链式法则如何运作？

在微分数学中，链式法则是用于管理复合函数的重要导数法则。使用链式法则只能找到复合函数的导数。

复合函数是写在另一个函数内部的函数。例如，函数 f(x) 和 g(x) 的复合函数是 f [g(x)] 其中 f(x) 是外部函数，g(x) 是内部函数，如 Cos(4x) 是一个复合函数，因为它结合了两个函数 Cos(x) 和 4x。外在规则、复合函数规则或函数规则的函数是链式规则的所有其他名称。

链式法则

链式法则表明复合函数f(g(x))的导数是f'(g(x))⋅ g'(x)。换句话说，正如我们之前讨论的，Cos(4x) 是一个复合函数，它可以写成 f(g(x)) 其中 f(x) = Cos(x) 和 g(x) = 4x。然后我们可以使用链式法则以及 Cos(x) 和 4x 的导数来计算 Cos(4x) 的导数。

链式法则的证明

根据莱布尼茨的微分符号，我们可以将导数视为分数，即对于 y = f(x)，f'(x) 可以视为 dy/dx。所以，

For a composite function, dy/dx = (dy/du) × (du/dx)

为了用更好的符号来证明这一点，让我们从导数的定义开始得出结果。

Given: y = f(u(x)).

From earlier, we know that, dy/dx = (dy/du) × (du/dx).

Then,

$\frac{dy}{dx}=\lim_{\delta{x} \to \infty}\frac{\delta{y}}{\delta{x}}$

$=\lim_{\delta{x} \to \infty}\frac{\delta{y}}{\delta{u}}\times\frac{\delta{u}}{\delta{x}}$

$=\lim_{\delta{x} \to \infty}\frac{\delta{y}}{\delta{u}}\times\lim_{\delta{x} \to \infty}\frac{\delta{u}}{\delta{x}}$

Now, we know that a function which is differentiable at a point c is also continuous at the point c i.e.

∆ u ⇢ 0 as  ∆ x ⇢ 0 

Then, 

$\frac{dy}{dx}=\lim_{\delta{x} \to \infty}\frac{\delta{y}}{\delta{u}}\times\lim_{\delta{x} \to \infty}\frac{\delta{u}}{\delta{x}}$

$\frac{dy}{du}\times\frac{du}{dx}$

= f'(g(x)). g'(x)

This is the Chain Rule. Hence, the Chain Rule has been proved.

使用链式法则区分函数的步骤和示例

让我们使用链式法则来获得函数Sin(x²) 的导数。

1. 检查函数是否为复合函数，即它包含函数中的函数。函数Sin(x2) 是复合函数。
2. 确定外部 f(x) 和内部函数 g(x)。在这种情况下，f(x) = Sin(x) 和 g(x) = x²。
3. 现在只求外函数的微分。在这种情况下，f'(x) = Cos (x)。
4. 现在只求内函数的微分。在这种情况下，g'(x) = 2x。
5. 在这里找到 f'(x) 和 g'(x) 的乘积，即 (2x)Cos(x)。

因此，我们使用链式法则 (x) 找到了 Sin(x ² ) 的导数，即 (2x)Cos。

示例问题

问题 1：求解，y(x) = (2x ² + 8) ²

解决方案：

Here, as you can see that y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is g(x)² and the inner function g(x) is 2x² + 8.

So, f'(g(x)) = 2g(x), here g(x) = 2x²+ 8. 

Therefore, f'(g(x)) = 2(2x² + 8) and g'(x) = 4x.

Now y'(x) = f'(g(x)).g'(x) 

= 2(2x² + 8)(4x) 

= 16x(x² + 4).

问题 2：求解，y(x) = Cos(4x)。

解决方案：

Here, as we know from earlier, y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is Cos(g(x)) and the inner function g(x) is 4x.

So, f'(g(x)) = -Sin(g(x)), here g(x) = 4x. 

Therefore, f'(g(x)) = -Sin(4x) and g'(x) = 4.

Now y'(x) = f'(g(x)).g'(x) 

= -(Sin(4x))(4) 

= -4Sin(4x).

问题 3：求解，y(x) = ln(x ² – 1)。

解决方案：

Here, as we know from earlier, y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is ln(g(x)) and the inner function g(x) is x² – 1.

So, f'(g(x)) = 1/(g(x)), here g(x) = x² – 1. Therefore, f'(g(x)) = 1/(x² – 1) and g'(x) = 2x.

Now y'(x) = f'(g(x)).g'(x) 

= (1/(x² – 1))(2x) 

= 2x/(x² – 1).

问题 4：求解 y(x) = (ln x) ² 。

解决方案：

Here, as we know from earlier, y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is (g(x))² and the inner function g(x) is ln x.

So, f'(g(x)) = 2g(x) , here g(x) = ln x. 

Therefore, f'(g(x)) = 2(ln x) and g'(x) = 1/x.

Now y'(x) = f'(g(x)).g'(x) 

= (2(ln x))(1/x) 

= 2(ln x)/(x).

问题 5：求解，y(x) = √(x ³ + 56)。

解决方案：

Here, as we know from earlier, y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is √(g(x)) and the inner function g(x) is x³ + 56.

So, f'(g(x)) = (1/2)(x³ + 56)^-1/2, here g(x) = x³ + 56. 

Therefore, f'(g(x)) = (1/2)(x³ + 56)^-1/2 and g'(x) = 3x²

Now y'(x) = f'(g(x)).g'(x) 

= ( (1/2)(x³ + 56)^-1/2) × ( 3x²) 

= [(3/2) x²] / (x³ + 56)1/2.

问题 6：求解，y(x) = ln √x。

解决方案：

Here, as we know from earlier, y(x) is a composite function. So it can be written as f(g(x)). The outer function f(g(x)) is ln(g(x)) and the inner function g(x) is √x.

So, f'(g(x)) = 1/g(x), here g(x) = √x. 

Therefore, f'(g(x)) = 1/(√x) and g'(x) = (1/(2√x)).

Now y'(x) = f'(g(x)).g'(x) 

= (1/(√x))(1/(2√x)) 

= 1/2x.