如何找到切线的斜率？

要找到切线的斜率，我们应该对切线和斜率有一个清晰的概念。斜率定义为 y 坐标差与 x 坐标差的比值。它由以下公式表示：

m =( _{y⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠⁠1} ) /( _{x⁠⁠⁠⁠⁠⁠⁠⁠⁠ 2 -} x⁠⁠⁠⁠⁠⁠⁠⁠ ₁ )

需要注意的是：

• tan θ 与 m 相同。斜率可以是正数或负数，具体取决于线是向上还是向下移动。
• 两条垂直线的斜率乘积为-1，平行线的斜率相同。
• 函数的导数给出了相对于自变量变化的速率变化。

切线的斜率

切线是在一点与曲线相接触的线。可能有切线稍后穿过曲线或在其他一些点接触曲线。但是，如果该线通过点 (a, f(a))（该点对于曲线和切线）并且切线具有斜率 f'(a)，其中 f'(a) 是函数f(x) 在点 a 处的导数。

The slope of the tangent line is the same as the derivate of the curve at some point. The formula for a tangent line whose slope is m and the point given is (x⁠⁠⁠⁠⁠⁠⁠₁, y⁠⁠⁠⁠⁠⁠⁠₁ ) is given by,

y – y⁠⁠⁠⁠⁠⁠⁠₁ = m × (x – x⁠⁠⁠⁠⁠⁠⁠₁ ) 

or 

y= mx + c 

where c is some constant

如何找到切线的斜率？

解决方案：

The slope of a tangent line can be found by finding the derivative of the curve f(x and finding the value of the derivative at the point where the tangent line and the curve meet. This gives us the slope

For example: Find the slope of the tangent line to the curve f(x) = x² at the point(1, 2). Also, find the equation of the tangent line. 

Let us find derivative of f(x):

f'(x) = dy/dx = d(x²) /dx = 2x

Value of slope at point(1, 2) is,

f'(x) = 2(1) = 2

The equation of tangent line is 

y – 2 = 2(x – 1) 

or 

y = 2x

类似问题

问题 1：求切线 6y = 3x + 5 的斜率。

解决方案：

Since we know the equation of a tangent line is of the form y= mx + c where m is the slope

We can write, 

y= (3x + 5 ) / 6

Therefore the value of the slope is 0.5.

问题 2：求给定两个点 (6, 7) 和 (8, 0) 的斜率。

解决方案：

Slope of any two points say (a, b) and (x, y) is given by,

m = (y-b) /(x-a) 

Therefore m = (0-7) /(8-6) = -3.5

问题 3：求曲线 y= 6x³ 的斜率。

解决方案：

The slope of curve is given by differentiation of the curve:

dy/dx = d(6x³) /dx = 18x²

问题 4：在给定 1 个方程为 y= 3x+8 的情况下，求 2 条相互垂直的直线的斜率

解决方案：

Let the slope of two perpendicular lines be m and n

m×n = -1

m=3

n= -1/3

问题 5：求曲线 f(x) = x⁴ 在点 (2, 1) 处的切线斜率。另外，求切线方程。

Solution:

Let us find the derivative of the curve as,

dy/dx = 4x³

At point (2, 1) value of dy/dx or slope m is,

m = 32

Equation of tangent line at point (2, 1) is,

y – 1 = 32(x – 2)