商规则公式

微积分是对连续变化的数学研究，类似于几何是对形状的研究，代数是对算术运算概括的研究。微积分和积分是两个主要学科；微积分关注瞬时变化率和曲线斜率，而积分微积分关注曲线下方或曲线之间的数量和面积的累积。微积分的基本定理将这两个学科联系起来，它们都采用了无限序列和无限级数收敛到明确定义的极限的基本原理。

商规则公式

在微积分中，商规则是一种技术，用于确定以商的形式提供的任何函数的导数，商的形式是通过将两个可微函数相除而得出的。商规则说，商的导数等于从分母乘以分子的导数减去分子乘以分母的导数得到的结果与分母的导数的平方之比。

如果我们有一个 u(x)/v(x) 类型的函数，我们可以使用商规则导数来获得该函数的导数。商规则的公式如下：

$\frac{d\left(\frac{u(x)}{v(x)}\right)}{dx}=[v(x) × u'(x) - u(x) × v'(x)]/[v(x)]^2$

where,

u(x) and v(x) are differentiable functions in R.

u'(x) and v'(x) are the derivatives of functions u(x) and v(x) respectively.

推导

Suppose a function f(x) = u(x)/v(x) is differentiable at x. We will prove the product rule formula using the definition of derivative or limits.

$f'(x)=\lim_{\Delta x\to0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

= $\lim_{\Delta x\to0} \frac{\frac{u(x+\Delta x)}{v(x+\Delta x)}-\frac{u(x)}{v(x)}}{\Delta x}$

= $\lim_{\Delta x\to0}\frac{u(x+\Delta x)v(x)-u(x)v(x+\Delta x)}{v(x+\Delta x)v(x)\Delta x}$

= $\frac{1}{[v(x)]^2}\lim_{\Delta x\to0}\frac{u(x+\Delta x)v(x)-u(x)v(x+\Delta x)}{\Delta x}$

= $\frac{1}{[v(x)]^2}\lim_{\Delta x\to0}\frac{u(x+\Delta x)v(x)-u(x)v(x)+u(x)v(x)-u(x)v(x+\Delta x)}{\Delta x}$

= $\frac{1}{[v(x)]^2}\lim_{\Delta x\to0}\frac{v(x)(u(x+\Delta x)-u(x))-u(x)(v(x+\Delta x)-v(x))}{\Delta x}$

= $\frac{1}{[v(x)]^2}\left[v(x)\lim_{\Delta x\to0}\frac{u(x+\Delta x)-u(x)}{\Delta x}-u(x)\lim_{\Delta x\to0}\frac{v(x+\Delta x)-v(x)}{\Delta x}\right]$

Put $\lim_{\Delta x\to0} \frac{u(x+\Delta x)-u(x)}{\Delta x}=u'(x)$ and $\lim_{\Delta x\to0}\frac{v(x+\Delta x)-v(x)}{\Delta x}=v'(x)$

= [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

This derives the formula for quotient rule.

示例问题

问题 1. 使用商规则求函数f(x) = 1/x 的导数。

解决方案：

We have, f(x) = 1/x. Here, u(x) = 1 and v(x) = x.

So, u'(x) = 0 and v'(x) = 1

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x (0) – 1 (1)/x²

= 1/x²

问题 2. 使用商规则求函数f(x) = 1/sin x 的导数。

解决方案：

We have, f(x) = 1/sin x. Here, u(x) = 1 and v(x) = sin x.

So, u'(x) = 0 and v'(x) = cos x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [sin x (0) – 1 (cos x)]/cos² x

= -cos x/ cos² x

= -1/cos x

= -sec x

问题 3. 使用商规则求函数f(x) = x/sin x 的导数。

解决方案：

We have, f(x) = x/sin x. Here, u(x) = x and v(x) = sin x.

So, u'(x) = 1 and v'(x) = cos x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [sin x (1) – x (cos x)]/cos² x

= (sin x – x cos x)/cos² x

问题 4. 使用商规则求函数f(x) = cos x/x 的导数。

解决方案：

We have, f(x) = cos x/x. Here, u(x) = cos x and v(x) = x.

So, u'(x) = -sin x and v'(x) = 1

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x (-sin x) – cos x (1)]/x²

= (-x sin x – cos x)/x²

问题 5. 使用商规则求函数f(x) = log x/x 的导数。

解决方案：

We have, f(x) = log x/x. Here, u(x) = log x and v(x) = x.

So, u'(x) = 1/x and v'(x) = 1

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x (1/x) – log x (1)]/x²

= (1 – log x)/x²

问题 6. 使用商法则求函数f(x) = (2x – 1)/x ²的导数。

解决方案：

We have, f(x) = (2x – 1)/x². Here, u(x) = 2x – 1 and v(x) = x².

So, u'(x) = 2 and v'(x) = 2x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x² (2) – (2x – 1) (2x)]/x⁴

= (2x² – 4x² + 2x)/x⁴

= (-2x² + 2x)/x⁴

= [-2x(x – 1)]/x⁴

= -2(x – 1)/x³

问题 7. 使用商规则求函数f(x) = log x/sin x 的导数。

解决方案：

We have, f(x) = log x/sin x. Here, u(x) = log x and v(x) = sin x.

So, u'(x) = 1/x and v'(x) = cos x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [sin x (1/x) – log x (cos x)]/sin² x

= [sin x/x – log x cos x]/sin² x