例 8.16 求 ^d

求 $\frac{dy}{dx}$ 当 $y$ 是

$$(a) (3x^3 - 2x^2 + 1)^5 \quad (b) \frac{1}{(5x^2 - 2)^7}$$

$$\text{Find } \frac{1}{dx} \text{ when } y \text{ is } (a) (3x^3 - 2x^2 + 1)^5 \quad (b) \frac{1}{(5x^2 - 2)^3} \ (c) (x^2 + 1)^3 \sqrt{x - 1} \quad (d) \frac{\sqrt{2x + 1}}{(x^2 + 1)^3}$$

解 (a) 引入中间变量 z = 3x ³ − 2x ² + 1，我们有

$$y = z^5 \quad \text{and} \quad z = 3x^3 - 2x^2 + 1$$

$$\frac{dy}{dz} = 5z^4 \quad \text{and} \quad \frac{dz}{dx} = 9x^2 - 4x$$

所以，根据链式法则，

$$\frac{\text{dy}}{\text{dx}} = \frac{\text{dy}}{\text{dz}} \frac{\text{dz}}{\text{dx}} = 5(3x^3 - 2x^2 + 1)^4(9x^2 - 4x) \ = 5x(9x - 4)(3x^3 - 2x^2 + 1)^4$$

(b) 引入中间变量 z = 5x² − 2，我们有

$$y = \frac{1}{z^7} = z^{-7} \quad \text{and} \quad z = 5x^2 - 2$$

$$\frac{dy}{dz} = -7z^{-8} = -\frac{7}{z^8} \quad \text{and} \quad \frac{dz}{dx} = 10x$$

所以，根据链式法则，

$$\frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = -\frac{7}{(5x^2 - 2)^8} 10x = -\frac{70x}{(5x^2 - 2)^8}$$

$$\frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = -\frac{7}{(5x^2 - 2)^8} 10x = -\frac{70x}{(5x^2 - 2)^8}$$ (c) 在此情形中我们处理的是乘积 $y = uv$ 其中 $u = (x^2 + 1)^3$ 以及 $v = \sqrt{(x - 1)}$ 。于是根据乘积法则

$$\frac{dy}{dx} = u \frac{dv}{dx} $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$$

To find $\frac{du}{dx}$ we introduce the intermediate variable $z = x^2 + 1$ giving $u = z^3$ and so, by the chain rule,

$$\frac{du}{dx} = \frac{du}{dz} \frac{dz}{dx} = 3(x^2 + 1)^2 2x = 6x(x^2 + 1)^2$$

the chain rule,

$$\frac{du}{dx} = \frac{du}{dz} \frac{dz}{dx} = 3(x^2 + 1)^2 2x = 6x(x^2 + 1)^2$$ Likewise, to find $\frac{dv}{dx}$ we introduce the intermediate variable $w = x - 1$ giving $v = \sqrt{w} = w^{1/2}$ and $w = x - 1$ , so by the chain rule

$$\frac{dv}{dx} = \frac{1}{2\sqrt{(x-1)}}$$$$