如何找到衍生品？

在日常生活中，人们经常会遇到诸如“患者对所提供的剂量有何反应？”等问题。或“与产品的生产成本相关的利润如何变化？”或“气体压力相对于其体积的变化率是多少？”或“速度相对于时间的变化率是多少？”还有很多。如果仔细观察，就会发现这些问题与一个变化量相对于另一个变化量的变化率有关。这种比率度量称为导数。

导数还可用于计算给定位置处曲线的切线方程和法线方程，以及函数的最大值和最小值及其估计值。例子，

• 生物学家使用导数来计算培养物中细菌的生长速率。
• 电气工程师使用它来描述电路中电流的变化。

点导数

Suppose a function f(x) is defined in a neighborhood of a.

If a and a + h belong to the domain set of a function f and 

$\lim\;_{h \to 0}\;\frac{f(a + h) - f(a)}{h}$

exists, then this limit is called the derivative of f(x) at x = a and is denoted by,

f‘ (a) or [df / dx]x = a

Thus, 

$f'(a) = \lim\;_{h \to a}\; \frac{f(a + h) - f(a)}{h}$

In general, the derivative of a function at any point x is given by,

$f'(x) = \lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h} = \frac{df}{dx}$

This is known as first principle of derivative.

注意：如果 f(x) 的导数存在，则称函数f 是可导的或可微的。微分是寻找导数的过程。

导数的几何解释

考虑下图，

Consider the function f(x) defined on an open interval (a, b). Let P be a point on the curve y = f(x). Let Q[(c – h), f(c – h)] and R be the point on either side of point P. Now,

Slope of chord PQ is, f(c – h) – f(c) / (-h).

The slope of chord PR can be written as, f(c + h) – f(c) / h

Now, we know that tangent to a curve at a point P is the limiting position of secant PQ when Q tends to P.

Similarly, it is also the limiting position of secant PR when R tends to P.

∴ as h⇢ 0 points Q and R both tend to P from left and right hand sides.

∴ The slope of tangent at point P is,

$P = \lim \;_{Q \to P} \;(Slope\ of\ secant\ PQ)$

$\lim \;_{h \to 0} \;(Slope\ of\ secant\ PQ)$

$\lim \;_{Q \to P} \; \frac{f(c + h) - f(c)}{h}$

If these limits exist and are equal, there is a unique tangent at point P.

The slope of tangent is denoted by dy / dx i.e., f'(x)

Thus,

$f'(x) = \lim\;_{h \to 0} \; \frac{f(c + h) - f(c)}{h}$

$= \lim\;_{h \to 0} \; \frac{f(c - h) - f(c)}{-h}$

标准函数的导数

• 常数函数

Let f(x) = k where k is any constant

∴ f(x + h) = k

Consider,

$\lim\;_{h \to 0}\; \frac{f(x + h) - f(x)}{h}$

$=\lim\;_{h \to 0}\; \frac{k - k}{h}$

= 0

∴ d(k) / dx = 0

• 电源函数

Let f(x) = xn

f(x + h) = (x + h)n , n∈ R

Consider,

$\lim\;_{h \to 0}\; \frac{f(x + h) - f(x)}{h}$

$=\lim\;_{h \to 0}\; \frac{(x + h)^n - x^n}{h}$

$= \lim\;_{h \to 0}\; \frac{[x^n \;+ \; nx^{n - 1}h \; + \; \frac{n(n-1)}{2!} x^{x-2}h^2 \; + .... + h^n] - x^n}{h}$

$= \lim\;_{h \to 0}\;\frac{h}{h}[nx^{n-1}\;+\;\frac{(n-1)}{2!}x^{n-2}h\;+......+\;h^{n-1}]$

$= \lim\;_{h \to 0}\; [nx^{n-1}\;+\;\frac{n(n-1)}{2!}x^{n-2}h\;+.....+h^{n-1}]$

=[ nxn-1 + 0 + …….+ 0]

= n xn-1

∴ d(xn) / dx = n xn-1

注意： d(xk) / dx = kxk-1，对于任何实数 k

• 三角函数

Let f(x) = sin x

f(x + h) = sin(x + h)

Here,

$\lim\;_{h \to 0}\; \frac{f(x + h) - f(x)}{h}$

$= \lim\;_{h \to 0}\; \frac {sin(x + h) - sin \;x}{h}$

$=\lim\;_{h \to 0}\; \frac{2cos(\frac{x + h + x}{2})sin\frac{x + h -x}{2}}{h}$

$=\frac{2}{2}\lim\;_{h \to 0} \;cos(\frac{2x + h}{2}) \lim\;_{h \to 0}\; \frac{sin \;h/2}{h/2})$

= cos (x + 0).1

= cos x

∴ d(sin x) / dx = cos x

Similarly,

• d(cos x) / dx = – sin x
• d(tan x) / dx = sec2 x
• d(cot x) / dx = – cosec2 x
• d(sec x) / dx = sec x . tan x
• d(cosec x) / dx = – cosec x . cot x

• 指数和对数函数

1. d(ax) / dx = ax . log a {a > 0 and a ≠ 1}
2. d(ex) / dx = ex
3. d(log x) / dx = 1 / x

区分规则

• 总和的导数（定理 1）

声明：如果 u 和 v 是 x 的可微函数，并且如果 y = u + v 则 dy / dx = du / dx + dv / dx。

证明：

Let δu, δv, and δy be the small increments in u, v, y respectively, corresponding to the increment δx in x

As δx ⇢ 0, δu⇢ 0, δv⇢ 0, δy ⇢ 0,

Since u and v are differentiable functions of x,

$\frac{du}{dx} = \lim\;_{\delta x \to0}\;\frac{\delta u}{\delta x} \; and \; \frac{dv}{dx}\lim\;_{\delta x \to0}\;\frac{\delta v}{\delta x}$

y = u + v ⇢ eq (1)

y + δy = (u + δu) + (v + δv) ⇢ eq (2)

Subtracting (1) from (2) we get,

δy = δu + δv

Dividing both sides by δx, 

δy / δx = δu / δx + δv / δx

Taking the limit as δx ⇢ 0, on both sides, 

$\lim\;_{\delta x \to 0}\;\frac{\delta y}{\delta x} = \;\lim\;_{\delta x \to 0}\;[\frac{\delta u}{\delta x} \;+\;\frac{\delta v}{\delta x} ]$

$\lim\;_{\delta x \to 0}\;\frac{\delta y}{\delta x} = \;\lim\;_{\delta x \to 0}\;\frac{\delta u}{\delta x} \;+\lim\;_{\delta x \to 0}\;\frac{\delta v}{\delta x}$

du / dx + dv / dx    

$\therefore \lim\;_{\delta x\to0}\; \frac{\delta y}{\delta x }$

∴ dy / dx = du / dx + dv / dx

Similarly,

• Derivative of difference (Theorem 2)

Statement: If u and v are differentiable functions of x and if y = u – v then dy / dx = du / dx – dv / dx.

• 乘积的导数（定理 3）

声明：如果 u 和 v 是 x 的可微函数并且如果 y = u v 那么

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

证明：

Let δu, δv, and δy be the small increments in u, v, y respectively, corresponding to the increment δx in x

As δx ⇢ 0, δu⇢ 0, δv⇢ 0, δy ⇢ 0,

Since u and v are differentiable functions of x,

$\frac{du}{dx} = \lim\;_{\delta x \to0}\;\frac{\delta u}{\delta x} \; and \; \frac{dv}{dx}\lim\;_{\delta x \to0}\;\frac{\delta v}{\delta x}$

y = uv ⇢ eq. (1)

y + δy = (u + δu) (v + δv)        

= uv + uδv + vδu + δuδv ⇢ eq(2)

Subtracting (1) from (2), we get

δy = uδv + vδu + δuδv

Dividing both the sides by δx, 

δy / δx = u(δv / δx) + v(δu / δx) + (δu / δx)δv

Taking limit as δx ⇢ 0, we get

$\lim\;_{\delta x \to 0}\; \frac{\delta y}{\delta x} = u\lim\;_{\delta x \to 0}\; \frac{\delta v}{\delta x}\; + \; v\lim\;_{\delta x \to 0}\; \frac{\delta u}{\delta x}\; + \; \lim\;_{\delta x \to 0}\;\frac{\delta u}{\delta x}.\lim\;_{\delta x\to 0}\;\delta v$

= u (dv / dx) + v(du / dx) + (du/dx) × 0

As R.H.S. exists and is equal to (dy / dx),

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

Thus, the derivative of product of two functions = first function × derivative of second function + second function × derivative of first function

Similarly,

• Derivative of quotient (Theorem 4)

Statement:  If u and v are differentiable functions of x and if y = u / v then 

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

示例问题

问题 1：利用导数第一原理求 3x + 4 的导数。

解决方案：

Let f(x) = 3x + 4

f(x + h) = 3(x + h) + 4

consider,

$\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}$

$\lim\;_{h \to 0} \; \frac{3(x + h)\; + \; 4 \; - \;(3x + 4)}{h}$

$\lim\;_{h \to 0}\; \frac{3h}{h}$

= 3

∴ f’ (x) = 3

问题 2：求导数

• 1 / x ²和
• x 因为 x。

使用导数第一原理。

解决方案：

• 1 / x²

Let f(x) = 1 / x²

f(x + h) = 1 / (x + h)²

Consider,

$\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}$

$\lim\;_{h \to 0} \; \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

$\lim\;_{h \to 0}\; \frac{x^2 - (x + h^2)}{h(x + h)^2 . x^2}$

$\lim\;_{h \to 0}\; \frac{x^2 - (x^2 + 2xh + h^2)}{h(x + h)^2 . x^2}$

$\lim\;_{h \to 0}\; \frac{-h(2x + h)}{h(x + h)^2 . x^2}$

$-\lim\;_{h \to 0}\; (2x + h)\;\;\lim\;_{h \to 0}\; \frac{1}{(x + h)^2 . x^2}$

= -2x(1 / x²x²)

= -2 / x³

∴ f'(x) = -2 / x³

• x cos x

Let f(x) = x cos x

f(x + h) = (x + h) cos (x + h)

consider,

$\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}$

$\lim\;_{h \to 0} \; \frac{(x + h) cos (x + h) - x cos x}{h}$

$\lim\;_{h \to 0} \; \frac{x[cos (x + h) - cos x]+ h\;cos(x + h)}{h}$

$\lim\;_{h \to 0} \; \frac{x(-2)sin(x + \frac{h}{2})\;sin\frac{h}{2}}{2x\frac{h}{2}}\; + \; \lim\;_{h \to 0}\; \frac{h\;cos(x + h)}{h}$

$-x\;\lim\;_{h \to 0}\; sin (x +\frac{h}{2}).\; \lim\;_{h \to 0}\; \frac{sin\frac{h}{2}}{\frac{h}{2}}\;+\; cos(x + 0)$

= -x sin x. 1 + cos x

= -x sin x + cos x

∴ f'(x) = -x sin x + cos x

问题3：区分以下内容

• 罪 x / 1 + 罪 x
• e ^x / 1 + 罪 x

解决方案：

• sin x / 1 + sin x

we have

y = sin x / 1 + sin x

$\frac{dy}{dx} = \frac{(1 + sin\;x) \frac{d}{dx}(sin\;x) - sin\;x \frac{d}{dx}(1 + sin\;x)}{(1+ sinx)^2}$

$=\frac{(1 + sin\;x).cos\;x - sin\;x. (0 + cos\;x)}{(1+ sinx)^2}$

$=\frac{cos\;x + sin\;x.cos\;x - sin\;x.cos\;x}{(1 + sin\;x)^2}$

= cos x / (1 + sin x)²

• e^x / 1 + sin x

y = e^x / 1 + sin x

$\frac{dy}{dx}\; = \frac{(1 + sin\;x)\frac{d}{dx}e^x\;-\;e^x\;\frac{d}{dx}(1 + sin\;x)}{(1 + sin\;x)^2}$

$=\frac{(1 + sin\;x).e^x\;-\;e^x.(0 + cos\;x)}{(1 + sin\;x)^2}$

$= \frac{e^x[1 + sin\;x - cos\;x]}{(1 + sin\;x)^2}$

问题 4：y = log 5(log7 x) 求 dy / dx

解决方案：

we have,

y  = log₅ (log₇x)

Using logarithmic property, we can write,

y = log( log₇x ) / log 5

$\frac{dy}{dx} = \frac{1}{log\;5}.\frac{d}{dx}[log(log_7x)]$

$= \frac{1}{log\;5}\;.\;\frac{1}{\frac{log\;x}{log\;7}}\;.\;\frac{d}{dx}(\frac{log\;x}{log\;7})$

$= \frac{1}{log\;5}\;.\;\frac{1}{\frac{log\;x}{log\;7}}\;.\;\frac{1}{log\;7}\frac{d}{dx}(log\;x)$

$= \frac{1}{log\;5\;.\;.log\;x}\;.\;\frac{1}{x}$

= 1 / x log 5. log x

问题 5：如果 y = $log[\frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25}-x}]$ 

解决方案：

y = $log[\frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25}-x}]$

y = $log\;(x + \sqrt{x^2 + 25})\; -\; log\;(\sqrt{x^2 + 25}-x)$

$\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2+25}}\;.\;\frac{d}{dx}(x + \sqrt{x^2+25})\;-\;\frac{1}{\sqrt{x^2+25}-x}\;.\;\frac{d}{dx}(\sqrt{x^2 + 25}-x)$

$= \frac{1}{x+\sqrt{x^2+25}}\;(1 \;+\;\frac{1}{2\sqrt{x^2+25}}\times2x)\;-\;\frac{1}{\sqrt{x^2+25}-x}\;(\frac{1\times2x}{2\sqrt{x^2+25}}\;-\;1)$

$= \frac{1}{x + \sqrt{x^2+25}}\;[\frac{\sqrt{x^2+25}\;+\;x}{\sqrt{x^2+25}}]\;+\;\frac{1}{\sqrt{x^2+25}-x}\;[\frac{\sqrt{x^2+25}-x}{\sqrt{x^2+25}}]$

$= \frac{1}{\sqrt{x^2+25}}\;+\;\frac{1}{\sqrt{x^2+25}}$

$= \frac{2}{\sqrt{x^2+25}}$

问题 6：如果 y = log(3x ² + 2x +1)，求 dy / dx

解决方案：

y = log (3x² + 2x +1)

$= \frac{dy}{dx} = \frac{1}{3x^2 + 2x + 1}\;\;\frac{d}{dx}(3x^2 + 2x + 1)$

$= \frac{1}{3x^2+2x+1}\;\;(6x +2)$

$= \frac{2(3x +1)}{3x^2 + 2x +1}$

问题7：利用导数第一原理求√sin x

解决方案：

Let f(x) = √sin x

f(x + h) = √sin (x + h)

consider,

$\lim\;_{h \to 0} \; \frac{f(x + h) - f(x)}{h}$

$= \lim\;_{h \to 0}\;\frac{\sqrt{sin(x + h)} \;-\;\sqrt{sin\;x} }{h}$

$= \lim\;_{h \to 0}\;\frac{\sqrt{sin(x + h)} \;-\;\sqrt{sin\;x} }{h} \; \times \;= \frac{\sqrt{sin(x + h)} \;+\;\sqrt{sin\;x} }{\sqrt{sin(x + h)} \;+\;\sqrt{sin\;x}}$

$= \;\lim\;_{h \to 0}\; \frac{sin(x + h) - sin\;x}{h[\sqrt{sin(x + h)} \;+\; \sqrt{sin\;x} ]}$

$= \lim\;_{h \to 0}\; \frac{2 cos\;(\frac{x + h+ x}{2})\;.\;sin\;(\frac{x + h - x}{2})}{h[\sqrt{sin(x + h)}\; +\;\sqrt{sin\;x} ]}$

$= \lim\;_{h \to 0}\; \frac{2\;cos\;(x +\frac{h}{2})\;.\;sin(\frac{h}{2})}{h[\sqrt{sin(x + h)\;+\;\sqrt{sin\;x}}]}$

$= \lim\;_{h \to 0}\; \frac{\;cos\;(x +\frac{h}{2})}{\sqrt{sin(x + h)\;+\;\sqrt{sin\;x}}}\;\times\;\frac{sin(\frac{h}{2})}{(h/2)}$

$=\frac{\lim\;_{h \to 0}\;cos\;(x +\frac{h}{2})}{\lim\;_{h\to 0}\;\sqrt{sin(x + h)\;\;+\sqrt{sin\;x}}}\;\times\;\lim\;_{h \to 0}\;\frac{sin(\frac{h}{2})}{(h / 2)}$

$=\frac{cos\;x}{\sqrt{sin\;x}\;+\;\sqrt{sin\;x}}\;\times1$

$.....[h \to 0, \frac{h}{2} \to 0 \;and\;\lim\;_{\theta \to 0}\;\frac{sin\;\theta}{\theta} =1]$

$= \frac{cos\;x}{2\sqrt{sin\;x}}$