三角函数的导数

函数f(x) 的导数是输入变化时函数值变化的速率。在这种情况下，x 称为自变量，f(x) 称为因变量。衍生物几乎在我们生活的方方面面都有应用。从火箭发射到我们的水箱设计，专业人士使用导数来测量变化率，有时还从数学角度分析物理现象。计算多项式的导数非常简单，三角函数及其导数需要额外注意。让我们看看如何详细计算它们。

衍生品

导数的正式和最基本的定义涉及限制。在几何上，对于函数f(x)，点 x 处的导数表示与该特定点处的曲线相切的直线的斜率。使用这种直觉，开发了使用极限的导数的正式定义。下图显示了导数的几何直觉。

在该图中，请注意连接点 (x, f(x)) 和 (x, f(x + h)) 的线与该图相割。但假设是 h 接近于零。因此，以这种方式，随着点 (x + h) 越来越靠近点 x，正割线慢慢开始变为切线。这就是为什么说导数只不过是曲线上切线的斜率。使用这个逻辑，给出了导数的第一个定义。

For a function f(x), at a point x = a. The derivative is defined as, 

$\lim_{x \to a}\frac{f(x) - f(a) }{x - a}$

It can also be written as, 

$\lim_{h \to 0}\frac{f(x + h) - f(x) }{h}$

这是导数的定义。我们还将导数表示为$\frac{df}{dx}$或 f'(x)。

导数的幂律

这是衍生品中最常用的规则。它说，

$\frac{dx^n}{dx} = nx^{n - 1}$

This can be derived through the limit definition of the derivatives. 

$\frac{d(x^n)}{dx} = \lim_{h \to 0}\frac{(x + h)^n - x^n}{h}$

(x + h)ⁿ can be opened through binomial expansion, 

$(x + h)^n = x^n + \frac{n!}{(n-1)!}x^{n-1}h + \frac{n!}{(n-2)!2!}h^2 + ....$

⇒$\frac{d(x^n)}{dx} = \lim_{h \to 0}\frac{(x^n + \frac{n!}{(n-1)!}x^{n-1}h + \frac{n!}{(n-2)!2!}h^2 + .... - x^n)}{h}$

⇒$\frac{d(x^n)}{dx} = \lim_{h \to 0}\frac{\frac{n!}{(n-1)!}x^{n-1}h + \frac{n!}{(n-2)!2!}h^2 + ....)}{h}$

⇒$\frac{d(x^n)}{dx} =\frac{n!}{(n-1)!}x^{n-1} + \lim_{h \to 0}\frac{( \frac{n!}{(n-2)!2!}h^2 + ....)}{h}$

⇒$\frac{d(x^n)}{dx} =nx^{n-1}$

问题 1：求 f(x) = √x 的导数。

解决方案：

$\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

⇒$\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0}\frac{\sqrt{x + h} - \sqrt{x}}{h}$

This is indeterminate form, using rationalization 

⇒$\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0}\frac{(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})}$

⇒$\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0} \frac{(x + h) -x}{h(\sqrt{x + h} + \sqrt{x})}$

⇒$\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}$

⇒$\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0} \frac{1}{(\sqrt{x + h} + \sqrt{x})}$

⇒$\frac{d(\sqrt{x})}{dx} = \frac{1}{2\sqrt{x}}$

三角函数的导数

首先通过极限定义计算三角函数的导数。在计算这些导数并查看它们的证明之前，有必要重新审视与三角函数相关的一些极限恒等式。

$\lim_{x \to 0}\frac{sin(x)}{x}$

 and 

$\lim_{x \to 0}\frac{cos(x) - 1}{x}$

让我们计算三个最重要的函数的导数。

1. $\frac{d(sin(x))}{dx} = cos(x)$
2. $\frac{d(cos(x))}{dx} = -sin(x)$
3. $\frac{d(tan(x))}{dx} = sec^2(x)$

证明正弦函数的导数

$\frac{d(sin(x))}{dx} = cos(x)$

Using the previously mentioned definition of the derivatives 

$\frac{df(x)}{dx} = \lim_{h \to 0}\frac{f(x + h) - f(x) }{h} \\ = \frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x + h) - sin(x) }{h}$

⇒$\frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x)cos(h) + sin(h)cos(x) - sin(x) }{h}$

⇒$\frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x)(cos(h)-1) + sin(h)cos(x)}{h} \\ = \frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x)(cos(h)-1)}{h}+ \lim_{h \to 0}\frac{sin(h)cos(x)}{h} \\ = \frac{dsin(x)}{dx} = sin(x)\lim_{h \to 0}\frac{(cos(h)-1)}{h}+ cos(x)\lim_{h \to 0}\frac{sin(h)}{h} \\$

Using the limit identities described above, 

⇒$\frac{d(sin(x))}{dx} = cos(x)$

证明余弦函数的导数

$\frac{d(cos(x))}{dx} = -sin(x)$

Using the previously mentioned definition of the derivatives 

$\frac{df(x)}{dx} = \lim_{h \to 0}\frac{f(x + h) - f(x) }{h}$

⇒$\frac{df(x)}{dx} = \lim_{h \to 0}\frac{cos(x + h) - cos(x) }{h}$

⇒$\frac{df(x)}{dx} = \lim_{h \to 0}\frac{cos(x)cos(h) - sin(x)sin(h) - cos(x) }{h}$

⇒$\frac{df(x)}{dx} = \lim_{h \to 0}\frac{cos(x)(cos(h) -1) -sin(x)sin(h)}{h}$

⇒$\frac{df(x)}{dx} = cos(x)\lim_{h \to 0}\frac{(cos(h) -1)}{h} -sin(x)\lim_{h \to 0}\frac{sin(h)}{h}$

Using the limit identities described above, 

⇒$\frac{d(sin(x))}{dx} = -sin(x)$

证明正切函数的导数

$\frac{d(tan(x))}{dx} = sec^2(x)$

Now that we have the derivatives for the sine and cosine functions. Derivatives of other functions can be calculated simply through Quotient and product rules. 

Quotient rule says, for a function f(x) = $\frac{h(x)}{g(x)}$, the derivative of this function is given by, 

$f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(g(x))^2}$

In this case tan(x) = $\frac{sin(x)}{cos(x)}$

Thus, h(x) = sin(x) and g(x) = cos(x). 

$f'(x) = \frac{cos(x)\frac{dsin(x)}{dx} - sin(x)\frac{dcos(x)}{dx} }{(cos(x))^2}$

⇒$f'(x) = \frac{cos(x)cos(x) + sin(x)sin(x) }{(cos(x))^2}$

⇒$f'(x) = \frac{1 }{(cos(x))^2}$

⇒$\frac{d(tan(x))}{dx} = sec^2(x)$

证明正割函数的导数

$\frac{d(sec(x))}{dx} = sec(x)tan(x)$

This can be proved easily through chain rule, 

$\frac{d(sec(x))}{dx} = \frac{d(\frac{1}{cos(x)})}{dx}$

⇒$\frac{d(sec(x))}{dx} = \frac{-1}{cos^2(x)}\frac{d(cos(x))}{dx}$

⇒$\frac{d(sec(x))}{dx} = \frac{1}{cos^2(x)}sin(x)$

⇒$\frac{d(sec(x))}{dx} = sec(x)tan(x)$

证明余割函数的导数

$\frac{d(cosec(x))}{dx} = -cosec(x)cot(x)$

This can be proved easily through chain rule, 

$\frac{d(cosec(x))}{dx} = \frac{d(\frac{1}{sin(x)})}{dx}$

⇒$\frac{d(cosec(x))}{dx} = \frac{-1}{sin^2(x)}\frac{d(sin(x))}{dx}$

⇒$\frac{d(cosec(x))}{dx} = \frac{-1}{sin^2(x)}cos(x)$

⇒$\frac{d(cosec(x))}{dx} = -cosec(x)cot(x)$

产品规则证明

有不同的方法来证明乘积规则，包括导数和对数微分的定义，让我们看看后一种方法，

y= (f(x)g(x))' = f'(x)g(x)+ f(x)g'(x)

This can easily be proved by taking the natural log,

ln y’= ln{(f(x)g(x))’} = lnf'(x) + lng'(x)

y’/y= f'(x)/f(x)+ g'(x)/g(x)

$y'=y[\frac{f'(x)g(x)+ g'(x)f(x)}{f(x)g(x)}]\\ y'=f(x)g(x)[\frac{f'(x)g(x)+ g'(x)f(x)}{f(x)g(x)}]\\ y'= f'(x)g(x)+ g'(x)f(x)$

Hence, proved.

让我们看一些示例问题

示例问题

问题 1：求函数f(x) 在 x = 0 处的导数。

f(x) = 罪² (x)

解决方案：

This derivative is the combination of power rule and chain rule. 

f'(x) = $\frac{dsin^2x}{dx}$

⇒f'(x) = $2sin(x)\frac{dsinx}{dx}$

Using the previous results for the derivative of sin(x). 

⇒f'(x) = 2sin(x)cos(x)

At x = 0 

f'(x) = 0

问题 2：求函数f(x) 在 x = 0 处的导数。

f(x) = 5sec(x) + 2cos(x)

解决方案：

This derivative is simple

f'(x) = $\frac{d(5sec(x) + 2cos(x))}{dx}$

⇒f'(x) = $\frac{d(5sec(x))}{dx} + \frac{d(2cos(x))}{dx}$

Using the previous results for the derivative of sin(x). 

⇒f'(x) = 5sec(x)tan(x) – 2sin(x). 

at x = 0 

f'(x) = 0

问题 3：求函数f(x) 在 x = 1 处的导数。

f(x) = x ² + x ⁴

解决方案：

This derivative is simple application of power rule 

f'(x) = $\frac{d(x^2 + x^4)}{dx}$

⇒f'(x) = $\frac{d(x^2)}{dx} + \frac{d(x^4)}{dx}$

Using the previous results for the derivative of sin(x). 

⇒f'(x) = 2x + 4x³

At x = 1 

f'(x) = 6

问题 4：求函数f(x) 的导数。

f(x) = $\frac{e^x + 1}{x}$

解决方案：

This derivative is simple application of quotient rule.

Quotient rule says, for a function f(x) = $\frac{h(x)}{g(x)}$, the derivative of this function is given by, 

$f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(g(x))^2}$

Here h(x) = e^x + 1 and g(x) = x

$f'(x) = \frac{x\frac{d(e^x + 1)}{dx} - (e^x + 1)\frac{dx}{dx}}{x^2}$

⇒$f'(x) = \frac{xe^x - (e^x + 1)}{x^2}$

问题 5：使用 Product 规则来区分给定的函数，

y = 5xsinx + 4x ² cosx

解决方案：

Differentiating using Product Rule,

y’= 5sinx + 5xcosx + 8xcosx – 4x²sinx

y’= 5sinx + 13xcosx – 4x²sinx