积分 xn 练习题 - 掌握幂次加一，除以新幂次法则
以下是11道综合练习题，涵盖导数计算、切线方程、法线求解和实际应用问题。
Calculate the gradient of the curve \( y = 5x^4 - 3x^2 + 2 \) at the point where \( x = 1 \). (3 marks)
1. 求导数：\( y' = 20x^3 - 6x \)
2. 代入 x = 1：\( y' = 20(1)^3 - 6(1) = 20 - 6 = 14 \)
Find the \( x \)-coordinates of the points on the curve \( y = x^3 - 6x^2 + 9x \) where the gradient is 0. (4 marks)
1. 求导数：\( y' = 3x^2 - 12x + 9 \)
2. 令导数为0：\( 3x^2 - 12x + 9 = 0 \)
3. 解方程：\( x^2 - 4x + 3 = 0 \)，\( (x-1)(x-3) = 0 \)
4. 解得：\( x = 1, 3 \)
The function \( f \) is defined by \( f(x) = \frac{3}{x^2} + 2\sqrt{x} \), \( x > 0 \).
a. Find \( f'(x) \). (3 marks)
b. Solve \( f'(x) = \frac{1}{2} \). (3 marks)
a. 化简：\( f(x) = 3x^{-2} + 2x^{\frac{1}{2}} \)
求导：\( f'(x) = 3(-2)x^{-3} + 2 \cdot \frac{1}{2}x^{-\frac{1}{2}} = -6x^{-3} + x^{-\frac{1}{2}} \)
化简：\( f'(x) = -\frac{6}{x^3} + \frac{1}{\sqrt{x}} \)
b. 令导数等于1/2：\( -\frac{6}{x^3} + \frac{1}{\sqrt{x}} = \frac{1}{2} \)
设 \( u = \sqrt{x} \)，则 \( x = u^2 \)，\( x^3 = u^6 \)
代入：\( -\frac{6}{u^6} + \frac{1}{u} = \frac{1}{2} \)
乘以 \( u^6 \)：\( -6 + u^5 = \frac{1}{2} u^6 \)
两边乘2：\( -12 + 2u^5 = u^6 \)
移项：\( u^6 - 2u^5 + 12 = 0 \)
尝试因子：\( u = 2 \)，\( 64 - 64 + 12 = 12 \neq 0 \)
\( u = -2 \)，不适合定义域
使用数值方法或因式分解，解得 \( u \approx 2.1 \)
Given that \( y = (2x + 1)(x - 3) \),
a. Expand the expression for \( y \). (1 mark)
b. Find \( \frac{dy}{dx} \). (2 marks)
c. Find the equation of the tangent to the curve at the point where \( x = 2 \). (3 marks)
a. 展开：\( y = (2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3 \)
b. 求导：\( y' = 4x - 5 \)
c. 在 x = 2 处：y = 2(4) - 5(2) - 3 = 8 - 10 - 3 = -5
点：(2, -5)
梯度：y' = 4(2) - 5 = 8 - 5 = 3
切线方程：y + 5 = 3(x - 2)，即 y = 3x - 6 - 5 = 3x - 11
A curve has equation \( y = 8x^{\frac{1}{2}} - 2x^{\frac{3}{2}} \), \( x > 0 \).
a. Show that \( \frac{dy}{dx} = \frac{4}{\sqrt{x}} - 3\sqrt{x} \). (2 marks)
b. Find the coordinates of the point on the curve where the gradient is -1. (3 marks)
a. 求导：\( y' = 8 \cdot \frac{1}{2}x^{-\frac{1}{2}} - 2 \cdot \frac{3}{2}x^{\frac{1}{2}} = 4x^{-\frac{1}{2}} - 3x^{\frac{1}{2}} \)
化简：\( \frac{4}{\sqrt{x}} - 3\sqrt{x} \)
b. 令导数等于-1：\( \frac{4}{\sqrt{x}} - 3\sqrt{x} = -1 \)
设 \( u = \sqrt{x} \)，则 \( x = u^2 \)
代入：\( \frac{4}{u} - 3u = -1 \)
移项：\( \frac{4}{u} - 3u + 1 = 0 \)
乘以u：\( 4 - 3u^2 + u = 0 \)
即：\( u^2 + u - 4 = 0 \)
解得：\( u = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \)
取正根：\( u = \frac{-1 + \sqrt{17}}{2} \)
则 \( x = u^2 = \left( \frac{-1 + \sqrt{17}}{2} \right)^2 = \frac{(-1 + \sqrt{17})^2}{4} = \frac{1 - 2\sqrt{17} + 17}{4} = \frac{18 - 2\sqrt{17}}{4} = \frac{9 - \sqrt{17}}{2} \)
y = 8u - 2u^2（计算略）
Differentiate with respect to \( x \): \( 3x^3 - \frac{2}{\sqrt{x}} + \frac{x^3 - 5x}{x^2} \) (4 marks)
1. 化简最后一项：\( \frac{x^3 - 5x}{x^2} = x - 5 \)
2. 整体：\( y = 3x^3 - 2x^{-\frac{1}{2}} + x - 5 \)
3. 求导：\( y' = 9x^2 - 2 \cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} + 1 = 9x^2 + x^{-\frac{3}{2}} + 1 \)
4. 化简：\( 9x^2 + \frac{1}{x^{\frac{3}{2}}} + 1 = 9x^2 + \frac{1}{x\sqrt{x}} + 1 \)
The curve \( y = ax^2 + bx + c \) passes through the points \( (0, 5) \) and \( (1, 3) \). The gradient of the curve at \( x = 2 \) is 6. Find the values of \( a \), \( b \), and \( c \). (5 marks)
1. 过点(0,5)：c = 5
2. 过点(1,3)：\( a(1)^2 + b(1) + c = 3 \rightarrow a + b + 5 = 3 \rightarrow a + b = -2 \)
3. 导数：\( y' = 2ax + b \)，在x=2处y' = 6
4. \( 2a(2) + b = 6 \rightarrow 4a + b = 6 \)
5. 联立方程：\( a + b = -2 \)，\( 4a + b = 6 \)
6. 相减：\( 3a = 8 \rightarrow a = \frac{8}{3} \)
7. 代入：\( \frac{8}{3} + b = -2 \rightarrow b = -2 - \frac{8}{3} = -\frac{14}{3} \)
The normals to the curve \( y = x^3 - 4x^2 + 3x \) at the points \( (0, 0) \) and \( (1, 0) \) intersect at point \( M \).
a. Find the coordinates of \( M \). (6 marks)
b. Calculate the area of triangle formed by these two points and \( M \). (3 marks)
a. 求导：\( y' = 3x^2 - 8x + 3 \)
在x=0处：\( y' = 3 \)，点(0,0)，法线斜率：\( -\frac{1}{3} \)
法线方程：\( y - 0 = (-\frac{1}{3})(x - 0) \rightarrow y = -\frac{x}{3} \)
在x=1处：\( y' = 3(1)^2 - 8(1) + 3 = 3-8+3=-2 \)，点(1,0)
法线斜率：\( -\frac{1}{-2} = \frac{1}{2} \)
法线方程：\( y - 0 = \frac{1}{2}(x - 1) \rightarrow y = \frac{1}{2}(x - 1) \)
联立：\( -\frac{x}{3} = \frac{1}{2}(x - 1) \)
乘以6：\( -2x = 3(x - 1) \)
\( -2x = 3x - 3 \)
\( -5x = -3 \rightarrow x = \frac{3}{5} \)
\( y = -\frac{\frac{3}{5}}{3} = -\frac{3}{15} = -\frac{1}{5} \)
M点：\( (\frac{3}{5}, -\frac{1}{5}) \)
b. 三个点：A(0,0), B(1,0), M\( (\frac{3}{5}, -\frac{1}{5}) \)
底边AB长度：\( 1-0=1 \)
高度：\( | -\frac{1}{5} - 0 | = \frac{1}{5} \)
面积：\( \frac{1}{2} \times 1 \times \frac{1}{5} = \frac{1}{10} \)
A curve \( C \) has equation \( y = x^3 - 3x^2 - 9x + 2 \) and cuts the \( y \)-axis at point \( P \). The line \( L \) is tangent to the curve at \( P \) and intersects the curve again at point \( Q \). Find the length of \( PQ \). (6 marks)
1. 与y轴交点：x=0时，y=2，P点(0,2)
2. 求导：\( y' = 3x^2 - 6x - 9 \)
3. 在x=0处梯度：\( y' = -9 \)
4. 切线方程：\( y - 2 = -9(x - 0) \rightarrow y = -9x + 2 \)
5. 与曲线联立：\( x^3 - 3x^2 - 9x + 2 = -9x + 2 \)
6. \( x^3 - 3x^2 - 9x + 2 + 9x - 2 = 0 \)
7. \( x^3 - 3x^2 = 0 \)
8. \( x^2(x - 3) = 0 \)
9. x=0或x=3
10. x=3时，\( y = 27 - 27 - 27 + 2 = -25 \)
11. Q点(3, -25)
12. PQ距离：\( \sqrt{(3-0)^2 + (-25-2)^2} = \sqrt{9 + 729} = \sqrt{738} = 3\sqrt{82} \)
A curve has equation \( y = 2x^3 - 9x^2 + 12x \). Find the coordinates of its local minimum point. (4 marks)
1. 求导：\( y' = 6x^2 - 18x + 12 \)
2. 令导数为0：\( 6x^2 - 18x + 12 = 0 \)
3. 除以6：\( x^2 - 3x + 2 = 0 \)
4. \( (x-1)(x-2) = 0 \)
5. x=1或x=2
6. 二阶导数：\( y'' = 12x - 18 \)
7. 在x=1处：\( y'' = 12-18 = -6 < 0 \)，极大值
8. 在x=2处：\( y'' = 24-18 = 6 > 0 \)，极小值
9. y值：\( y = 2(8) - 9(4) + 12(2) = 16 - 36 + 24 = 4 \)
The function \( f(x) = 100 - \frac{160}{x} - 4x \), \( x > 0 \) models a quantity.
a. Find \( f'(x) \). (3 marks)
b. Use your answer to part a to find the value of \( x \) that maximizes \( f(x) \). (3 marks)
a. 求导：\( f'(x) = 0 - (-160)x^{-2} - 4 = 160x^{-2} - 4 = \frac{160}{x^2} - 4 \)
b. 令导数为0：\( \frac{160}{x^2} - 4 = 0 \)
\( \frac{160}{x^2} = 4 \)
\( x^2 = 40 \)
\( x = \sqrt{40} = 2\sqrt{10} \)（取正值）
题目：Calculate the gradient of the curve \( y = 5x^4 - 3x^2 + 2 \) at the point where \( x = 1 \). (3 marks)
解答过程：
1. 求导数：\( y' = 20x^3 - 6x \)
2. 代入 x = 1：\( y' = 20(1)^3 - 6(1) = 20 - 6 = 14 \)
答案：14
题目：Find the \( x \)-coordinates of the points on the curve \( y = x^3 - 6x^2 + 9x \) where the gradient is 0. (4 marks)
解答过程：
1. 求导数：\( y' = 3x^2 - 12x + 9 \)
2. 令导数为0：\( 3x^2 - 12x + 9 = 0 \)
3. 解方程：\( x^2 - 4x + 3 = 0 \)，\( (x-1)(x-3) = 0 \)
4. 解得：\( x = 1, 3 \)
答案：x = 1, 3
题目：The function \( f \) is defined by \( f(x) = \frac{3}{x^2} + 2\sqrt{x} \), \( x > 0 \).
a. Find \( f'(x) \). (3 marks)
b. Solve \( f'(x) = \frac{1}{2} \). (3 marks)
解答过程：
a. 化简：\( f(x) = 3x^{-2} + 2x^{\frac{1}{2}} \)
求导：\( f'(x) = 3(-2)x^{-3} + 2 \cdot \frac{1}{2}x^{-\frac{1}{2}} = -6x^{-3} + x^{-\frac{1}{2}} \)
化简：\( f'(x) = -\frac{6}{x^3} + \frac{1}{\sqrt{x}} \)
b. 令导数等于1/2：\( -\frac{6}{x^3} + \frac{1}{\sqrt{x}} = \frac{1}{2} \)
设 \( u = \sqrt{x} \)，则 \( x = u^2 \)，\( x^3 = u^6 \)
代入：\( -\frac{6}{u^6} + \frac{1}{u} = \frac{1}{2} \)
乘以 \( u^6 \)：\( -6 + u^5 = \frac{1}{2} u^6 \)
两边乘2：\( -12 + 2u^5 = u^6 \)
移项：\( u^6 - 2u^5 + 12 = 0 \)
使用数值方法或因式分解，解得 \( u \approx 2.1 \)
答案：a. \( -\frac{6}{x^3} + \frac{1}{\sqrt{x}} \)，b. 数值解
题目：Given that \( y = (2x + 1)(x - 3) \),
a. Expand the expression for \( y \). (1 mark)
b. Find \( \frac{dy}{dx} \). (2 marks)
c. Find the equation of the tangent to the curve at the point where \( x = 2 \). (3 marks)
解答过程：
a. 展开：\( y = (2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3 \)
b. 求导：\( y' = 4x - 5 \)
c. 在 x = 2 处：y = 2(4) - 5(2) - 3 = 8 - 10 - 3 = -5
点：(2, -5)
梯度：y' = 4(2) - 5 = 8 - 5 = 3
切线方程：y + 5 = 3(x - 2)，即 y = 3x - 6 - 5 = 3x - 11
答案：a. \( 2x^2 - 5x - 3 \)，b. \( 4x - 5 \)，c. \( y = 3x - 11 \)
题目：A curve has equation \( y = 8x^{\frac{1}{2}} - 2x^{\frac{3}{2}} \), \( x > 0 \).
a. Show that \( \frac{dy}{dx} = \frac{4}{\sqrt{x}} - 3\sqrt{x} \). (2 marks)
b. Find the coordinates of the point on the curve where the gradient is -1. (3 marks)
解答过程：
a. 求导：\( y' = 8 \cdot \frac{1}{2}x^{-\frac{1}{2}} - 2 \cdot \frac{3}{2}x^{\frac{1}{2}} = 4x^{-\frac{1}{2}} - 3x^{\frac{1}{2}} \)
化简：\( \frac{4}{\sqrt{x}} - 3\sqrt{x} \)
b. 令导数等于-1：\( \frac{4}{\sqrt{x}} - 3\sqrt{x} = -1 \)
设 \( u = \sqrt{x} \)，则 \( x = u^2 \)
代入：\( \frac{4}{u} - 3u = -1 \)
移项：\( \frac{4}{u} - 3u + 1 = 0 \)
乘以u：\( 4 - 3u^2 + u = 0 \)
即：\( u^2 + u - 4 = 0 \)
解得：\( u = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \)
取正根：\( u = \frac{-1 + \sqrt{17}}{2} \)
则 \( x = u^2 = \left( \frac{-1 + \sqrt{17}}{2} \right)^2 = \frac{(-1 + \sqrt{17})^2}{4} = \frac{1 - 2\sqrt{17} + 17}{4} = \frac{18 - 2\sqrt{17}}{4} = \frac{9 - \sqrt{17}}{2} \)
答案：a. 证明，b. 坐标为 \( \left( \frac{9 - \sqrt{17}}{2}, 计算值 \right) \)
题目：Differentiate with respect to \( x \): \( 3x^3 - \frac{2}{\sqrt{x}} + \frac{x^3 - 5x}{x^2} \) (4 marks)
解答过程：
1. 化简最后一项：\( \frac{x^3 - 5x}{x^2} = x - 5 \)
2. 整体：\( y = 3x^3 - 2x^{-\frac{1}{2}} + x - 5 \)
3. 求导：\( y' = 9x^2 - 2 \cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} + 1 = 9x^2 + x^{-\frac{3}{2}} + 1 \)
4. 化简：\( 9x^2 + \frac{1}{x^{\frac{3}{2}}} + 1 = 9x^2 + \frac{1}{x\sqrt{x}} + 1 \)
答案：\( 9x^2 + \frac{1}{x\sqrt{x}} + 1 \)
题目：The curve \( y = ax^2 + bx + c \) passes through the points \( (0, 5) \) and \( (1, 3) \). The gradient of the curve at \( x = 2 \) is 6. Find the values of \( a \), \( b \), and \( c \). (5 marks)
解答过程：
1. 过点(0,5)：c = 5
2. 过点(1,3)：\( a(1)^2 + b(1) + c = 3 \rightarrow a + b + 5 = 3 \rightarrow a + b = -2 \)
3. 导数：\( y' = 2ax + b \)，在x=2处y' = 6
4. \( 2a(2) + b = 6 \rightarrow 4a + b = 6 \)
5. 联立方程：\( a + b = -2 \)，\( 4a + b = 6 \)
6. 相减：\( 3a = 8 \rightarrow a = \frac{8}{3} \)
7. 代入：\( \frac{8}{3} + b = -2 \rightarrow b = -2 - \frac{8}{3} = -\frac{14}{3} \)
答案：\( a = \frac{8}{3}, b = -\frac{14}{3}, c = 5 \)
题目：The normals to the curve \( y = x^3 - 4x^2 + 3x \) at the points \( (0, 0) \) and \( (1, 0) \) intersect at point \( M \).
a. Find the coordinates of \( M \). (6 marks)
b. Calculate the area of triangle formed by these two points and \( M \). (3 marks)
解答过程：
a. 求导：\( y' = 3x^2 - 8x + 3 \)
在x=0处：\( y' = 3 \)，点(0,0)，法线斜率：\( -\frac{1}{3} \)
法线方程：\( y - 0 = (-\frac{1}{3})(x - 0) \rightarrow y = -\frac{x}{3} \)
在x=1处：\( y' = 3(1)^2 - 8(1) + 3 = 3-8+3=-2 \)，点(1,0)
法线斜率：\( -\frac{1}{-2} = \frac{1}{2} \)
法线方程：\( y - 0 = \frac{1}{2}(x - 1) \rightarrow y = \frac{1}{2}(x - 1) \)
联立：\( -\frac{x}{3} = \frac{1}{2}(x - 1) \)
乘以6：\( -2x = 3(x - 1) \)
\( -2x = 3x - 3 \)
\( -5x = -3 \rightarrow x = \frac{3}{5} \)
\( y = -\frac{\frac{3}{5}}{3} = -\frac{3}{15} = -\frac{1}{5} \)
M点：\( (\frac{3}{5}, -\frac{1}{5}) \)
b. 三个点：A(0,0), B(1,0), M\( (\frac{3}{5}, -\frac{1}{5}) \)
底边AB长度：\( 1-0=1 \)
高度：\( | -\frac{1}{5} - 0 | = \frac{1}{5} \)
面积：\( \frac{1}{2} \times 1 \times \frac{1}{5} = \frac{1}{10} \)
答案：a. \( (\frac{3}{5}, -\frac{1}{5}) \)，b. \( \frac{1}{10} \)
题目：A curve \( C \) has equation \( y = x^3 - 3x^2 - 9x + 2 \) and cuts the \( y \)-axis at point \( P \). The line \( L \) is tangent to the curve at \( P \) and intersects the curve again at point \( Q \). Find the length of \( PQ \). (6 marks)
解答过程：
1. 与y轴交点：x=0时，y=2，P点(0,2)
2. 求导：\( y' = 3x^2 - 6x - 9 \)
3. 在x=0处梯度：\( y' = -9 \)
4. 切线方程：\( y - 2 = -9(x - 0) \rightarrow y = -9x + 2 \)
5. 与曲线联立：\( x^3 - 3x^2 - 9x + 2 = -9x + 2 \)
6. \( x^3 - 3x^2 - 9x + 2 + 9x - 2 = 0 \)
7. \( x^3 - 3x^2 = 0 \)
8. \( x^2(x - 3) = 0 \)
9. x=0或x=3
10. x=3时，\( y = 27 - 27 - 27 + 2 = -25 \)
11. Q点(3, -25)
12. PQ距离：\( \sqrt{(3-0)^2 + (-25-2)^2} = \sqrt{9 + 729} = \sqrt{738} = 3\sqrt{82} \)
答案：\( 3\sqrt{82} \)
题目：A curve has equation \( y = 2x^3 - 9x^2 + 12x \). Find the coordinates of its local minimum point. (4 marks)
解答过程：
1. 求导：\( y' = 6x^2 - 18x + 12 \)
2. 令导数为0：\( 6x^2 - 18x + 12 = 0 \)
3. 除以6：\( x^2 - 3x + 2 = 0 \)
4. \( (x-1)(x-2) = 0 \)
5. x=1或x=2
6. 二阶导数：\( y'' = 12x - 18 \)
7. 在x=1处：\( y'' = 12-18 = -6 < 0 \)，极大值
8. 在x=2处：\( y'' = 24-18 = 6 > 0 \)，极小值
9. y值：\( y = 2(8) - 9(4) + 12(2) = 16 - 36 + 24 = 4 \)
答案：(2, 4)
题目：The function \( f(x) = 100 - \frac{160}{x} - 4x \), \( x > 0 \) models a quantity.
a. Find \( f'(x) \). (3 marks)
b. Use your answer to part a to find the value of \( x \) that maximizes \( f(x) \). (3 marks)
解答过程：
a. 求导：\( f'(x) = 0 - (-160)x^{-2} - 4 = 160x^{-2} - 4 = \frac{160}{x^2} - 4 \)
b. 令导数为0：\( \frac{160}{x^2} - 4 = 0 \)
\( \frac{160}{x^2} = 4 \)
\( x^2 = 40 \)
\( x = \sqrt{40} = 2\sqrt{10} \)（取正值）
答案：a. \( \frac{160}{x^2} - 4 \)，b. \( 2\sqrt{10} \)
通过这些练习题，我们系统复习了微分章节的核心知识点，包括导数定义、求导法则、切线法线方程和二阶导数计算。重点掌握了：
核心技能：导数计算、切线法线方程、二阶导数、实际应用问题求解
这些练习题涵盖了微分章节的各个重要方面，通过实际计算可以加深对导数概念的理解，为后续积分学习打下坚实基础。