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微积分(1):预备知识->导数的应用

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预备知识

三角恒等式

\[sin^2 + cos^2 = 1\]

证明:

​ 在三角形中斜边为1,直边分别为sin(θ) 和 cos(θ)

​ 由勾股定理得此公式

\[1 + tan^2 = sec^2\]

证明: 公式(1)等式两边同时除以cos

\[1 + cot^2 = csc^2\]

证明:

​ 公式(1)等式两边同时除以sin

和角公式

\[cos(A+B) = cosAcosB-sinAsinB\\ sin(A+B)=sinAcosB+cosAsinB\]

Proof:

preview

​ The figure shows

​ \(b=\frac{sin(β)}{cos(\alpha)}\)

​ \(a=sin(\alpha)\times(cos(\beta)-b\times sin(\alpha))\)

​ \(sin(\alpha+\beta)=a+b\)

​ Integrate the reletionship and get formula(4)

Proof:

​ The figure shows

​ \(b=\frac{sin(β)}{cos(\alpha)}\)

​ \(a=sin(\alpha)\times(cos(\beta)-b\times sin(\alpha))\)

​ \(cos(\alpha)=\frac{a}{tan(\alpha)}\)

​ Integate the relationship and get formula(5)

余弦定理

\[cos(A) = \frac{b^2+c^2-a^2}{2bc}\]

Proof:

​ \(\vec{a}^2=(\vec{b}-\vec{c})^2\\\quad=b^2+c^2-2bccos(A)\)

​ Integrate and get formula(6)

极限与连续

极限的定义

定义 极限的正式定义

f(x)定义在可能不包含\(x_0\)的开区间上, 当x趋于\(x_0\)时f(x)趋于极限L,记为
\(\lim_{x \to x_0}{f(x)}=L\)
如果,对任何数\(\epsilon>0\),存在相应的数\(\delta>0\)使得对所有满足\(0<|x-x_0|<\delta\)的x,有 \(|f(x)-L|<\epsilon\)

定义 右侧极限和左侧极限

设f(x)定义在(a,b)上, a<b,如果在区间(a,b)内趋于a时f(x)任意接近地趋近于L,f在a有右侧极限,记作
\(\lim_{x \to a^+}f(x)=L\)
设f(x)定义在(c,a)上, c<a,如果在区间(c,a)内趋于a时f(x)任意接近地趋近于L,f在a有左侧极限,记作
\(\lim_{x \to a^-}f(x)=L\)

定理 单侧极限和双侧极限的关系

当\(x \to c\)时函数f(x)有极限当且仅当f的左侧极限和右侧极限存在且相等:
\(\lim_{x \to c}=L \Leftrightarrow \lim_{x \to c-}=L 且\lim_{x \to c+}=L\)

定义 无穷极限

x趋于\(x_0\)时f(x)趋于无穷,记作
\(\lim_{x \to c}=\infty\) 如果对于任何正实数B存在相应的\(\delta>0\),使得对一起满足\(0<|x-x_0|<\delta\)的x,有f(x)>B

x趋于\(x_0\)时f(x)趋于负无穷,记作
\(\lim_{x \to c}=-\infty\) 如果对于任何正实数B存在相应的\(\delta>0\),使得对一起满足\(0<|x-x_0|<\delta\)的x,有f(x)>-B

定义 水平渐近线和垂直渐近线

直线\(y=b\)是函数\(y=f(x)\)图形的水平渐近线,如果有
\(\lim_{x \to \infty}f(x)=b或\lim_{x \to -\infty}f(x)=b\)
直线\(x=a\)是函数\(y=f(x)\)图形的水平渐近线,如果有
\(\lim_{x \to a^+}f(x)=\pm\infty或\lim_{x \to a^-}f(x)=\pm\infty\)

极限的性质

加减乘除

定理 极限法则

如果L,M,c,k为实数,且 \(\lim_{x \to x_0}f(x)=L\)和\(\lim_{x \to x_0}g(x)=M\)
\(\begin{align} &\lim_{x \to c}(f(x)+g(x))=L+M\\ &\lim_{x \to c}(f(x)-g(x))=L-M\\ &\lim_{x \to c}(f(x) \times g(x))=L \times M\\ &\lim_{x \to c}(k \cdot f(x))=k \cdot f(x)\\ &\lim_{x \to c}(\frac{f(x)}{g(x)})=\frac{L}{M},M \neq 0\\ &\lim_{x \to c}(f(x))^\frac{r}{s}=L^\frac{r}{s} \end{align}\)

求极限

多项式

定理 使用代入法求多项式极限

如果\(P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0\),那么 \(\lim_{x \to c}P(x)=P(c)\)

有理函数

定理 代入法求有理函数的极限

如果P(x)和Q(x)都是多项式且Q(c)\(\neq\)0,那么 \(\lim_{x \to c}\frac{P(x)}{Q(x)}=\frac{P(c)}{Q(c)}\)

夹逼定理

定理 夹逼定理

如果在包含c的某个开区间中\(x=c\)处除外的所有x,有\(g(x) \leq f(x) \leq h(x)\). 又假设 \(\lim_{x \to c}g(x)=\lim_{x \to c}h(x)=L\) 那么 \(\lim_{x \to c}f(x)=L\)

三角函数

定理 sin的极限 \(\lim_{\theta \to 0}\frac{sin \theta}{\theta}=1\)

Proof:

img

​ The figure shows \(\frac{1}{2}sin\theta < \frac{1}{2}\theta < \frac{1}{2}tan\theta\) ​ Sort the expression and get \(1>\frac{sin\theta}{\theta}>cos\theta\) ​ Also because \(\lim_{x \to 0}cos\theta=1\) ​ Thus \(\lim_{x \to 0}{\frac{sin\theta}{\theta}}=1\)

定理 cos的极限 \(\lim_{x \to 0}\frac{1-cos (x)}{x}=0\)

Proof:
\(\begin{align} \lim_{x \to 0}\frac{cos (x)-1}{x}&=\lim_{x \to 0}\frac{1-cos^2(x)}{x}\times\frac{1}{1+cos(x)}\\ &=\lim_{x \to 0}sin(x)\times\frac{sin(x)}{x}\times \frac{1}{1+cos(x)}\\ &=0 \times1\times\frac{1}{1+1}\\ &=0 \end{align}\)

不定式

定理 L’Hopital法则

假定\(f(x_0)=g(x_0)=0或\pm\infty\) \(\lim_{x \to x_0}\frac{f(x)}{g(x)}=\lim_{x \to x_0}\frac{f'(x)}{g'(x)}\)

Proof:

Let \(F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]\) That \(F(a)=F(b)=0\)
According to Median Value Theorem,there is figure ‘c’ let \(F'(c)=0\)
Which is \(\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\)
Then compelet the proof of Cauthy Median Value Theorem

Let x right of \(x_0\)

Accourding to CMVT, there is c between x and \(x_o\) let \(\frac{f'(c)}{g'(c)}=\frac{f(x)-f(x_0)}{g(x)-g(x_0)}\)
because \(f(x_0)=g(x_0)=0\) \(\frac{f'(c)}{g'(c)}=\frac{f(x)}{g(x)}\)
because \(x_0<c<x\) \(x_0 \to x \Rightarrow c \to x\)
Thus \(\lim_{x \to x_0+}\frac{f(x)}{g(x)}=\lim_{x \to x_0+}\frac{f'(x)}{g'(x)}\)
The same \(\lim_{x \to x_0-}\frac{f(x)}{g(x)}=\lim_{x \to x_0-}\frac{f'(x)}{g'(x)}\)
Thus \(\lim_{x \to x_0}\frac{f(x)}{g(x)}=\lim_{x \to x_0}\frac{f'(x)}{g'(x)}\)
then complete proof of the \(\frac{0}{0}\)form of L’Hopital Theorem

Let \(F(x)=\frac{1}{f(x)},G(x)=\frac{1}{g(x)}\)

That \(\lim_{x \to x_0}F(x)=\infty,\lim_{x \to x_0}G(x)=\infty\)

Suppose \(\lim_{x \to x_0}\frac{F(x)}{G(x)}=\lim_{x \to x_0}\frac{F'(x)}{G'(x)}\)

\[\lim_{x \to x_0}{\frac{g(x)}{f(x)}}=\lim_{x \to x_0}{\frac{-\frac{1}{f^2(x)}f'(x)}{-\frac{1}{g^2(x)}g'(x)}}=\lim_{x \to x_0}\frac{g^2(x)f'(x)}{f^2(x)g'(x)}\] \[\lim_{x \to x_0}{\frac{g(x)}{f(x)}}=\lim_{x \to x_0}{\frac{g'(x)}{f'(x)}}\]

Which is the \(\frac{0}{0}\)form of L’hopital Theorem

Then we have proof the suppose, which is \(\lim_{x \to x_0}\frac{F(x)}{G(x)}=\lim_{x \to x_0}\frac{F'(x)}{G'(x)}\) The \(\frac{\infty}{\infty}\)form of L’Hopital Theorem

连续性的定义

定义 在一点的连续性

内点: 函数f(x)在定义域的内点c处是连续的,如果 \(\lim_{x \to c}f(x)=f(x)\) 端点: 函数载器定义域的左端点a或右端点b是连续的,如果 \(\lim_{x \to a^+}f(x)=f(a)\quad\lim_{x \to b^-}f(x)=f(b)\)

连续性的性质

加减乘除

定理 连续函数的性质

如果函数f和g在x=c连续,下列函数在x=c连续 \(\begin{align} &f + g \\ &f - g \\ &f \cdot g \\ &\frac{f}{g},倘若g(c)\neq0\\ &f \circ g,倘若f在g(c)连续 \end{align}\)

连续函数的中值定理

定理 连续函数的中值定理

在闭区间[a,b]上连续的函数一定取到f(a)和f(b)之间的每一个值

连续性和可导性

定理 可导性蕴含着连续性

如果f在x=c有导数,那么f在x=c连续

导数

定义

定义 导函数

函数f(x)关于变量x的导数是函数f’,它在x处的值为 \(f'(x)=\lim_{h\to0}{\frac{f(x+h)-f(x)}{h}}\) 如果该极限存在

导数的性质

常数倍

\[\frac{\mathrm{d}}{\mathrm{d}x}(cu)=c\frac{\mathrm{d}}{\mathrm{d}x}\]

\[\frac{\mathrm{d}}{\mathrm{d}x}(u+v)=\frac{\mathrm{d}u}{\mathrm{d}x}+\frac{\mathrm{d}v}{\mathrm{d}x}\]

\[\frac{\mathrm{d}}{\mathrm{d}x}(uv)=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}\]

Proof:
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(uv)&=\lim_{\Delta x\to 0}\frac{\Delta (uv)}{\Delta x}\\ &=\lim_{\Delta x\to 0}(u\frac{\Delta v}{\Delta x}+v\frac{\Delta u}{\Delta x}+\frac{\Delta u \Delta v}{\Delta x})\\ &=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x} \end{align}\)

\[\frac{\mathrm{d}}{\mathrm{d}x}(\frac{u}{v})=\frac{v\frac{\mathrm{d}u}{\mathrm{d}x}-u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\]

Proof:

Regard \(\frac{u}{v}\) as \(u \times \frac{1}{v}\)

According to the Product Rule of the Derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(u \times \frac{1}{v})&=u(-\frac{1}{v^2})(\frac{\mathrm{d}v}{\mathrm{d}x})+\frac{1}{v}\frac{\mathrm{d}u}{\mathrm{d}x}\\ &=\frac{v\frac{\mathrm{d}u}{\mathrm{d}x}-u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \end{align}\)

复合

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}\]

Proof:

\(\frac{\mathrm{d}}{\mathrm{d}x}f(g(x))=f'(g(x))g'(x)\) is another form of \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}\)

Let \(h(x)=f(g(x))\)
Let \(\epsilon=g(x+\Delta x)-g(x)\)
That \(x \to 0 \Rightarrow \epsilon \to 0\) Then
\(\begin{align} h'(x)&=\lim_{\Delta x \to 0}\frac{f(g(x + \Delta x))-f(g(x))}{\Delta x}\\ &=\lim_{\Delta x \to 0}\frac{f(g(x + \Delta x))-f(g(x))}{g(x + \Delta x)-g(x)}\times\frac{g(x + \Delta x)-g(x)}{\Delta x}\\ &=\lim_{\Delta x \to 0}\frac{f(g(x) + \epsilon)-f(g(x))}{\epsilon}\times\frac{g(x + \Delta x)-g(x)}{\Delta x}\\ &=f'(g(x))g'(x) \end{align}\)

求导数

常函数

\[\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(c)=0\]

幂函数

\[\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=nx^{n-1}\]

Proof: \(\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{\ln{x}\,n}=nx^{n-1}\)

自然对数

\[\mathrm{e}=\lim_{h \to 0^+}(1+h)^{1/h}\]

Proof:

let \(F(x)=\int_{1}^{x}\frac{1}{t}\mathrm(d)t\)

that
\(F(1)=0\\ \lim_{x \to \infty}F(x)=\infty\\ \frac{\mathrm{d}}{\mathrm{d}x}F(x)=\frac{1}{x}\)
According to Intermediate Value Theorem

There is \(\mathrm{}e\) let \(F(e)=1\)

let \(G(x)=F(x^a)\)

that
\(G'(x)=a\frac{1}{x}=aF'(x)\\ G(x)=aF(x)+C\)
let \(x=1 \Rightarrow C=0\) \(aF(x)=F(x^a)\)
let \(x=\mathrm{e},a=x\)

Thus \(F(\mathrm{e})=1\)
\(F(\mathrm{e}^x)=x\\ F(x)=\log_{\mathrm{e}}x=\ln(x)\\ \frac{\mathrm{d}}{\mathrm{d}x}ln(x)=\frac{1}{x}\\ \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)=\mathrm{e}^x\) let \(y=(1+h)^{1/h}\)

According to the L’Hopital Theorem
\(\lim_{h \to 0^+}\ln(y)=\lim_{h \to 0^+}\frac{\ln(1+h)}{h}=1\)
Thus \(\ln(\mathrm{e})=1\)
\(\mathrm{e}=\lim_{h \to 0^+}(1+h)^{1/h}\)

对数

\[\frac{\mathrm{d}}{\mathrm{d}x}\log_{b}(x)=\frac{1}{x\ln(b)}\\ \frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{x}\]

Proof:

According to the definition of derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\log_b(x)=&\lim_{h \to 0}\frac{\log_b(x+h)-\log_b(x)}{h}\\ =&\lim_{h \to 0}\log_b(\frac{x+h}{x})^{1/h}\\ =&\frac{1}{x}\log_b\mathrm{e}\\ =&\frac{1}{x\ln(b)} \end{align}\)
Then
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\ln(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\log_{\mathrm{e}}(x)\\ &=\frac{1}{ln(\mathrm{e})x}\\ &=\frac{1}{x} \end{align}\)

自然指数

\[\frac{\mathrm{d}}{\mathrm{d}x}(b^x)=b^x\ln(b)\\ \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)=\mathrm{e}^x\]

Proof:
\(\begin{align} y&=b^x\\ \log_by&=x\\ \frac{\mathrm{d}}{\mathrm{d}x}log_by&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}\frac{1}{\ln(b)y}&=1\\ \frac{\mathrm{d}}{\mathrm{d}x}(b^x)&=b^x\ln(b) \end{align}\)

Then

\[\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)&=e^x\ln(\mathrm{e})\\ &=\mathrm{e}^x \end{align}\]

三角函数

\[\frac{\mathrm{d}}{\mathrm{d}x}sin(x)=cos(x)\]

Proof:

According to the definition of the derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}sin(x)&=\lim_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\\ &=\lim_{h \to 0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\\ &=\lim_{h \to 0}sin(x)\frac{cos(h)-1}{h}+cos(x)\frac{sin(h)}{h}\\ &=sin(x) \times 0+cos(x) \times 1\\ &=cos(x) \end{align}\)

\[\frac{\mathrm{d}}{\mathrm{d}x}cos(x)=-sin(x)\]

Proof:

According to the definition of the derivative
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}cos(x)&=\lim_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\\ &=\lim_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\\ &=\lim_{h \to 0}cos(x)\frac{cos(h)-1}{h}-sin(x)\frac{sin(h)}{h}\\ &=cos(x) \times 0-sin(x) \times 1\\ &=-sin(x) \end{align}\)

\[\begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}tan(x)=sec^2(x)\\ &\frac{\mathrm{d}}{\mathrm{d}x}sec(x)=sec(x)tan(x)\\ &\frac{\mathrm{d}}{\mathrm{d}x}csc(x)=-csc(x)cot(x)\\ &\frac{\mathrm{d}}{\mathrm{d}x}cot(x)=-csc^2(x) \end{align}\]

反三角函数

\[\begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}tan^{-1}(x)=\frac{1}{1+x^2}\\ &\frac{\mathrm{d}}{\mathrm{d}x}csc^{-1}(x)=-\frac{1}{|x|\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}sec^{-1}(x)=\frac{1}{|x|\sqrt{1-x^2}}\\ &\frac{\mathrm{d}}{\mathrm{d}x}cot^{-1}(x)=-\frac{1}{1+x^2} \end{align}\]

Proof:
\(\begin{align} y&=sin^{-1}(x)\\ sin(y)&=x\\ \frac{\mathrm{d}y}{\mathrm{d}x}cos(y)&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{\sqrt{1-sin(y)^2}}\\ \frac{\mathrm{d}}{\mathrm{d}x}sin^{-1}(x)&=\frac{1}{\sqrt{1-x^2}} \end{align}\)
As the same
\(\begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}cos(x)=-sin(x) &\qquad &-\frac{1}{sin(y)}=-\frac{1}{\sqrt{1-cos(y)^2}} \\ &\frac{\mathrm{d}}{\mathrm{d}x}tan(x)=sec^2(x) &\qquad &-\frac{1}{sec^2(y)}=\frac{1}{1+tan(y)^2} \\ &\frac{\mathrm{d}}{\mathrm{d}x}sec(x)=sec(x)tan(x) &\qquad &\frac{1}{sec(y)tan(y)}=\frac{1}{|sec(y)|\sqrt{1-sec(y)^2}} \\ &\frac{\mathrm{d}}{\mathrm{d}x}csc(x)=-csc(x)cot(x) &\qquad &-\frac{1}{csc(y)cot(y)}=-\frac{1}{|csc(y)|\sqrt{1-csc(y)^2}} \\ &\frac{\mathrm{d}}{\mathrm{d}x}cot(x)=-csc^2(x) &\qquad &-\frac{1}{csc^2(y)}=-\frac{1}{1+cot(y)^2} \\ \end{align}\)

双曲函数

定义 双曲函数
\(\begin{aligned} &\cosh x=\frac{e^{x}+e^{-x}}{2}\\ &\sinh x=\frac{e^{x}-e^{-x}}{2}\\ &\tanh x=\frac{\sinh x}{\cosh x}\\ &\operatorname{coth} x=\frac{\cosh x}{\sinh x}\\ &\operatorname{sech} x=\frac{1}{\cosh x}\\ &\operatorname{csch} x=\frac{1}{\sin x}\\ \end{aligned}\)

性质
\(\begin{align} &\sinh 2x =\sinh x \cosh x\\ &\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\\ &\cosh ^{2} x=\frac{\cosh 2 x+1}{2}\\ &\sinh ^{2} x=\frac{\cos h 2 x-1}{2}\\ &\cosh ^{2} x-\sinh ^{2} x=1\\ &\tanh ^{2} \alpha=1-\operatorname{sech}^{2} x\\ &\operatorname{coth}^{2} x=1+\csc h^{2} x\\ &\operatorname{sech} ^{-1} x=\cosh ^{-1} \frac{1}{x}\\ &\operatorname{csch} ^{-1} x=\sinh ^{-1} \frac{1}{x}\\ &\coth ^{-1} x=\tanh ^{-1} \frac{1}{x} \end{align}\)

导数
\(\frac{d}{d x} \sinh x=\operatorname{cosh} x \\ \frac{d}{d x} \cosh x=\sinh x \\ \frac{d}{d x} \operatorname{tanh} x=\operatorname{sech}^{2} x \\ \frac{d}{d x} \operatorname{coth} x=-\operatorname{csch}^{2} x \\ \frac{d}{d x} \operatorname{sech} x=-\operatorname{sech} x \tan x \\ \frac{d}{d x} \operatorname{csch} x=-\operatorname{csch} x \operatorname{coth} x\\ \frac{d}{d x} \sinh ^{-1} x =\frac{1}{\sqrt{1+x^{2}}} \\ \frac{d}{d x} \cosh ^{-1} x =\frac{1}{\sqrt{x^{2}-1}},x>1 \\ \frac{d}{a x} \tanh ^{-1} x =\frac{1}{1-x^{2}},|x|<1 \\ \frac{d}{d x} \coth^{-1} x =\frac{1}{1-x^{2}},|x|>1 \\ \frac{d}{d x} \operatorname{sech} ^{-1} x =-\frac{1}{x \sqrt{1-x^{2}}},0<x<1 \\ \frac{d}{d x} \operatorname{csch}^{-1} x =-\frac{-1}{|x|\sqrt{1+x^{2}}},x\ne0\)

隐函数

解题 隐函数求导

  1. 处理方程 y=f(x) 方程两边对x求导数
  2. 并项\(\frac{\mathrm{d}y}{\mathrm{d}x}\)到一边
  3. 提出因子\(\frac{\mathrm{d}y}{\mathrm{d}x}\)
  4. 解\(\frac{\mathrm{d}y}{\mathrm{d}x}\)

相关变化率

定义 相关变化率

如果Q为某个量,那么Q的变化率为\(\frac{\mathrm{d}Q}{\mathrm{d}t}\)

解题 隐函数求导

  1. 对涉及变化率的问题进行建模
  2. 变量对\(t\)进行求导,\(\frac{\mathrm{d}Q}{\mathrm{d}t}\)使用已知条件v(变化率)进行替换
  3. 对方程进行求解变量之间的相关变化率

导数的应用

物理中的应用

速度 \(v=\frac{\mathrm{d}s}{\mathrm{d}t}\)

加速度 \(a=\frac{\mathrm{d}v}{\mathrm{d}t}\)

急推 \(z=\frac{\mathrm{d}a}{\mathrm{d}t}\)

函数的极值

函数的极值

如果函数\(f\)在定义域\(c\)点取到局部最小值或局部最大值,那么 \(f'(c)=0 或 f'(c)不存在\)

中值定理

罗尔定理

假设函数\(f\)在闭区间\([a,b]\)内连续,在开区间\((a,b)\)可导,如果\(f(a)=f(b)\),那么在开区间\((a,b)\)内至少存在一点\(c\),使得\(f'(c)=0\).

中值定理

假设函数\(f\)在闭区间\([a,b]\)内连续,在开区间\((a,b)\)可导,如果\(f(a)=f(b)\),那么在开区间\((a,b)\)内至少存在一点\(c\),使得 \(f'(c)=\frac{f(b)-f(a)}{b-a}\)

Proof:

let \(g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)\)

that \(g(a)=g(b)=f(a)\) According to Rolle Theorem

There is c let \(g'(x)=0\) which is \(g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}\\ f'(c)=\frac{f(b)-f(a)}{b-a}\)

最优化

解决最优化问题

  1. 分析问题,数学建模
  2. 求导使用函数极值解决最优化问题

线性化

使用线性化进行估值 \(f(a+\Delta x) \approx f(a)+f'(a)\Delta x\)

Newton法

Newton法

\(a\)为\(f(x)=0\)的近似解 \(b=a-\frac{f(a)}{f'(a)}\) \(b\)为更优解

积分

定义

定义 定积分作为黎曼和的极限

设f时定义在区间[a,b]的一个函数, 对于[a,b]的任意划分P,设\(c_k\)是在子区间\([x_{k-1},x_k]\)上任意选取的数.

如果存在一个数I,使得不论划分P怎样和\(c_k\)如何选取,都有 \(\lim_{||P|| \to 0}\sum_{k=1}^{n}f(c_k)\Delta x_k = I\) 则称f在[a,b]上是可积的,而I称为f在区间[a,b]上的定积分 \(I=\int_a^bf(x)\mathrm{d}x\)

性质

定积分性质
\(\begin{align} &\int_a^bf(x)\mathrm{d}x=-\int_b^af(x)\mathrm{d}x\\ &\int_a^af(x)\mathrm{d}x=0\\ &\int_a^bf(x)\mathrm{d}x=\int_a^cf(x)\mathrm{d}x+\int_c^bf(x)\mathrm{d}x\\ &\int_a^bcf(x)\mathrm{d}x=c\int_a^bf(x)\mathrm{d}x\\ &\int_a^b(f(x)+g(x))\mathrm{d}x=\int_a^bf(x)\mathrm{d}x+\int_a^bg(x)\mathrm{d}x \end{align}\)

微积分基本定理

微积分第一基本定理(不定积分为反导数) \(\frac{d}{d x}\int_a^xf(t)\mathrm{d}t=f(x)\)

Proof:
\(\begin{align} \frac{d}{d x}\int_a^xf(t)\mathrm{d}t&=\lim_{h \to 0}\frac{\int_a^{x+h}f(t)\mathrm{d}t-\int_a^{x}f(t)\mathrm{d}t}{h}\\ &=\lim_{h \to 0}\frac{\int_h^{x+h}f(t)\mathrm{d}t}{h}\\ &=\lim_{h \to 0}\frac{f(x)h}{h}\\ &=f(x) \end{align}\)

微积分第二基本定理 \(\int_a^bf(x)\mathrm{d}x=F(b)-F(a)\)

Proof:

let \(G(x)=\int_a^xf(t)\mathrm{d}t\), G is one of anti-derivative of f
let \(F(x)=G(x)+C\), F is any of anti-derivative of f
\(\begin{align} F(b)-F(a)&=[G(b)+C]-[G(a)+C]\\ &=G(b)-G(a)\\ &=\int_a^bf(t)dt-\int_a^af(t)dt\\ &=\int_a^bf(t)dt \end{align}\)

求积分方法

积分公式

\[\begin{aligned} &\int d u=u+c\\ &\int k d u=k u+c\\ &\int du + dv=\int d u+\int d v\\ &\int u^{n} d u=\frac{u^{n+1}}{n+1}+C\\ &\int \frac{1}{u} d u=\ln |u|+c\\ &\int \sin u d u=-\cos u+c\\ &\int \cos u d u=\sin u+c\\ &\int \sec u d u=\ln |\sec x+\tan x|+c\\ &\int \csc u d u=-\ln |\csc x+\cot x|+c\\ &\int \sec ^{2} u d u=\tan u+c\\ &\int \csc ^{2} u d u=-\cot u+c\\ &\int \sec u \tan u d u=\sec u+c\\ &\int \csc u \cot u d u=-\csc u+c\\ &\int \tan u d u=\ln |\sec u|+C\\ &\int \cot u d u=\ln |\sin u|+C\\ &\int e^{u} d u=e^{u}+c \\ &\int a^{u} d u=\frac{a^{u}}{\ln a}+c\\ &\int \sinh u d u=\cosh u d x+c\\ &\int \cosh u d u=\sinh u+c\\ &\int \frac{d u}{\sqrt{a^{2}-u^{2}}}=\sin ^{-1}\left(\frac{u}{a}\right)+c\\ &\int \frac{d u}{\sqrt{u^{2}-a^{2}}}=\cos h^{-1}\left(\frac{u}{a}\right)+c \\ &\int \frac{d u}{\sqrt{a^{2}+u^{2}}}=\sinh ^{-1}\left(\frac{u}{a}\right)+c \\ &\int \frac{d u}{a^{2}+u^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{u}{a}\right)+c\\ &\int \frac{d u}{u \sqrt{u^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1}\left(\frac{u}{a}\right)+C \end{aligned}\]

变量替换

替换积分法

当f和g’是连续函数时,为求积分\(\int f(g(x))g'(x)dx\)

  1. 做替换\(u=g(x)\),则\(du=g'(x)dx\),得到积分\(\int f(u)du\)
  2. 对u积分
  3. 使用g(x)替代u

定积分的变量替换 \(\int _a^bf(g(x))\cdot g'(x)dx=\int_{g(a)}^{g(b)}f(u)du\)

分部积分

分部积分公式 \(\int u\mathrm{d}v=uv-\int v\mathrm{d}u\)

Proof:
\(\begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(uv)&=u\frac{\mathrm{d}v}{\mathrm{d}x}+v\frac{\mathrm{d}u}{\mathrm{d}x}\\ \int(\frac{\mathrm{d}}{\mathrm{d}x}(uv)\mathrm{d}x&=\int(u\frac{\mathrm{d}v}{\mathrm{d}x})+\int (v\frac{\mathrm{d}u}{\mathrm{d}x})\\\int u\mathrm{d}v&=uv-\int v\mathrm{d}u \end{align}\)

列表积分

对于形如\(\int f(x)g(x)dx\)的积分, 其中f(x)可以反复求导得到0, g(x)可以简单地重复积分


列表积分2

部分分式

形如\(\int \frac{dx}{x^2+px+q}\)

\(\Delta>0\):
\(\frac{1}{x^2+px+q}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}\\\) 其中
\(A_1(x-a_2)+A_2(x-a_1)=1\)
\(\Delta=0\):
\(\int \frac{dx}{x^2+px+q}=\int \frac{dx}{(x-a)^2}=-\frac{1}{x-a}+C\) \(\Delta<0\):

配方 \(x^2+px+q=(x+\frac{p}{2})^2-\frac{\Delta}{4}\)
化简
\(\begin{align} \int \frac{dx}{x^2+px+q}=&-\frac{4}{\Delta}\int \frac{dx}{1+(\frac{2x+p}{\sqrt{-\Delta}})^2}\\ =&\frac{2}{\sqrt{-\Delta}}\int\frac{du}{u^2+1}\\ =&\frac{2}{\sqrt{-\Delta}}tan^{-1}(\frac{2x+p}{\sqrt{-\Delta}})+C \end{align}\)

形如\(\int\frac{mx+n}{x^2+px+q}dx\)

\(\Delta>0\):
\(\frac{mx+n}{x^2+px+q}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}\)
\(\Delta=0\):
\(\frac{mx+n}{x^2+px+q}=\frac{A}{x-a}+\frac{B}{(x-a)^2}\)
\(\Delta<0\): 换元后长除
\(u=x^2+px+q\\ du=2x+p\\ \int\frac{mx+n}{x^2+px+q}dx=\int (r\frac{2x+p}{x^2+px+q}+s\frac{1}{x^2+px+q})dx\\ mx+n=r(2x+q)+s\)

三角换元

三角替换

  1. \(x=atan\theta\) \(a^2+x^2=a^2sec^2\theta\)
  2. \(x=asin\theta\) \(a^2-x^2=a^2cos^2\theta\)
  3. \(x=asec\theta\) \(x^2-a^2=a^2tan^2\theta\)

估算积分

  1. 黎曼和

    在区间[a,b],均分为n等分 \(h=\frac{b-a}{n}\\ \int_a^bf(x)dx\approx h\sum_{j=1}^{n}f(c_j)\)

  2. 梯形法则

    在区间[a,b],均分为n等分 \(h=\frac{b-a}{n}\\ \int_a^bf(x)dx\approx \frac{h}{2}\sum_{j=1}^{n}(f(x_{j-1})+f(x_j))\)

  3. 辛普森法则(二次曲线逼近)

    在区间[a,b],均分为n等分 \(h=\frac{b-a}{n}\\ \int_a^bf(x)dx\approx \frac{h}{3}\sum_{j=1}^{\frac{1}{2}n}(f(x_{2j-2})+4f(x_{2j-1})+f(x_{2j}))\)

    Proof: \(A=\frac{h}{3}(y_0+4y_1+y_2)\)

积分的应用

计算体积

切片法

旋转轴为y=h \(V=\int_a^b\pi (y-h)^2 dx\)

壳法

旋转轴为x=h \(V=\int_a^b2\pi(x-h)ydx\)

椎体体积

椎体体积 \(V=\int_0^hA(x)dx=\frac{1}{3}Ah\)

Proof:
\(\begin{align} \frac{x}{l}=\frac{h}{L} \Rightarrow \frac{A}{A(x)}=&(\frac{L}{l})^2 \Rightarrow A(x)=\frac{Ax^2}{h^2}\\ V=&\int _0^hA(x)dx\\ =&\int _0^h\frac{A}{h^2}x^2dx\\ =&\frac{1}{3}Ah \end{align}\)

计算弧长

\[L=\int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx\\ L=\int_{t_0}^{t_1} \sqrt{(\frac{dy}{dt})^2+(\frac{dx}{dt})^2}dx\\ L=\int_{\theta_0}^{\theta_1} \sqrt{f(\theta)^2+f'(\theta)^2}dx\]

计算旋转体表面积

\[S=\int_a^b2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx\\ S=\int_{t_0}^{t_1}2\pi y\sqrt{(\frac{dy}{dt})^2+(\frac{dx}{dt})^2}dt\]