π/2 = y，则 x → π/2 = y → 0

因此，

lim_{x → π/2} tan 2x/(x - π/2) = lim_{y → 0} tan 2(y + π/2)/y

= lim_{y → 0} tan (2y + π)/y

= lim_{y → 0} tan (2y)/y

= lim_{y → 0} sin 2y/y cos 2y

乘以并除以 2

= lim_{y → 0} sin 2y/2y × 2/cos 2y

= lim_{2y → 0} sin 2y/2y × lim_{y → 0} 2/cos 2y

我们知道

lim_{x → 0} sin x/x = 1

因此，

lim_{2y → 0} sin 2y/2y × lim_{y → 0} 2/cos 2y

= 1 × lim_{y → 0} 2/cos 2y

现在，代入 y = 0，

lim_{y → 0} 2/cos 2y = 2/cos 2(0) = 2/cos 0

= 2

23. 求 lim_{x → 0} f(x) 和 lim_{x → 1} f(x)，其中 。

解决方案

lim_{x → 0-} f(x) = lim_{x → 0} (2x + 3)

代入 x = 0，

lim_{x → 0} (2x + 3) = 2(0) + 3 = 0 + 3 = 3

并且

lim_{x → 0+} f(x) = lim_{x → 0} 3(x + 1)

代入 x = 0，

lim_{x → 0} 3(x + 1) = 3(0 + 1) = 3(1) = 3

因此，lim_{x → 0-} f(x) = lim_{x → 0+} f(x) = lim_{x → 0} f(x) = 3

lim_{x → 1-} f(x) = lim_{x → 1} 3(x + 1)

代入 x = 1，

lim_{x → 1} 3(x + 1) = 3(1 + 1) = 3(2) = 6

并且

lim_{x → 1+} f(x) = lim_{x → 1} 3(x + 1)

代入 x = 1，

lim_{x → 1} 3(x + 1) = 3(1 + 1) = 3(2) = 6

因此，lim_{x → 1-} f(x) = lim_{x → 1+} f(x) = lim_{x → 1} f(x) = 6

24. 求 lim_{x → 1} f(x)，其中

解决方案

lim_{x → 1-} f(x) = lim_{x → 1} (x²- 1)

代入 x = 1，

lim_{x → 1} (x²- 1) = 1² - 1 = 1 - 1 = 0

并且

lim_{x → 1+} f(x) = lim_{x → 1} (-x²- 1)

代入 x = 1，

lim_{x → 1} (-x²- 1) = (-1² - 1) = (-1 - 1) = -2

现在，lim_{x → 1-} f(x) ≠ lim_{x → 1+} f(x)

因此，lim_{x → 1} f(x) 不存在。

25. 求 lim_{x → 0} f(x)，其中

解决方案

lim_{x → 0-} f(x) = lim_{x → 0} (|x|/x)

我们知道当 x 为负数时，|x| = -x。因此，

lim_{x → 0} (|x|/x) = lim_{x → 0} (-x/x)

= lim_{x → 0} (-1)

= -1

并且

lim_{x → 0+} f(x) = lim_{x → 0} (|x|/x)

我们知道当 x 为正数时，|x| = x。因此，

lim_{x → 0} (|x|/x) = lim_{x → 0} (x/x)

= lim_{x → 0} (1)

= 1

现在，lim_{x → 0-} f(x) ≠ lim_{x → 0+} f(x)

因此，lim_{x → 0} f(x) 不存在。

26. 求 lim_{x → 0} f(x)，其中

解决方案

lim_{x → 0-} f(x) = lim_{x → 0} (x/|x|)

我们知道当 x 为负数时，|x| = -x。因此，

lim_{x → 0} (x/|x|) = lim_{x → 0} (x/-x)

= lim_{x → 0} (-1)

= -1

并且

lim_{x → 0+} f(x) = lim_{x → 0} (x/|x|)

我们知道当 x 为正数时，|x| = x。因此，

lim_{x → 0} (x/|x|) = lim_{x → 0} (x/x)

= lim_{x → 0} (1)

= 1

现在，lim_{x → 0-} f(x) ≠ lim_{x → 0+} f(x)

因此，lim_{x → 0} f(x) 不存在。

27. 求 lim_{x → 5} f(x)，其中 f(x) = |x| - 5。

解决方案

lim_{x → 5-} f(x) = lim_{x → 5} (|x| - 5)

= lim_{x → 5} (x - 5)

代入 x = 5，

= lim_{x → 5} (x - 5) = 5 - 5

= 0

并且

lim_{x → 5+} f(x) = lim_{x → 5} (|x| - 5)

= lim_{x → 5} (x - 5)

代入 x = 5，

= lim_{x → 5} (x - 5) = 5 - 5

= 0

因此，lim_{x → 5-} f(x) = lim_{x → 5+} f(x) = lim_{x → 5} f(x) = 0。

28. 假设 且若 lim_{x → 1} f(x) = f(1) 则 a 和 b 的可能值是多少？

解决方案

lim_{x → 1-} f(x) = lim_{x → 1} (a + bx)

代入 x = 1，

lim_{x → 1} (a + bx) = (a + b(1))

= a + b

并且

lim_{x → 1+} f(x) = lim_{x → 1} (b - ax)

代入 x = 1，

lim_{x → 1} (b - ax) = (b - a(1))

= b - a

并且

lim_{x → 1} f(x) = f(1) = 4

现在，lim_{x → 1-} f(x) = lim_{x → 1+} f(x) = lim_{x → 1} f(x) = 4。因此，

a + b = b - a = 4

a + b = 4 且 b - a = 4

两方程相加

a + b + b - a = 4 + 4

2b = 8

b = 4

所以，

a + 4 = 4

a = 0

因此，a 和 b 的可能值分别为 0 和 4。

29. 设 a₁, a₂, . . ., a_n 为固定的实数，并定义函数

f(x) = (x - a₁) (x - a₂) … (x - a_n)。

lim_{x → a1}f(x) 是什么？对于某些 a ≠ a₁, a₂, . . ., a_n，计算 lim_{x → a} f(x)？

解决方案

lim_{x → a1}= lim_{x → a1}[(x - a₁) (x - a₂) … (x - a_n)]

= [lim_{x → a1}(x - a₁)] [lim_{x → a1}(x - a₂)] … [lim_{x → a1}(x - a_n)]

代入 x = a₁，

[lim_{x → a1}(x - a₁)] [lim_{x → a1}(x - a₂)] … [lim_{x → a1}(x - a_n)]

= (a₁ - a₁) (a₁ - a₂) … (a₁ - a_n)

= 0(a₁ - a₂) … (a₁ - a_n) = 0

因此，lim_{x → a1}f(x) = 0。

lim_{x → a} = lim_{x → a} [(x - a₁) (x - a₂) … (x - a_n)]

= [lim_{x → a} (x - a₁)] [lim_{x → a} (x - a₂)] … [lim_{x → a} (x - a_n)]

代入 x = a，

[lim_{x → a} (x - a₁)] [lim_{x → a} (x - a₂)] … [lim_{x → a} (x - a_n)]

= (a - a₁) (a - a₂) … (a - a_n)

因此，lim_{x → a} f(x) = (a - a₁) (a - a₂) … (a - a_n)。

30. 若 对于 a 的何值，lim_{x → a} f(x) 存在？

解决方案

当 a < 0 时

lim_{x → a-} f(x) = lim_{x → a-} (|x| + 1)

我们知道当 x 为负数时，|x| = -x。因此，

lim_{x → a-} (|x| + 1) = lim_{x → a} (-x + 1)

代入 x = a，

lim_{x → a} (-x + 1) = -a + 1

并且

lim_{x → a+} f(x) = lim_{x → a+} (|x| + 1)

我们知道当 x 为负数时，|x| = -x。因此，

lim_{x → a+} (|x| + 1) = lim_{x → a} (-x + 1)

代入 x = a，

lim_{x → a} (-x + 1) = -a + 1

现在，lim_{x → a-} f(x) = lim_{x → a+} f(x) = lim_{x → a} f(x) = -a + 1。因此，

lim_{x → a} f(x) 存在于 x = a；a < 0。

当 a = 0 时

lim_{x → 0-} f(x) = lim_{x → 0-} (|x| + 1)

我们知道当 x 为负数时，|x| = -x。因此，

lim_{x → 0-} (|x| + 1) = lim_{x → 0} (-x + 1)

代入 x = 0，

lim_{x → 0} (-x + 1) = -0 + 1

= 1

并且

lim_{x → 0+} f(x) = lim_{x → 0+} (|x| - 1)

我们知道当 x 为正数时，|x| = x。因此，

lim_{x → 0+} (|x| - 1) = lim_{x → 0} (x - 1)

代入 x = 0，

lim_{x → 0} (x - 1) = 0 - 1

= -1

现在，lim_{x → 0-} f(x) ≠ lim_{x → 0+} f(x)。因此，lim_{x → 0} f(x) 不存在。

当 a > 0 时

lim_{x → a-} f(x) = lim_{x → a-} (|x| - 1)

我们知道当 x 为正数时，|x| = x。因此，

lim_{x → a-} (|x| - 1) = lim_{x → a} (x - 1)

代入 x = a，

lim_{x → a} (x - 1) = a - 1

并且

lim_{x → a+} f(x) = lim_{x → a+} (|x| - 1)

我们知道当 x 为正数时，|x| = x。因此，

lim_{x → a+} (|x| - 1) = lim_{x → a} (x - 1)

代入 x = a，

lim_{x → a} (x - 1) = a - 1

现在，lim_{x → a-} f(x) = lim_{x → a+} f(x) = lim_{x → a} f(x) = a - 1。因此，

lim_{x → a} f(x) 存在于 x = a；a > 0。

31. 如果函数 f(x) 满足 lim_{x → 1} (f(x) - 2)/(x² - 1) = π，计算 lim_{x → 1} f(x)。

解决方案

f(x) 满足 lim_{x → 1} (f(x) - 2)/(x² - 1) = π。所以，

lim_{x → 1} (f(x) - 2)/lim_{x → 1} (x² - 1) = π

lim_{x → 1} (f(x) - 2) = π lim_{x → 1} (x² - 1)

代入 x = 1，

lim_{x → 1} (f(x) - 2) = π lim_{x → 1} (1² - 1)

lim_{x → 1} (f(x) - 2) = π lim_{x → 1} (1 - 1)

lim_{x → 1} (f(x) - 2) = π lim_{x → 1} 0

lim_{x → 1} f(x) - lim_{x → 1} 2 = 0

lim_{x → 1} f(x) - 2 = 0

lim_{x → 1} f(x) = 2

32. 若 对于哪些整数 m 和 n，lim_{x → 0} f(x) 和 lim_{x → 1} f(x) 都存在？

解决方案

lim_{x → 0-} f(x) = lim_{x → 0} (mx²+ n)

代入 x = 0，

lim_{x → 0} (mx²+ n) = m(0) + n

= 0 + n

= n

并且

lim_{x → 0+} f(x) = lim_{x → 0} (nx + m)

代入 x = 0，

lim_{x → 0} (nx + m) = n(0) + m

= 0 + m

= m

现在，如果 lim_{x → 0-} f(x) = lim_{x → 0+} f(x)，即 n = m，则 lim_{x → 0} f(x) 存在。

lim_{x → 1-} f(x) = lim_{x → 1} (nx + m)

代入 x = 1，

lim_{x → 1} (nx + m) = n(1) + m

= n + m

并且

lim_{x → 1+} f(x) = lim_{x → 1} (nx³ + m)

代入 x = 1，

lim_{x → 1} (nx + m) = n(1)³ + m

= n(1) + m

= n + m

因此，lim_{x → 1+} f(x) = lim_{x → 1-} f(x) = lim_{x → 1} f(x)。

因此，对于 m 和 n 的任何整数值，lim_{x → 1} f(x) 存在。

练习 13.2

1. 求 x² - 2 在 x = 10 处的导数。

解决方案

设 x² - x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

对于 x = 10

f'(10) = lim_{h → 0} (f(10 + h) - f(10))/h

= lim_{h → 0} ((10 + h)² - 2 - (10² - 2))/h

= lim_{h → 0} (100 + h² + 20h - 2 - 100 + 2)/h

= lim_{h → 0} (h² + 20h)/h

= lim_{h → 0} h(h + 20)/h

= lim_{h → 0} (h + 20)

当 h = 0 时取极限，

lim_{h → 0} (h + 20) = 0 + 20 = 20

2. 求 x 在 x = 1 处的导数。

解决方案

设 x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

对于 x = 1

f'(1) = lim_{h → 0} (f(1 + h) - f(1))/h

= lim_{h → 0} (1 + h - 1)/h

= lim_{h → 0} h/h

= lim_{h → 0} 1

= 1

3. 求 99x 在 x = 100 处的导数。

解决方案

设 99x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

对于 x = 100

f'(100) = lim_{h → 0} (f(100 + h) - f(100))/h

= lim_{h → 0} (99(100 + h) - 99(100))/h

= lim_{h → 0} (9900 + 99h - 99100)/h

= lim_{h → 0} 99h/h

= lim_{h → 0} 99

= 99

4. 从第一原理求以下函数的导数。

(i) x³- 27 (ii) (x - 1)(x - 2)

(iii) 1/x² (iv) (x + 1)/(x - 1)

解决方案

(i) 设 x³- 27 = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

= lim_{h → 0} ((x + h)³ - 27 - (x³- 27))/h

= lim_{h → 0} (x³+ h³ + 3x²h + 3xh² - 27 - x³+ 27)/h

= lim_{h → 0} (h³ + 3x²h + 3xh²)/h

= lim_{h → 0} h(h² + 3x² + 3xh)/h

= lim_{h → 0} (h² + 3x² + 3xh)

当 h = 0 时取极限，

lim_{h → 0} (h² + 3x² + 3xh) = (0² + 3x²+ 3x(0))

= 0 + 3x²+ 0

= 3x²

(ii) 设 (x - 1)(x - 2) = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

= lim_{h → 0} ((x + h - 1)(x + h - 2) - (x - 1)(x - 2))/h

= lim_{h → 0} (x² + xh - 2x + xh + h² - 2h - x - h + 2 - (x² - x - 2x + 2))/h

= lim_{h → 0} (x² + 2xh - 3x + h² - 3h + 2 - x² + x + 2x - 2)/h

= lim_{h → 0} (2xh + h² - 3h)/h

= lim_{h → 0} h(2x + h - 3)/h

= lim_{h → 0} (2x + h - 3)

当 h = 0 时取极限，

lim_{h → 0} (2x + h - 3) = 2x + 0 - 3

= 2x - 3

(iii) 设 1/x² = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

= lim_{h → 0} (1/(x + h)² - 1/x²)/h

= lim_{h → 0} (x² - (x + h)²)/x²(x + h)²h

= lim_{h → 0} (x² - x² - h² - 2xh)/x²(x + h)²h

= lim_{h → 0} (-h² - 2xh)/x²(x + h)²h

= lim_{h → 0} h(-h - 2x)/x²(x + h)²h

= lim_{h → 0} (-h - 2x)/x²(x + h)²

当 h = 0 时取极限，

lim_{h → 0} (-h - 2x)/x²(x + h)²

= (-0 - 2x)/x²(x + 0)² = (-2x)/x²(x²)

= -2/x³

(iv) 设 (x + 1)/(x - 1) = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

= lim_{h → 0} ((x + h + 1)/(x + h - 1) - (x + 1)/(x - 1))/h

= lim_{h → 0} ((x + h + 1)(x - 1) - (x + 1)(x + h - 1))/h(x + h - 1)(x - 1)

= lim_{h → 0} (x² - x + xh - h + x - 1 - x² - xh + x - x - h + 1)/h(x + h - 1)(x - 1)

= lim_{h → 0} (-2h)/h(x + h - 1)(x - 1)

= lim_{h → 0} (-2)/(x + h - 1)(x - 1)

当 h = 0 时取极限，

lim_{h → 0} (-2)/(x + h - 1)(x - 1) = (-2)/(x + 0 - 1)(x - 1)

= -2/(x - 1)(x - 1)

= -2/(x - 1)²

5. 对于函数

f(x) = x¹⁰⁰/100 + x⁹⁹/99 + … + x²/2 + x + 1。

证明 f'(1) = 100f'(0)。

解决方案

f(x) = x¹⁰⁰/100 + x⁹⁹/99 + … + x²/2 + x + 1

两边求导

d/dx (f(x)) = d/dx [x¹⁰⁰/100 + x⁹⁹/99 + … + x²/2 + x + 1]

f'(x) = d/dx (x¹⁰⁰/100) + d/dx (x⁹⁹/99) + … + d/dx (x²/2) + d/dx (x) + d/dx

我们知道

d/dx (xⁿ) = nx^{n - 1}

因此，

f'(x) = 100x⁹⁹/100 + 99x⁹⁸/99 + … + 2x/2 + 1 + 0

f'(x) = x⁹⁹+ x⁹⁸ + … + x + 1

当 x = 0 时

f'(0) = 0⁹⁹+ 0⁹⁸ + … + 0 + 1

= 0 + 0 + … + 0 + 1

= 1

当 x = 1 时

f'(1) = 1⁹⁹+ 1⁹⁸ + … + 1 + 1

= 1 + 1 + … + 1 + 1

= 100 × 1

= 100 f'(0)

因此，证明完毕。

6. 对于某个固定的实数 a，求 xⁿ + ax^{n - 1} + a²x^{n - 2} + … + a^{n - 1}x + aⁿ 的导数。

解决方案

设 xⁿ + ax^{n - 1} + a²x^{n - 2} + … + a^{n - 1}x + aⁿ = f(x)

两边求导

d/dx (f(x)) = d/dx [xⁿ + ax^{n - 1} + a²x^{n - 2} + … + a^{n - 1}x + aⁿ]

f'(x) = d/dx (xⁿ) + a d/dx (x^{n - 1}) + a² d/dx (x^{n - 2}) + … + a^{n - 1} d/dx (x) + aⁿ d/dx

我们知道

d/dx (xⁿ) = nx^{n - 1}

因此，

f'(x) = nx^{n - 1} + a (n - 1) x^{n - 2} + a² (n - 2) x^{n - 3} + … + a^{n - 1} (1) + aⁿ (0)

f'(x) = nx^{n - 1} + a (n - 1) x^{n - 2} + a² (n - 2) x^{n - 3} + … + a^{n - 1}

7. 对于某些常数 a 和 b，求以下函数的导数

(i) (x - a)(x - b)

(ii) (ax² + b)²

(iii) (x - a)/(x - b)

解决方案

(i) 设 (x - a)(x - b) = f(x)

f(x) = x² - (a + b)x + ab

两边求导

d/dx (f(x)) = d/dx [x² - (a + b)x + ab]

f'(x) = d/dx (x²) - d/dx ((a + b)x) + d/dx (ab)

f'(x) = d/dx (x²) - (a + b) d/dx (x) + (ab) d/dx (1)

我们知道

d/dx (xⁿ) = nx^{n - 1}

因此，

f'(x) = 2x - (a + b)(1) + ab (0)

f'(x) = 2x - a - b

(ii) 设 (ax² + b)² = f(x)

两边求导

d/dx (f(x)) = d/dx [(ax² + b)²]

f'(x) = d/dx [a²x⁴ + b² + 2abx²]

f'(x) = d/dx (a²x⁴) + d/dx (b²) + d/dx (2abx²)

f'(x) = a² d/dx (x⁴) + b² d/dx (1) + 2ab d/dx (x²)

我们知道

d/dx (xⁿ) = nx^{n - 1}

因此，

f'(x) = a² (4x³)+ b² (0) + 2ab (2x)

f'(x) = 4a²x³ + 4abx

f'(x) = 4ax (ax² + b)

(iii) 设 (x - a)/(x - b) = f(x)

两边求导

d/dx (f(x)) = d/dx [(x - a)/(x - b)]

f'(x) = [(x - b) d/dx (x - a) - (x - a) d/dx (x - b)]/(x - b)²

f'(x) = [(x - b) {d/dx (x) - d/dx (a)} - (x - a) {d/dx (x) - d/dx (b)}]/(x - b)²

f'(x) = [(x - b) (1 - 0) - (x - a) (1 - 0)]/(x - b)²

f'(x) = [(x - b) - (x - a)]/(x - b)²

f'(x) = [x - b - x + a)]/(x - b)²

8. 对于某个常数 a，求 (xⁿ - aⁿ)/(x - a) 的导数。

解决方案

设 (xⁿ - aⁿ)/(x - a) = f(x)

两边求导

d/dx (f(x)) = d/dx [(xⁿ - aⁿ)/(x - a)]

f'(x) = [(x - a) d/dx (xⁿ - aⁿ) - (xⁿ - aⁿ) d/dx (x - a)]/(x - a)²

f'(x) = [(x - a) {d/dx (xⁿ) - d/dx (aⁿ)} - (xⁿ - aⁿ) {d/dx (x) - d/dx (a)}]/(x - a)²

f'(x) = [(x - a) (nx^{n - 1} - 0) - (xⁿ - aⁿ) (1 - 0)]/(x - a)²

f'(x) = [(x - a) (nx^{n - 1}) - (xⁿ - aⁿ)]/(x - a)²

f'(x) = [nxⁿ - anx^{n - 1} - xⁿ + aⁿ]/(x - a)²

9. 求以下函数的导数

(i) 2x - 3/4 (ii) (5x³ + 3x + 1)(x - 1)

(iii) x^-3(5 + 3x) (iv) x⁵(3 - 6x^-9)

(v) x^-4(3 - 4x^-5) (vi) 2/(x + 1) - x²/(3x - 1)

解决方案

(i) 设 2x - 3/4 = f(x)

两边求导

d/dx (f(x)) = d/dx (2x - 3/4)

f'(x) = 2 d/dx (x) - 3/4 d/dx (1)

f'(x) = 2(1) - 3/4 (0)

f'(x) = 2 - 0

f'(x) = 2

(ii) 设 (5x³ + 3x + 1)(x - 1) = f(x)

两边求导

d/dx (f(x)) = d/dx [(5x³ + 3x + 1)(x - 1)]

f'(x) = (x - 1) d/dx (5x³ + 3x + 1) + (5x³ + 3x + 1) d/dx (x - 1)

f'(x) = (x - 1) [5 d/dx (x³) + 3 d/dx (x) + d/dx (1)] + (5x³ + 3x + 1) [d/dx (x) - d/dx (1)]

f'(x) = (x - 1) [5(3x²) + 3(1) + 0] + (5x³ + 3x + 1) [1 - 0]

f'(x) = (x - 1) [15x² + 3] + (5x³ + 3x + 1) (1)

f'(x) = 15x³ + 3x - 15x² - 3 + 5x³ + 3x + 1

f'(x) = 20x³ - 15x² + 6x - 2

(iii) 设 x^-3(5 + 3x) = f(x)

两边求导

d/dx (f(x)) = d/dx [x^-3(5 + 3x)]

f'(x) = x^-3 d/dx (5 + 3x) + (5 + 3x) d/dx (x^-3)

f'(x) = x^-3 [5 d/dx (1) + 3 d/dx (x)] + (5 + 3x) (-3x^-4)

f'(x) = x^-3 [5(0) + 3(1)] + (-15x^-4 - 9x^-3)

f'(x) = x^-3 (0 + 3) + (-15x^-4 - 9x^-3)

f'(x) = 3x^-3 - 15x^-4 - 9x^-3

f'(x) = -15x^-4 - 6x^-3

f'(x) = -15/x⁴ - 6/x³

f'(x) = 3/x³ [-5/x - 2]

f'(x) = 3/x³ [(-5 - 2x)/x]

f'(x) = -3(5 + 2x)/x⁴

(iv) 设 x⁵(3 - 6x^-9) = f(x)

两边求导

d/dx (f(x)) = d/dx [x⁵(3 - 6x^-9)]

f'(x) = x⁵ d/dx (3 - 6x^-9) + (3 - 6x^-9) d/dx (x⁵)

f'(x) = x⁵ [3 d/dx (1) - 6 d/dx (x^-9)] + (3 - 6x^-9) (5x⁴)

f'(x) = x⁵ [3(0) - 6 (-9x^-10)] + (3 - 6x^-9) (5x⁴)

f'(x) = x⁵ [0 + 54x^-10] + (15x⁴ - 30x^-5)

f'(x) = 54x^-5 + 15x⁴ - 30x^-5

f'(x) = 24x^-5 + 15x⁴

f'(x) = 24/x⁵ + 15x⁴

(v) 设 x^-4(3 - 4x^-5) = f(x)

两边求导

d/dx (f(x)) = d/dx [x^-4(3 - 4x^-5)]

f'(x) = x^-4 d/dx (3 - 4x^-5) + (3 - 4x^-5) d/dx (x^-4)

f'(x) = x^-4 [3 d/dx (1) - 4 d/dx (x^-5)] + (3 - 4x^-5) (-4x^-5)

f'(x) = x^-4 [3(0) - 4 (-5x^-6)] + (-12x^-5 + 16x^-10)

f'(x) = x^-4 [0 + 20x^-6] + (-12x^-5 + 16x^-10)

f'(x) = 20x^-10 - 12x^-5 + 16x^-10

f'(x) = 36x^-10 - 12x^-5

f'(x) = 36/x¹⁰ - 12/x⁵

(vi) 设 2/(x + 1) - x²/(3x - 1) = f(x)

两边求导

d/dx (f(x)) = d/dx [2/(x + 1) - x²/(3x - 1)]

f'(x) = d/dx [2/(x + 1)] - d/dx [x²/(3x - 1)]

f'(x) = [(x + 1) d/dx (2) + 2 d/dx (x + 1)]/(x + 1)² - [(3x - 1) d/dx (x²) - x² d/dx (3x - 1)]/(3x - 1)²

f'(x) = [(x + 1)(0) + 2(d/dx (x) + d/dx (1))]/(x + 1)² - [(3x - 1)(2x) - x² (3 d/dx (x) - d/dx(1))]/(3x - 1)²

f'(x) = [0 + 2(1 + 0)]/(x + 1)² - [(6x² - 2x) - x² (3(1) - 0)]/(3x - 1)²

f'(x) = [2]/(x + 1)² - [6x² - 2x - 3x²]/(3x - 1)²

f'(x) = 2/(x + 1)² - [3x² - 2x]/(3x - 1)²

f'(x) = 2/(x + 1)² - x(3x - 2)]/(3x - 1)²

10. 从第一原理求 cos x 的导数。

解决方案

设 cos x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (cos (x + h) - cos x)/h

f'(x) = lim_{h → 0} (-2 sin ((x + h + x)/2) sin ((x + h - x)/2))/h

f'(x) = lim_{h → 0} (-2 sin ((2x + h)/2) sin h/2)/h

f'(x) = lim_{h → 0} -sin ((2x + h)/2) × lim_{h → 0} sin h/2/(h/2)

取极限，

f'(x) = -sin ((2x + 0)/2) × 1

f'(x) = -sin (2x/2)

f'(x) = -sin x

11. 求以下函数的导数

(i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x
(iv) cosec x (v) 3 cot x + 5 cosec x
(vi) 5 sin x - 6 cos x + 7 (vii) 2 tan x - 7 sec x

解决方案

(i) 设 sin x cos x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (sin (x + h) cos (x + h) - sin x cos x)/h

乘以并除以 2

f'(x) = lim_{h → 0} (2sin (x + h) cos (x + h) - 2sin x cos x)/2h

f'(x) = lim_{h → 0} (sin 2(x + h) - sin 2x)/2h

f'(x) = lim_{h → 0} (2 cos ((2x + 2h + 2x)/2) sin ((2x + 2h - 2x)/2)/2h

f'(x) = lim_{h → 0} (2 cos ((4x + 2h)/2) sin (2h/2))/2h

f'(x) = lim_{h → 0} (cos (2x + h) sin h)/h

f'(x) = lim_{h → 0} cos (2x + h) × lim_{h → 0} (sin h/h)

f'(x) = lim_{h → 0} cos (2x + h) × 1

取极限，

f'(x) = cos (2x + 0)

f'(x) = cos 2x

(ii) 设 sec x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (sec (x + h) - sec x)/h

f'(x) = lim_{h → 0} (1/cos (x + h) - 1/cos x)/h

f'(x) = lim_{h → 0} (cos x - cos (x + h))/h cos x cos (x + h)

f'(x) = lim_{h → 0} [-2 sin ((x + x + h)/2) sin ((x - x - h)/2)]/h cos x cos (x + h)

f'(x) = lim_{h → 0} [-2 sin ((2x + h)/2) sin (-h/2)]/h cos x cos (x + h)

f'(x) = lim_{h → 0} [2 sin ((2x + h)/2) sin (h/2)]/h cos x cos (x + h)

f'(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × 1

取极限，

f'(x) = sin ((2x + 0)/2)/cos x cos (x + 0)

f'(x) = sin (2x/2)/cos x cos x

f'(x) = sin x/cos x × 1/cos x

f'(x) = tan x sec x

(iii) 设 5 sec x + 4 cos x = f(x)

f'(x) = 5 (sec x )' + 4 (cos x)'

设 sec x = f₁(x)

根据第一原理

f'₁(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'₁(x) = lim_{h → 0} (sec (x + h) - sec x)/h

f'₁(x) = lim_{h → 0} (1/cos (x + h) - 1/cos x)/h

f'₁(x) = lim_{h → 0} (cos x - cos (x + h))/h cos x cos (x + h)

f'₁(x) = lim_{h → 0} [-2 sin ((x + x + h)/2) sin ((x - x - h)/2)]/h cos x cos (x + h)

f'₁(x) = lim_{h → 0} [-2 sin ((2x + h)/2) sin (-h/2)]/h cos x cos (x + h)

f'₁(x) = lim_{h → 0} [2 sin ((2x + h)/2) sin (h/2)]/h cos x cos (x + h)

f'₁(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'₁(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × 1

取极限，

f'₁(x) = sin ((2x + 0)/2)/cos x cos (x + 0)

f'₁(x) = sin (2x/2)/cos x cos x

f'₁(x) = sin x/cos x × 1/cos x

f'₁(x) = tan x sec x

设 cos x = f₂(x)

根据第一原理

f'₂(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'₂(x) = lim_{h → 0} (cos (x + h) - cos x)/h

f'₂(x) = lim_{h → 0} (-2 sin ((x + h + x)/2) sin ((x + h - x)/2))/h

f'₂(x) = lim_{h → 0} (-2 sin ((2x + h)/2) sin h/2)/h

f'₂(x) = lim_{h → 0} -sin ((2x + h)/2) × lim_{h → 0} sin h/2/(h/2)

取极限，

f'₂(x) = -sin ((2x + 0)/2) × 1

f'₂(x) = -sin (2x/2)

f'₂(x) = -sin x

因此，

f'(x) = 5(tan x sec x) + 4(-sin x)

f'(x) = 5 tan x sec x - 4 sin x

(iv) 设 cosec x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (cosec (x + h) - cosec x)/h

f'(x) = lim_{h → 0} (1/sin (x + h) - 1/sin x)/h

f'(x) = lim_{h → 0} (sin x - sin (x + h))/h sin x sin (x + h)

f'(x) = lim_{h → 0} [2 cos ((x + x + h)/2) sin ((x - x - h)/2)]/h sin x sin (x + h)

f'(x) = lim_{h → 0} [2 cos ((2x + h)/2) sin (-h/2)]/h sin x sin (x + h)

f'(x) = lim_{h → 0} [-2 cos ((2x + h)/2) sin (h/2)]/h sin x sin (x + h)

f'(x) = -lim_{h → 0} [cos ((2x + h)/2)]/sin x sin (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'(x) = -lim_{h → 0} [cos ((2x + h)/2)]/sin x sin (x + h) × 1

取极限，

f'(x) = -cos ((2x + 0)/2)/sin x sin (x + 0)

f'(x) = -cos (2x/2)/sin x sin x

f'(x) = -cos x/sin x × 1/sin x

f'(x) = -cot x cosec x

(v) 设 3 cot x + 5 cosec x = f(x)

f'(x) = 3 (cot x)' + 5 (cosec x)'

设 cot x = f₁(x)

根据第一原理

f₁'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f₁'(x) = lim_{h → 0} (cot (x + h) - cot x)/h

f₁'(x) = lim_{h → 0} (cos (x + h)/sin (x + h) - cos x/sin x)/h

f₁'(x) = lim_{h → 0} (sin x cos (x + h) - cos x sin (x + h))/h sin x sin (x + h)

f₁'(x) = lim_{h → 0} (sin (x - x - h))/h sin x sin (x + h)

f₁'(x) = lim_{h → 0} (sin (-h))/h sin x sin (x + h)

f₁'(x) = lim_{h → 0} (-sin h)/h sin x sin (x + h)

f₁'(x) = -lim_{h → 0} (sin h)/h sin x sin (x + h)

f₁'(x) = -lim_{h → 0} (sin h)/h × lim_{h → 0} 1/[sin x sin (x + h)]

f₁'(x) = -1 × lim_{h → 0} 1/[sin x sin (x + h)]

取极限，

f₁'(x) = -1 × 1/sin x sin (x + 0)

f₁'(x) = -1 × 1/sin x sin x

f₁'(x) = -1/sin² x

f₁'(x) = -cosec² x

设 cosec x = f₂(x)

根据第一原理

f₂'(x) = lim_{h → 0} (f₂(x + h) - f₂(x))/h

f'₂(x) = lim_{h → 0} (cosec (x + h) - cosec x)/h

f₂'(x) = lim_{h → 0} (1/sin (x + h) - 1/sin x)/h

f₂'(x) = lim_{h → 0} (sin x - sin (x + h))/h sin x sin (x + h)

f₂'(x) = lim_{h → 0} [2 cos ((x + x + h)/2) sin ((x - x - h)/2)]/h sin x sin (x + h)

f₂'(x) = lim_{h → 0} [2 cos ((2x + h)/2) sin (-h/2)]/h sin x sin (x + h)

f₂'(x) = lim_{h → 0} [-2 cos ((2x + h)/2) sin (h/2)]/h sin x sin (x + h)

f₂'(x) = -lim_{h → 0} [cos ((2x + h)/2)]/sin x sin (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f₂'(x) = -lim_{h → 0}[cos ((2x + h)/2)]/sin x sin (x + h) × 1

取极限，

f₂'(x) = -cos((2x + 0)/2)/sin x sin (x + 0)

f₂'(x) = -cos(2x/2)/sin x sin x

f₂'(x) = -cos x/sin x × 1/sin x

f₂'(x) = -cot x cosec x

因此，

f'(x) = 3 (-cosec²x) + 5 (-cot x cosec x)

f'(x) = -3 cosec²x - 5 cot x cosec x

(vi) 设 5 sin x - 6 cos x + 7 = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (5 sin (x + h) - 6 cos (x + h) + 7 - (5 sin x - 6 cos x + 7))/h

f'(x) = lim_{h → 0} (5 sin (x + h) - 6 cos (x + h) + 7 - 5 sin x + 6 cos x - 7))/h

f'(x) = lim_{h → 0} [5 {sin (x + h) - sin x} - 6 {cos (x + h) - cos x}]/h

f'(x) = 5 lim_{h → 0} [sin (x + h) - sin x]/h - 6 lim_{h → 0} [cos (x + h) - cos x]/h

f'(x) = 5lim_{h → 0} [2 cos ((x + h + x)/2) sin ((x + h - x)/2)]/h - 6 lim_{h → 0} [-2 sin ((x + h + x)/2) sin ((x + h - x)/2)]/h

f'(x) = 5 lim_{h → 0} [2 cos ((2x + h)/2) sin (h/2)]/h + 6 lim_{h → 0} [2 sin ((2x + h)/2) sin (h/2)]/h

f'(x) = 5 lim_{h → 0} [cos ((2x + h)/2)] lim_{h/2 → 0} [sin (h/2)]/(h/2) + 6 lim_{h → 0} [sin ((2x + h)/2)]
lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'(x) = 5lim_{h → 0} [cos ((2x + h)/2)] × 1 + 6 lim_{h → 0} [sin ((2x + h)/2)] × 1

取极限，

f'(x) = 5 [cos ((2x + 0)/2)] + 6 [sin ((2x + 0)/2)]

f'(x) = 5 (cos (2x/2)) + 6 (sin (2x/2))

f'(x) = 5 cos x + 6 sin x

(vii) 设 2 tan x - 7 sec x = f(x)

f'(x) = 2 (tan x)' - 7 (sec x)'

设 tan x = f₁(x)

根据第一原理

f₁'(x) = lim_{h → 0} (f₁(x + h) - f₁(x))/h

f₁'(x) = lim_{h → 0} (tan (x + h) - tan x)/h

f₁'(x) = lim_{h → 0} (sin (x + h)/cos (x + h) - sin x/cos x)/h

f₁'(x) = lim_{h → 0} (cos x sin (x + h) - cos (x + h) sin x)/h cos x cos (x + h)

乘以并除以 2

f₁'(x) = lim_{h → 0} (2 cos x sin (x + h) - 2 cos (x + h) sin x)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin (x + h + x) - sin (x - x - h) - sin (x + h + x) + sin (x + h - x))/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin (2x + h) - sin (-h) - sin (2x + h) + sin h)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin h + sin h)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (2 sin h)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin h)/h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin h)/h × lim_{h → 0} 1/cos x cos(x + h)

f₁'(x) = 1 × lim_{h → 0} 1/cos x cos (x + h)

取极限，

f₁'(x) = 1/cos x cos (x + 0)

f₁'(x) = 1/cos x cos x

f₁'(x) = sec² x

设 sec x = f₂(x)

根据第一原理

f'₂(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'₂(x) = lim_{h → 0} (sec (x + h) - sec x)/h

f'₂(x) = lim_{h → 0} (1/cos (x + h) - 1/cos x)/h

f'₂(x) = lim_{h → 0} (cos x - cos (x + h))/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [-2 sin ((x + x + h)/2) sin ((x - x - h)/2)]/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [-2 sin ((2x + h)/2) sin (-h/2)]/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [2 sin ((2x + h)/2) sin (h/2)]/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'₂(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × 1

取极限，

f'₂(x) = sin ((2x + 0)/2)/cos x cos (x + 0)

f'₂(x) = sin (2x/2)/cos x cos x

f'₂(x) = sin x/cos x × 1/cos x

f'₂(x) = tan x sec x

因此，

f'(x) = 2(sec² x) - 7(tan x sec x)

f'(x) = 2 sec² x - 7 tan x sec x

杂项练习

1. 从第一原理求以下函数的导数

(i) -x (ii) (-x)^-1 (iii) sin (x + 1) (iv) cos (x - π/8)

解决方案

(i) 设 -x = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (-(x + h) - (-x))/h

f'(x) = lim_{h → 0} (-x - h + x))/h

f'(x) = lim_{h → 0} (-h)/h

f'(x) = lim_{h → 0} (-1)

f'(x) = -1

(ii) 设 (-x)^-1 = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (-(x + h)^-1 - (-x)^-1)/h

f'(x) = lim_{h → 0} (-1/(x + h) + 1/x)/h

f'(x) = lim_{h → 0} (-x + x + h)/hx(x + h)

f'(x) = lim_{h → 0} (h)/hx(x + h)

f'(x) = lim_{h → 0} 1/x(x + h)

f'(x) = 1/x(x)

f'(x) = 1/x²

(iii) 设 sin (x + 1) = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (sin (x + h + 1) - sin (x + 1))/h

f'(x) = lim_{h → 0} [2 cos ((x + h + 1 + x + 1)/2) sin ((x + h + 1 - x - 1)/2)]/h

f'(x) = lim_{h → 0} [2 cos ((2x + h + 2)/2) sin (h/2)]/h

h → 0

所以，h/2 → 0

f'(x) = lim_{h → 0} [cos ((2x + h + 2)/2)] × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'(x) = lim_{h → 0} [cos ((2x + h + 2)/2)] × 1

f'(x) = lim_{h → 0} [cos ((2x + h + 2)/2)] × 1

f'(x) = cos ((2x + 2)/2)

f'(x) = cos (x + 1)

(iv) 设 cos (x - π/8) = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (cos (x + h - π/8) - cos (x - π/8))/h

f'(x) = lim_{h → 0} (cos (x + h - π/8) - cos (x - π/8))/h

f'(x) = lim_{h → 0} (-2 sin ((x + h - π/8 + x - π/8)/2) sin ((x + h - π/8 - x + π/8)/2))/h

f'(x) = lim_{h → 0} (-2 sin ((2x + h - π/4)/2) sin (h/2))/h

f'(x) = lim_{h → 0} [-sin ((2x + h - π/4)/2)] × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'(x) = lim_{h → 0} [-sin ((2x + h - π/4)/2)] × 1

f'(x) = -sin ((2x + 0 - π/4)/2)

f'(x) = -sin (2(x - π/8)/2)

f'(x) = -sin (x - π/8)

求以下函数的导数（a、b、c、d、p、q、r 和 s 为固定的非零常数，m 和 n 为整数）

2. (x + a)

解决方案

设 (x + a) = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} ((x + h + a) - (x + a))/h

f'(x) = lim_{h → 0} (x + h + a - x - a)/h

f'(x) = lim_{h → 0} h/h

f'(x) = lim_{h → 0} 1

f'(x) = 1

3. (px + q) (r/x + s)

解决方案

设 (px + q) (r/x + s) = f(x)

两边求导

d/dx (f(x)) = d/dx [(px + q) (r/x + s)]

f'(x) = (px + q) d/dx (r/x + s) + (r/x + s) d/dx (px + q)

f'(x) = (px + q) d/dx (r/x + s) + (r/x + s) d/dx (px + q)

f'(x) = (px + q) [r d/dx (x^-1) + d/dx (s)] + (r/x + s) [p d/dx (x) + d/dx (q)]

f'(x) = (px + q) [(-rx^-2) + 0] + (r/x + s) [p(1) + 0]

f'(x) = -prx^-1 - qrx^-2 + pr/x + ps

f'(x) = -qrx^-2 + ps

f'(x) = ps - qr/x²

4. (ax + b)(cx + d)²

解决方案

设 (ax + b)(cx + d)² = f(x)

两边求导

d/dx (f(x)) = d/dx [(ax + b)(cx + d)²]

f'(x) = (ax + b) d/dx [(cx + d)²] + (cx + d)² d/dx (ax + b)

f'(x) = (ax + b) [2(cx + d)¹ d/dx (cx + d)] + (cx + d)² [d/dx (ax) + d/dx (b)]

f'(x) = (ax + b) [2c(cx + d) {d/dx (x) + d/dx (d)}] + (cx + d)² [a d/dx (x) + d/dx (b)]

f'(x) = (ax + b) [2c(cx + d) {1 + 0}] + (cx + d)² [a(1) + 0]

f'(x) = (ax + b) [2c(cx + d)] + a(cx + d)²

f'(x) = 2c(acx² + adx + bcx + bd) + a(c²x² + d² + 2cdx)

f'(x) = 2ac²x² + 2acdx + 2bc²x + 2bcd + ac²x² + ad² + 2acdx

f'(x) = 3ac²x² + ad² + 2bc²x + 4acdx + 2bcd

5. (ax + b)/(cx + d)

解决方案

设 (ax + b)/(cx + d) = f(x)

两边求导

d/dx (f(x)) = d/dx [(ax + b)/(cx + d)]

f'(x) = [(cx + d) d/dx (ax + b) - (ax + b) d/dx (cx + d)]/(cx + d)²

f'(x) = [(cx + d) [a d/dx (x) + d/dx (b)] - (ax + b) [c d/dx (x) + d/dx (d)]]/(cx + d)²

f'(x) = [(cx + d) (a(1) + 0) - (ax + b) (c(1) + 0)]/(cx + d)²

f'(x) = [a(cx + d) - c(ax + b)]/(cx + d)²

f'(x) = [acx + ad - acx - bc]/(cx + d)²

f'(x) = [ad - bc]/(cx + d)²

6. (1 + 1/x)/(1 - 1/x)

解决方案

设 (1 + 1/x)/(1 - 1/x) = f(x)

两边求导

d/dx (f(x)) = d/dx [(1 + 1/x)/(1 - 1/x)]

f'(x) = [(1 - 1/x) d/dx (1 + 1/x) - (1 + 1/x) d/dx (1 - 1/x)]/(1 - 1/x)²

f'(x) = [(1 - 1/x) {d/dx (1) + d/dx (x^-1)} - (1 + 1/x) {d/dx (1) - d/dx (x^-1)}]/(1 - 1/x)²

f'(x) = [(1 - 1/x) {0 + (-x^-2)} - (1 + 1/x) {0 - (-x^-2)}]/(1 - 1/x)²

f'(x) = [(1 - x^-1)(-x^-2) - (1 + x^-1)(x^-2)]/(1 - 1/x)²

f'(x) = [-x^-2 + x^-3 - x^-2 - x^-3]/(1 - 1/x)²

f'(x) = [-2x^-2]/[(x - 1)²/(x²)]

f'(x) = x²[-2x^-2]/(x - 1)²

f'(x) = -2/(x - 1)²

7. 1/(ax² + bx + c)

解决方案

设 1/(ax² + bx + c) = f(x)

两边求导

d/dx (f(x)) = d/dx [1/(ax² + bx + c)]

f'(x) = [(ax² + bx + c) d/dx (1) - d/dx (ax² + bx + c)]/(ax² + bx + c)²

f'(x) = [(ax² + bx + c) (0) - a {d/dx (x²) + b d/dx (x) + d/dx(c)}]/(ax² + bx + c)²

f'(x) = [-a {(2x) + b(1) + 0}](ax² + bx + c)²

f'(x) = -[2ax + ab](ax² + bx + c)²

8. (ax + b)/(px² + qx + r)

解决方案

设 (ax + b)/(px² + qx + r) = f(x)

两边求导

d/dx (f(x)) = d/dx [(ax + b)/(px² + qx + r)]

f'(x) = [(px²+ qx + r) d/dx (ax + b) - (ax + b) d/dx (px² + qx + r)]/(px² + qx + r)²

f'(x) = [(px²+ qx + r) {a d/dx (x) + d/dx (b)} - (ax + b) {p d/dx (x²) + q d/dx (x) + d/dx (r)}]/(px² + qx + r)²

f'(x) = [(px²+ qx + r) {a(1) + 0} - (ax + b) {p(2x) + q(1) + 0}]/(px² + qx + r)²

f'(x) = [(px²+ qx + r)a - (ax + b) {2px + q}]/(px² + qx + r)²

f'(x) = [apx²+ aqx + ar - 2apx² - aqx - 2bpx - bq]/(px² + qx + r)²

f'(x) = [-apx² + ar - 2bpx - bq]/(px² + qx + r)²

9. (px² + qx + r)/(ax + b)

解决方案

令 (px² + qx + r)/(ax + b) = f(x)

两边求导

d/dx (f(x)) = d/dx [(px² + qx + r)/(ax + b)]

f'(x) = [(ax + b) d/dx (px² + qx + r) - (px²+ qx + r) d/dx (ax + b)]/(ax + b)²

f'(x) = [(ax + b) {p d/dx (x²) + q d/dx (x) + d/dx (r)} - (px²+ qx + r) {a d/dx (x) + d/dx (b)}]/(ax + b)²

f'(x) = [(ax + b) {p(2x) + q(1) + 0} - (px²+ qx + r) {a(1) + 0}]/(ax + b)²

f'(x) = [(ax + b) {2px + q} - (px²+ qx + r)a]/(ax + b)²

f'(x) = [2apx² + aqx + 2bpx + bq - apx² - aqx - ar]/(ax + b)²

f'(x) = [apx² - ar + 2bpx + bq]/(ax + b)²

10. a/x⁴ - b/x² + cos x

解决方案

令 a/x⁴ - b/x² + cos x = f(x)

两边求导

d/dx (f(x)) = d/dx [a/x⁴ - b/x² + cos x]

f'(x) = d/dx (a/x⁴) - d/dx (b/x²) + d/dx (cos x)

f'(x) = [x⁴ d/dx (a) - a d/dx (x⁴)]/x⁸ - [x²d/dx (b) - b d/dx (x²)]/x⁴ + (-sin x)

f'(x) = [x⁴ (0) - a(4x³)]/x⁸ - [x²(0) - b(2x)]/x⁴ - sin x

f'(x) = [-4ax³]/x⁸ - [-2bx]/x⁴ - sin x

f'(x) = -4a/x⁵ + 2b/x³ - sin x

11. 4√x - 2

解决方案

令 4√x - 2 = f(x)

两边求导

d/dx (f(x)) = d/dx [4√x - 2]

f'(x) = 4 d/dx (√x) - d/dx (2)

f'(x) = 4 d/dx (x^1/2) - 0

f'(x) = 4 (1/2 × x^-1/2)

f'(x) = 2x^-1/2

f'(x) = 2/√x

12. (ax + b)ⁿ

解决方案

令 (ax + b)ⁿ = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} [(a(x + h) + b)ⁿ - (ax + b)ⁿ]/h

f'(x) = lim_{h → 0} [(ax + b + ah)ⁿ - (ax + b)ⁿ]/h

f'(x) = lim_{h → 0} [{(ax + b) (1 + ah/(ax + b))}ⁿ - (ax + b)ⁿ]/h

f'(x) = lim_{h → 0} [{(ax + b) (1 + ah/(ax + b))}ⁿ - (ax + b)ⁿ]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ {(1 + ah/(ax + b))ⁿ - 1}]/h

使用二项式定理

f'(x) = lim_{h → 0} [(ax + b)ⁿ {(1ⁿ + ⁿC₁ 1^{n - 1} (ah/(ax + b))¹ + … + ⁿC_n 1⁰ (ah/(ax + b))ⁿ) - 1}]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ {1 + n!/1!(n - 1)! × (ah)/(ax + b) + ⁿC₂ (ah)²/(ax + b)² + … + (ah)ⁿ/(ax + b)ⁿ - 1}]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ {1 + n(ah)/(ax + b) + ⁿC₂ (ah)²/(ax + b)² + … + (ah)ⁿ/(ax + b)ⁿ - 1}]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ {n(ah)/(ax + b) + ⁿC₂ (ah)²/(ax + b)² + … + (ah)ⁿ/(ax + b)ⁿ}]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ {h(an)/(ax + b) + ⁿC₂ h² (a)²/(ax + b)² + … + hⁿ (a)ⁿ/(ax + b)ⁿ}]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ h {(an)/(ax + b) + ⁿC₂ h (a)²/(ax + b)² + … + h^{n - 1} (a)ⁿ/(ax + b)ⁿ}]/h

f'(x) = lim_{h → 0} [(ax + b)ⁿ {(an)/(ax + b) + ⁿC₂ h (a)²/(ax + b)² + … + h^{n - 1} (a)ⁿ/(ax + b)ⁿ}]

f'(x) = (ax + b)ⁿ {(an)/(ax + b) + ⁿC₂ 0 (a)²/(ax + b)² + … + 0^{n - 1} (a)ⁿ/(ax + b)ⁿ}

f'(x) = (ax + b)ⁿ (an)/(ax + b) + 0

f'(x) = an(ax + b)^{n - 1}

13. (ax + b)ⁿ (cx + d)^m

解决方案

令 (ax + b)ⁿ (cx + d)^m = f(x)

f'(x) = (cx + d)^m ((ax + b)ⁿ)' + (ax + b)ⁿ ((cx + d)^m)'

令 (ax + b)ⁿ = f₁(x) 且 (cx + d)^m = f₂(x)

根据第一原理

f'₁(x) = lim_{h → 0} (f₁(x + h) - f₁(x))/h

f'₁(x) = lim_{h → 0} [(a(x + h) + b)ⁿ - (ax + b)ⁿ]/h

f'₁(x) = lim_{h → 0} [(ax + b + ah)ⁿ - (ax + b)ⁿ]/h

f'₁(x) = lim_{h → 0} [{(ax + b) (1 + ah/(ax + b))}ⁿ - (ax + b)ⁿ]/h

f'₁(x) = lim_{h → 0} [{(ax + b) (1 + ah/(ax + b))}ⁿ - (ax + b)ⁿ]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {(1 + ah/(ax + b))ⁿ - 1}]/h

使用二项式定理

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {(1ⁿ + ⁿC₁ 1^{n - 1} (ah/(ax + b))¹ + … + ⁿC_n 1⁰ (ah/(ax + b))ⁿ) - 1}]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {1 + n!/1!(n - 1)! × (ah)/(ax + b) + ⁿC₂ (ah)²/(ax + b)² + … + (ah)ⁿ/(ax + b)ⁿ - 1}]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {1 + n(ah)/(ax + b) + ⁿC₂ (ah)²/(ax + b)² + … + (ah)ⁿ/(ax + b)ⁿ - 1}]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {n(ah)/(ax + b) + ⁿC₂ (ah)²/(ax + b)² + … + (ah)ⁿ/(ax + b)ⁿ}]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {h(an)/(ax + b) + ⁿC₂ h² (a)²/(ax + b)² + … + hⁿ (a)ⁿ/(ax + b)ⁿ}]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ h {(an)/(ax + b) + ⁿC₂ h (a)²/(ax + b)² + … + h^{n - 1} (a)ⁿ/(ax + b)ⁿ}]/h

f'₁(x) = lim_{h → 0} [(ax + b)ⁿ {(an)/(ax + b) + ⁿC₂ h (a)²/(ax + b)² + … + h^{n - 1} (a)ⁿ/(ax + b)ⁿ}]

f'₁(x) = (ax + b)ⁿ {(an)/(ax + b) + ⁿC₂ 0 (a)²/(ax + b)² + … + 0^{n - 1} (a)ⁿ/(ax + b)ⁿ}

f'₁(x) = (ax + b)ⁿ (an)/(ax + b) + 0

f'₁(x) = an(ax + b)^{n - 1}

类似地，f₂^'(x) = mc(cx + d)^{m - 1}

因此，

f'(x) = (cx + d)^m an(ax + b)^{n - 1} + (ax + b)ⁿ mc(cx + d)^{m - 1}

f'(x) = (cx + d)^{m - 1} (ax + b)^{n - 1} [(cx + d)an + (ax + b)mc]

14. sin (x + a)

解决方案

令 sin (x + a) = f(x)

根据第一原理

f'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'(x) = lim_{h → 0} (sin (x + h + a) - sin (x + a))/h

f'(x) = lim_{h → 0} [2 cos ((x + h + a + x + a)/2) sin ((x + h + a - x - a)/2)]/h

f'(x) = lim_{h → 0} [2 cos ((2x + h + 2a)/2) sin (h/2)]/h

f'(x) = lim_{h → 0} cos ((2x + h + 2a)/2) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'(x) = lim_{h → 0} cos ((2x + h + 2a)/2) × 1

f'(x) = cos ((2x + 0 + 2a)/2)

f'(x) = cos (2(x + a)/2)

f'(x) = cos (x + a)

15. cosec x cot x

解决方案

令 cosec x cot x = f(x)

f'(x) = cosec x (cot x)' + cot x (cosec x)'

设 cot x = f₁(x)

根据第一原理

f₁'(x) = lim_{h → 0} (f(x + h) - f(x))/h

f₁'(x) = lim_{h → 0} (cot (x + h) - cot x)/h

f₁'(x) = lim_{h → 0} (cos (x + h)/sin (x + h) - cos x/sin x)/h

f₁'(x) = lim_{h → 0} (sin x cos (x + h) - cos x sin (x + h))/h sin x sin (x + h)

f₁'(x) = lim_{h → 0} (sin (x - x - h))/h sin x sin (x + h)

f₁'(x) = lim_{h → 0} (sin (-h))/h sin x sin (x + h)

f₁'(x) = lim_{h → 0} (-sin h)/h sin x sin (x + h)

f₁'(x) = -lim_{h → 0} (sin h)/h sin x sin (x + h)

f₁'(x) = -lim_{h → 0} (sin h)/h × lim_{h → 0} 1/[sin x sin (x + h)]

f₁'(x) = -1 × lim_{h → 0} 1/[sin x sin (x + h)]

f₁'(x) = -1 × 1/sin x sin (x + 0)

f₁'(x) = -1 × 1/sin x sin x

f₁'(x) = -1/sin² x

f₁'(x) = -cosec² x

设 cosec x = f₂(x)

根据第一原理

f₂'(x) = lim_{h → 0} (f₂(x + h) - f₂(x))/h

f'₂(x) = lim_{h → 0} (cosec (x + h) - cosec x)/h

f₂'(x) = lim_{h → 0} (1/sin (x + h) - 1/sin x)/h

f₂'(x) = lim_{h → 0} (sin x - sin (x + h))/h sin x sin (x + h)

f₂'(x) = lim_{h → 0} [2 cos ((x + x + h)/2) sin ((x - x - h)/2)]/h sin x sin (x + h)

f₂'(x) = lim_{h → 0} [2 cos ((2x + h)/2) sin (-h/2)]/h sin x sin (x + h)

f₂'(x) = lim_{h → 0} [-2 cos ((2x + h)/2) sin (h/2)]/h sin x sin (x + h)

f₂'(x) = -lim_{h → 0} [cos ((2x + h)/2)]/sin x sin (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f₂'(x) = -lim_{h → 0}[cos ((2x + h)/2)]/sin x sin (x + h) × 1

f₂'(x) = -cos((2x + 0)/2)/sin x sin (x + 0)

f₂'(x) = -cos(2x/2)/sin x sin x

f₂'(x) = -cos x/sin x × 1/sin x

f₂'(x) = -cot x cosec x

因此，

f'(x) = cosec x (-cosec² x) + cot x (-cot x cosec x)

f'(x) = -cosec³ x - cot² x cosec x

f'(x) = -cosec x (cosec² x + cot²)

16. cos x/(1 + sin x)

解决方案

令 cos x/(1 + sin x) = f(x)

f'(x) = [(1 + sin x) d/dx (cos x) - cos x d/dx (1 + sinx)]/(1 + sin x)²

f'(x) = [(1 + sin x) (-sin x) - cos x {d/dx (1) + d/dx (sin x)}]/(1 + sin x)²

f'(x) = [-sin x - sin²x - cos x {0 + cos x}]/(1 + sin x)²

f'(x) = [-sin x - sin²x - cos² x]/(1 + sin x)²

f'(x) = [-sin x - (sin²x + cos² x)]/(1 + sin x)²

f'(x) = [-sin x - 1]/(1 + sin x)²

f'(x) = -[1 + sin x]/(1 + sin x)²

f'(x) = -1/(1 + sin x)

17. (sin x + cos x)/(sin x - cos x)

解决方案

令 (sin x + cos x)/(sin x - cos x) = f(x)

f'(x) = [(sin x - cos x) d/dx (sin x + cos x) - (sin x + cos x) d/dx (sin x - cos x)]/(sin x - cos x)²

f'(x) = [(sin x - cos x) {d/dx (sin x) + d/dx (cos x)} - (sin x + cos x) {d/dx (sin x) - d/dx (cos x)}]/(sin x - cos x)²

f'(x) = [(sin x - cos x) {cos x - sin x} - (sin x + cos x) {cos x + sin x}]/(sin x - cos x)²

f'(x) = [-(sin x - cos x)² - (sin x + cos x)²]/(sin x - cos x)²

f'(x) = [-sin² x - cos² x + 2 sin x cos x - sin² x - cos² x - 2 sin x cos x]/(sin x - cos x)²

f'(x) = [-2 sin² x - 2 cos² x]/(sin x - cos x)²

f'(x) = [-2(sin² x + cos² x)]/(sin x - cos x)²

f'(x) = [-2(1)]/(sin x - cos x)²

f'(x) = -2/(sin x - cos x)²

18. (sec x - 1)/(sec x + 1)

解决方案

令 (sec x - 1)/(sec x + 1) = f(x)

f(x) = (1/cos x - 1)/(1/cos x + 1)

f(x) = [(1 - cos x)/cos x]/[(1 + cos x)/cos x]

f(x) = (1 - cos x)/(1 + cos x)

f'(x) = [(1 + cos x) d/dx (1 - cos x) - (1 - cos x) d/dx (1 + cos x)]/(1 + cos x)²

f'(x) = [(1 + cos x) {d/dx (1) - d/dx (cos x)} - (1 - cos x) {d/dx (1) + d/dx (cos x)}]/(1 + cos x)²

f'(x) = [(1 + cos x) {0 + sin x} - (1 - cos x) {0 - sin x}]/(1 + cos x)²

f'(x) = [(1 + cos x)(sin x) + (1 - cos x)(sin x)]/(1 + cos x)²

f'(x) = [sin x + sinx cos x + sin x - sin x cos x]/(1 + cos x)²

f'(x) = (2 sin x)/(1 + cos x)²

f'(x) = (2 sin x)/(1 + 1/sec x)²

f'(x) = (2 sin x)/[(sec x + 1)²/sec²x]

f'(x) = (2 sin x sec² x)/(sec x + 1)²

f'(x) = (2 tan x secx)/(sec x + 1)²

19. sinⁿ x

解决方案

令 y = sinⁿ x

当 n = 1 时，y = sin x

dy/dx = d/dx (sin x) = cos x

当 n = 2 时，y = sin² x

dy/dx = d/dx (sin x sin x)

dy/dx = sin x d/dx (sin x) + sin x d/dx (sin x)

dy/dx = sin x (cos x) + sin x (cos x)

dy/dx = 2 sin x cos x

当 n = 3 时，y = sin³ x

dy/dx = d/dx (sin² x sin x)

dy/dx = sin² x d/dx (sin x) + sin x d/dx (sin² x)

dy/dx = sin² x (cos x) + sin x d/dx (sin x sin x)

dy/dx = sin² x cos x + sin x (2 sin x cos x)

dy/dx = sin² x cos x + 2 sin² x cos x

dy/dx = 3 sin² x cos x

因此，我们可以得出结论

d/dx (sinⁿ x) = n sin^{n - 1} x cos x

假设这对于 n = k 成立

d/dx (sin^k x) = k sin^{k - 1} x cos x … (I)

现在，对于 n = k + 1

d/dx (sin^{k + 1} x) = d/dx (sin^k x sin x)

d/dx (sin^{k + 1} x) = sin^k x d/dx (sin x) + sin x d/dx (sin^k x)

使用方程 (I)，我们得到

d/dx (sin^{k + 1} x) = sin^k x (cos x) + sin x (k sin^{k - 1} x cos x)

d/dx (sin^{k + 1} x) = sin^k x cos x + k sin^k x cos x

d/dx (sin^{k + 1} x) = (k + 1) sin^k x cos x

因此，对于 n = k + 1 也成立。

故，

d/dx (sinⁿ x) = n sin^{n - 1} x cos x

20. (a + b sin x)/(c + d cos x)

解决方案

令 (a + b sin x)/(c + d cos x) = f(x)

f'(x) = [(c + d cos x) d/dx (a + b sin x) - (a + b sin x) d/dx (c + d cos x)]/(c + d cos x)²

f'(x) = [(c + d cos x) {d/dx (a) + b d/dx (sin x)} - (a + b sin x) {d/dx (c) + d d/dx (cos x)}]
/(c + d cos x)²

f'(x) = [(c + d cos x) {0 + b(cos x)} - (a + b sin x) {0 + d(-sin x)}]/(c + d cos x)²

f'(x) = [(c + d cos x) b cos x + (a + b sin x) d sin x]/(c + d cos x)²

f'(x) = [bc cos x + bd cos² x + ad sin x + bd sin² x]/(c + d cos x)²

f'(x) = [bc cos x + ad sin x + bd(cos² x + sin² x)]/(c + d cos x)²

f'(x) = [bc cos x + ad sin x + bd(1)]/(c + d cos x)²

f'(x) = (bc cos x + ad sin x + bd)/(c + d cos x)²

21. sin (x + a)/cos x

解决方案

令 sin (x + a)/cos x = f(x)

f'(x) = [cos x d/dx (sin (x + a)) - sin (x + a) d/dx (cos x)]/cos² x

f'(x) = [cos x d/dx (sin (x + a)) - sin (x + a) d/dx (cos x)]/cos² x

令 sin (x + a) = g(x)

根据第一原理

g'(x) = lim_{h → 0} (g(x + h) - g(x))/h

g'(x) = lim_{h → 0} (sin (x + h + a) - sin (x + a))/h

g'(x) = lim_{h → 0} [2 cos ((x + h + a + x + a)/2) sin ((x + h + a - x - a)/2)]/h

g'(x) = lim_{h → 0} [2 cos ((2x + h + 2a)/2) sin (h/2)]/h

g(x) = lim_{h → 0} cos ((2x + h + 2a)/2) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

g'(x) = lim_{h → 0} cos ((2x + h + 2a)/2) × 1

g'(x) = cos ((2x + 0 + 2a)/2)

g'(x) = cos (2(x + a)/2)

g'(x) = cos (x + a)

因此，

f'(x) = [cos x (cos (x + a)) - sin (x + a) (-sin x)]/cos² x

f'(x) = [cos x cos (x + a) + sin (x + a) sin x]/cos² x

f'(x) = [cos (x + a - x)]/cos² x

f'(x) = cos a/cos² x

22. x⁴(5 sin x - 3 cos x)

解决方案

令 x⁴(5 sin x - 3 cos x) = f(x)

f'(x) = x⁴ d/dx (5 sin x - 3 cos x) + (5 sin x - 3 cos x) d/dx (x⁴)

f'(x) = x⁴ (5 d/dx (sin x) - 3 d/dx (cos x)) + (5 sin x - 3 cos x) (4x³)

f'(x) = x⁴ (5 cos x + 3 sin x) + 20x³ sin x - 12x³ cos x

f'(x) = 5x⁴ cos x + 3x⁴ sin x + 20x³ sin x - 12x³ cos x

f'(x) = x³(5x cos x + 3x sin x + 20 sin x - 12 cos x)

23. (x² + 1) cos x

解决方案

令 (x² + 1) cos x = f(x)

f'(x) = (x² + 1) d/dx (cos x) + cos x d/dx (x² + 1)

f'(x) = (x² + 1)(-sin x) + cos x (2x + 0)

f'(x) = -x² sin x - sin x + 2x cos x

24. (ax² + sin x)(p + q cos x)

解决方案

令 (ax² + sin x)(p + q cos x) = f(x)

f'(x) = (ax² + sin x) d/dx (p + q cos x) + (p + q cos x) d/dx (ax² + sin x)

f'(x) = (ax² + sin x)(0 - q sin x) + (p + q cos x)(a d/dx (x²) + cos x)

f'(x) = -aqx² sin x - sin² x + (p + q cos x)(2ax + cos x)

25. (x + cos x)(x - tan x)

解决方案

令 (x + cos x)(x - tan x) = f(x)

f'(x) = (x + cos x) d/dx (x - tan x) + (x - tan x) d/dx (x + cos x)

f'(x) = (x + cos x)(1 - d/dx (tan x)) + (x - tan x)(1 + d/dx (cos x))

令 tan x = g(x)

根据第一原理

g'(x) = lim_{h → 0} (g(x + h) - g(x))/h

g'(x) = lim_{h → 0} (tan (x + h) - tan x)/h

g'(x) = lim_{h → 0} (sin (x + h)/cos (x + h) - sin x/cos x)/h

g'(x) = lim_{h → 0} (cos x sin (x + h) - cos (x + h) sin x)/h cos x cos (x + h)

乘以并除以 2

g'(x) = lim_{h → 0} (2 cos x sin (x + h) - 2 cos (x + h) sin x)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin (x + h + x) - sin(x - x - h) - sin(x + h + x) + sin (x + h - x))/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin (2x + h) - sin(-h) - sin (2x + h) + sin h)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin h + sin h)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (2 sin h)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin h)/h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin h)/h × lim_{h → 0} 1/cos x cos (x + h)

g'(x) = 1 × lim_{h → 0} 1/cos x cos (x + h)

g'(x) = 1/cos x cos (x + 0)

g'(x) = 1/cos x cos x

g'(x) = sec²x

因此，

f'(x) = (x + cos x) (1 - sec² x) + (x - tan x)(1 - sin x)

f'(x) = (x + cos x)(-tan² x) + (x - tan x)(1 - sin x)

26. (4x + 5 sin x)/(3x + 7 cos x)

解决方案

令 (4x + 5 sin x)/(3x + 7 cos x) = f(x)

f'(x) = [(3x + 7 cos x) d/dx (4x + 5 sin x) - (4x + 5 sin x) d/dx (3x + 7 cos x)]/(3x + 7 cos x)²

f'(x) = [(3x + 7 cos x) (4 + 5(cos x)) - (4x + 5 sin x) (3 + 7(-sin x))]/(3x + 7 cos x)²

f'(x) = [(3x + 7 cos x)(4 + 5 cos x) - (4x + 5 sin x)(3 - 7 sin x)]/(3x + 7 cos x)²

f'(x) = [12x + 28 cos x + 15x cos x + 35 cos² x - 12x + 28x sin x - 15 sin x + 35 sin² x]/(3x + 7 cos x)²

f'(x) = [28 cos x + 15x cos x + 35 (cos²x + sin² x) + 28x sin x - 15 sin x]/(3x + 7 cos x)²

f'(x) = [35 + 28 cos x + 15x cos x + 28x sin x - 15 sin x]/(3x + 7 cos x)²

27. (x² cos π/4)/sin x

解决方案

令 (x² cos π/4)/sin x = f(x)

f'(x) = cos π/4 × [sin x d/dx (x²) - x² d/dx (sin x)]/sin² x

f'(x) = cos π/4 × [2x sin x - x² cos x]/sin² x

f'(x) = x cos π/4 × [2 sin x - x cos x]/sin² x

28. x/(1 + tan x)

解决方案

令 x/(1 + tan x) = f(x)

f'(x) = [(1 + tan x) d/dx (x) - x d/dx (1 + tan x)]/(1 + tan x)²

f'(x) = [(1 + tan x)(1) - x (0 + d/dx (tan x))]/(1 + tan x)²

令 tan x = g(x)

根据第一原理

g'(x) = lim_{h → 0} (g(x + h) - g(x))/h

g'(x) = lim_{h → 0} (tan (x + h) - tan x)/h

g'(x) = lim_{h → 0} (sin (x + h)/cos (x + h) - sin x/cos x)/h

g'(x) = lim_{h → 0} (cos x sin (x + h) - cos (x + h) sin x)/h cos x cos (x + h)

乘以并除以 2

g'(x) = lim_{h → 0} (2 cos x sin (x + h) - 2 cos (x + h) sin x)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin (x + h + x) - sin(x - x - h) - sin(x + h + x) + sin (x + h - x))/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin (2x + h) - sin(-h) - sin (2x + h) + sin h)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin h + sin h)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (2 sin h)/2h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin h)/h cos x cos (x + h)

g'(x) = lim_{h → 0} (sin h)/h × lim_{h → 0} 1/cos x cos (x + h)

g'(x) = 1 × lim_{h → 0} 1/cos x cos (x + h)

g'(x) = 1/cos x cos (x + 0)

g'(x) = 1/cos x cos x

g'(x) = sec²x

因此，

f'(x) = [1 + tan x - x sec² x]/(1 + tan x)²

29. (x + sec x) (x - tan x)

解决方案

令 (x + sec x)(x - tan x) = f(x)

f'(x) = (x + sec x) d/dx (x - tan x) + (x - tan x) d/dx (x + sec x)

f'(x) = (x + sec x)(1 - d/dx (tan x)) + (x - tan x)(1 + d/dx (sec x))

设 tan x = f₁(x)

根据第一原理

f₁'(x) = lim_{h → 0} (f₁(x + h) - f₁(x))/h

f₁'(x) = lim_{h → 0} (tan (x + h) - tan x)/h

f₁'(x) = lim_{h → 0} (sin (x + h)/cos (x + h) - sin x/cos x)/h

f₁'(x) = lim_{h → 0} (cos x sin (x + h) - cos (x + h) sin x)/h cos x cos (x + h)

乘以并除以 2

f₁'(x) = lim_{h → 0} (2 cos x sin (x + h) - 2 cos (x + h) sin x)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin (x + h + x) - sin (x - x - h) - sin (x + h + x) + sin (x + h - x))/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin (2x + h) - sin (-h) - sin (2x + h) + sin h)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin h + sin h)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (2 sin h)/2h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin h)/h cos x cos (x + h)

f₁'(x) = lim_{h → 0} (sin h)/h × lim_{h → 0} 1/cos x cos(x + h)

f₁'(x) = 1 × lim_{h → 0} 1/cos x cos (x + h)

f₁'(x) = 1/cos x cos (x + 0)

f₁'(x) = 1/cos x cos x

f₁'(x) = sec² x

设 sec x = f₂(x)

根据第一原理

f'₂(x) = lim_{h → 0} (f(x + h) - f(x))/h

f'₂(x) = lim_{h → 0} (sec (x + h) - sec x)/h

f'₂(x) = lim_{h → 0} (1/cos (x + h) - 1/cos x)/h

f'₂(x) = lim_{h → 0} (cos x - cos (x + h))/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [-2 sin ((x + x + h)/2) sin ((x - x - h)/2)]/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [-2 sin ((2x + h)/2) sin (-h/2)]/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [2 sin ((2x + h)/2) sin (h/2)]/h cos x cos (x + h)

f'₂(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × lim_{h/2 → 0} [sin (h/2)]/(h/2)

f'₂(x) = lim_{h → 0} [sin ((2x + h)/2)]/cos x cos (x + h) × 1

f'₂(x) = sin ((2x + 0)/2)/cos x cos (x + 0)

f'₂(x) = sin (2x/2)/cos x cos x

f'₂(x) = sin x/cos x × 1/cos x

f'₂(x) = tan x sec x

因此，

f'(x) = (x + sec x)(1 - sec² x) + (x - tan x)(1 + sec x tan x)

30. x/sinⁿ x

解决方案

令 x/sinⁿ x = f(x)

f'(x) = [sinⁿ x d/dx (x) - x d/dx (sinⁿ x)]/sin²ⁿ x

我们知道 d/dx (sinⁿ x) = n sin^{n - 1} x cos x。因此，

f'(x) = [sinⁿ x (1) - x (n sin^{n - 1} x cos x)]/sin²ⁿ x

f'(x) = [sinⁿ x - nx sin^{n - 1} x cos x]/sin²ⁿ x

f'(x) = sin^{n -1} x (sin x - nx cos x)/sin²ⁿ x

f'(x) = (sin x - nx cos x)/sin^{n + 1} x