差商

差商公式是函数导数定义的一部分。函数的导数是通过将变量 h 变为 0 时的极限应用于函数的差商来获得的。让我们看一下差商公式及其推导。

差商公式

在单变量微积分中，差商是公式中的项，当 h 接近零时，产生函数f 的导数。差商公式用于计算连接两个位置的直线的斜率。它也用于导数定义。

函数y = f(x) 的差商公式由下式给出，

$\frac{f(x+h)-f(x)}{h}$

where,

f (x + h) is evaluated by substituting x as x + h in f(x),

f(x) is the given function.

推导

Consider the function y = f(x) and a secant line that passes through two points on the curve (x, f(x)) and (x + h, f(x + h)). It is depicted as a curve below:

Using the slope formula $m=\frac{y_2-y_1}{x_2-x_1}$, the slope of the secant line is,

$m=\frac{f(x+h)-f(x)}{(x+h)-x}$

$m=\frac{f(x+h)-f(x)}{h}$

This proves the difference quotient formula.

示例问题

问题 1. 求函数f(x) = x – 3 的差商。

解决方案：

Use the difference quotient formula for f(x) = x – 3.

D = [ f(x + h) – f(x) ] / h

= [ (x + h) – 3 – (x – 3) ] / h

= [ x + h – 3 – x + 3 ] / h

= h/ h

= 1

问题 2. 求函数f(x) = 4x – 1 的差商。

解决方案：

Use the difference quotient formula for f(x) = 4x – 1.

D = [ f(x + h) – f(x) ] / h

= [ 4(x + h) – 1 – (4x – 1) ] / h

= [ 4x + 4h – 1 – 4x + 1 ] / h

= 4h/ h

= 4

问题 3. 求函数f(x) = 7x – 2 的差商。

解决方案：

Use the difference quotient formula for f(x) = 7x – 2.

D = [ f(x + h) – f(x) ] / h

= [ 7(x + h) – 2 – (7x – 2) ] / h

= [ 7x + 7h – 2 – 7x + 2 ] / h

= 7h/ h

= 7

问题 4. 求函数f(x) = x ² – 4 的差商。

解决方案：

Use the difference quotient formula for f(x) = x² – 4.

D = [ f(x + h) – f(x) ] / h

= [ (x + h)² – 4 – (x² – 4) ] / h

= [ x² + h² + 2xh – 4 – x² + 4 ] / h

= (h² + 2xh)/ h

= h (h + 2x)/h

= h + 2x

问题 5. 求函数f(x) = 3x ² – 5 的差商。

解决方案：

Use the difference quotient formula for f(x) = 3x² – 5.

D = [ f(x + h) – f(x) ] / h

= [ 3(x + h)² – 5 – (3x² – 5) ] / h

= [ 3(x² + h² + 2xh) – 3x² + 5 ] / h

= (3x² + 3h² + 6xh – 5 – 3x² + 5)/h 

= (3h² + 6xh)/h

= 3h (h + 2x)/h

= 3(h + 2x)

问题 6. 求函数f(x) = x/2 的差商。

解决方案：

Use the difference quotient formula for f(x) = x/2.

D = [ f(x + h) – f(x) ] / h

= [ (x + h)/2 – x/2 ] / h

= [ (x + h – x)/2 ] / h

= (h/2) / h

= 1/2

问题 7. 求函数f(x) = log x 的差商。

解决方案：

Use the difference quotient formula for f(x) = log x.

D = [ f(x + h) – f(x) ] / h

= [ log(x + h) – log x ] / h

Use the quotient property log a – log b = log (a/b).

= log [ (x + h)/h ]/ h