导数

定义

1.3

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\(\frac{\mathrm{d}y}{\mathrm{d}x}\)