导数例题编练

First

$$\begin{align} 设f( x ) = x^{3} + 2\cos{x} + ln3,\quad求f ( x )' 和f( \frac { π } { 2 } ) ' \\ \\ f( x ) ' = ( x^{3} ) ' + (2\cos{x})' + ( ln3)' \\ \\ ( x^{3} ) ' = \lim_ { Δx \to0 } \frac { ( x +Δx) ^ 3 - x^{3} }{ Δx } \\ \\ 得:\quad \lim_ { Δx \to 0 }\frac { ( 3x^{2}Δx+3xΔx ^ {2} +Δx^{3})Δx}{Δx} \\ \\ =\lim_ { Δx \to0} 3x^{2}+3xΔx+Δx^{2} \\ \\ \because \lim_ {Δx\to0} Δx = 0 \\ \\ \therefore \lim_{ Δx\to0} 3x^{2} \\ \\ \because ( \cos{x}) ' = - \sin{x}, \quad \therefore ( 2\cos{x}) '= - 2\sin{x} \\ \\ ( ln3) ' = \lim_ { Δx \to 0} \frac { \ln{3} - \ln{3} } { Δx}=0 \\ \\ f'(x)= 3x^{2} -2\sin{x} \\ \\ f'(\frac{ π } { 2})=3( \frac { π } { 2} ) ^ { 2 }-2 \sin{\frac{ π }{2}} \\ \\ =\frac { 3π ^ { 2} }{ 4} - 2\sin{\frac{ π } { 2 }} \\ \\ \because \sin{\frac { π } {2}}=1 \\ \\ \therefore f'(\frac{ π } { 2})= \frac { 3π ^ { 2} }{ 4} -2 \\ \\ f'(x)= 3x^{2} -2\sin{x} \\ f'(\frac{ π } { 2})= \frac { 3π ^ { 2} }{ 4 } -2 \end{align}$$

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Second

$$已知: y=\ln (1+e^{x})-x, \quad 求 y'$$

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$$y'=[\ln (1+e^{x})]'-(x)'$$

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$$设 u=\ln (1+e^{x})$$

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根据复合函数求导原则: $$\ y'=(\ln u)' \cdot [(1)'+(e^{x})'] -(x)'$$

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$$首先: (x)'=1$$

$$\\ 根据常见求导公式得出:$$

$$\\ (\ln u)'=\frac{1}{u}=\frac{1}{1+e^{x}}$$

$$\\ [(1)'+(e^{x})']=0+e^{x}$$

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$$综合: (\ln u)' \cdot [(1)'+(e^{x})'] +(x)' \iff \frac{1}{1+e^{x}} \cdot e^{x} - 1 \\ $$

$$y' = \frac{e^{x}}{1+e^{x}} - 1$$

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Third

$$y=\sqrt{x^{2}+x}, \quad y'=?$$

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$$设u=x^{2}+x, 则y=\sqrt{u}$$

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根据复合函数求导法则: $$\ $$

$$y'=(\sqrt{u})' \cdot (u)'$$

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$$(u)' = (x^{2})'+(x)'=2x+1$$

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$$(\sqrt{u})' = \frac{1}{2\sqrt{u}} = \frac{1}{2\sqrt{x^{2}+x}}$$

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$$y'=\frac{2x+1}{2\sqrt{x^{2}+x}}$$

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Parametric Equation

$$参数方程: \begin{cases} \text{} x= \cos{t} \\ \text{} y= \sin{t} + t \end{cases} \quad, 求 \frac{dy}{dx}$$

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$$\frac{dx}{dt} = (\cos{t})'=-\sin{t}$$

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$$\frac{dy}{dt} = (\sin{t}+t)'=\cos{t}+t'$$

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$$注: t 在此不是常数,而是函数: (t)=t_{0} \\ \frac{d}{dt}(t)=\lim_{ h \to 0 } \frac{t_{0}+h-t_{0}}{h} = 1$$

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$$\therefore \frac{dy}{dt}=\cos{t}+1$$

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$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

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$$\therefore \frac{dy}{dx}=\frac{\cos{t}+1}{-\sin{t}}$$

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$$\frac{dy}{dx}=\frac{\cos{t}}{-\sin{t}}+\frac{1}{-\sin{t}}$$

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$$\frac{dy}{dx}=-\cot{t}-\csc{t}$$

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Fifth

$$y=2^{x}+x^{x}, \quad y'?$$

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$$y'=(2^{x})'+(x^{x})' \\ 根据常见导数公式: \quad (2^{x})'=2^{x}\ln{2}$$

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$$x^{x}转为e为底的指数: \quad e^{x\ln{x}}, \quad e^{x\ln{x}}为复合函数$$

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$$由链式法则: \quad (x^{x})'=(e^{x\ln{x}})'=(e^{x\ln{x}})' \cdot (x\ln{x})'$$

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$$据导数乘法公式: \quad (x\ln{x})'=(x)'\ln{x}+x(\ln{x})'$$

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$$\Rightarrow \ln{x}+x \cdot \frac{1}{x} = \ln{x}+1$$

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$$\therefore (x^{x})'=e^{x\ln{x}}(\ln{x}+1)=x^{x}(\ln{x}+1)$$

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$$y'= 2^{x}\ln{2} + x^{x}(\ln{x}+1)$$

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Sixth

$$\begin{align} 求下列分段函数的导: \\ f(x)=\begin{cases} \text{} x^{2}\sin{\frac{1}{x}}+2x, \quad x \ne 0 \\ \text{} 0, \quad x = 0 \end{cases} \\ \\ 第一,当x=0时,用定义法(x=0为分段函数之分段点): \\ f'(0)=\lim_{ x \to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0}\frac{x^{2}\sin{\frac{1}{x}}+2x-0}{x} \\ \lim_{x \to 0}\frac{x^{2}\sin{\frac{1}{x}}+2x}{x} =\lim_{x \to 0}\frac{x(x\sin{\frac{1}{x}}+2)}{x} =\lim_{x \to 0}(x\sin{\frac{1}{x}}+2) \\ \lim_{x \to 0}x\sin{\frac{1}{x}}+\lim_{x \to 0}2=0+2=2 \\ \\ 第二,当 x\ne0 时用公式法: \\ f'(x)=(x^{2}\sin{\frac{1}{x}}+2x)' \\ 依导数乘法公式: \quad (x^{2})'\sin\frac{1}{x}+x^{2}(\sin\frac{1}{x})'+(2x)' \\ \\ 其中 \sin\frac{1}{x} 符合复合函数 \\ \therefore (\sin\frac{1}{x})'=(\sin\frac{1}{x})' \cdot (\frac{1}{x})' =\cos{\frac{1}{x}} \cdot (-x^{-2}) \\ \\ f'(x)=2x\sin\frac{1}{x}+x^{2}\cos\frac{1}{x}(-x^{-2})+2 \\ \\ f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}+2 \\ \\ 综上: \quad f'(x)=\begin{cases} \text{} 2x\sin\frac{1}{x}-\cos\frac{1}{x}+2, \quad x \ne 0 \\ \text{} 2, \quad x = 0 \end{cases} \end{align}$$

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Seventh

$$\begin{align} 求导数: y=\sqrt{x+\cos{x}} \ ,\quad 并求其在x=0处的微分 \\ \\ 据链式法则: \ y'=(\sqrt{x+\cos{x}})' \cdot (x+\cos{x})' \\ \\ \frac{1}{2\sqrt{x+\cos{x}}} \cdot (1-\sin{x}) =\frac{1-\sin{x}}{2\sqrt{x+\cos{x}}} \\ \\ 求其在x=0处的微分, \ 依公式: dy=f'(x_{0})dx 将x=0代入: \\ \\ y'=\frac{1-\sin{0}}{2\sqrt{0+\cos{0}}} \cdot dx =\frac{1-0}{2\sqrt{0+1}}\cdot dx=\frac{1}{2}dx \\ \\ 综上: \ y'=\frac{1-\sin{x}}{2\sqrt{x+\cos{x}}} \ , \ 其在x=0处的微分为\frac{1}{2}dx \end{align}$$

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Ninth

$$\begin{align} 已知\lim_{x \to 0} \frac{f(5x)-f(0)}{2x}=1, \ f'(0)=? \\ \\ 注: 利用导数定义求函数某一点之导数 \\ \\ 思路: 令x_{0}=0, 则\Delta x=5x, 因此2x须转化为5x, 使其符合f'(0)的导数定义之形式 \\ \\ \lim_{x \to 0} f(0+5x)-f(0)=2x \cdot \frac{5}{2} \cdot \frac{2}{5} \\ \\ \lim_{x \to 0} f(0+5x)-f(0)=5x \cdot \frac{2}{5} \\ \\ \lim_{x \to 0} \frac{f(0+5x)-f(0)}{5x}=\frac{2}{5} \\ \\ \because \lim_{x \to 0} \frac{f(0+5x)-f(0)}{5x}符合f'(0)的导数定义之形式 \\ \\ \therefore f'(0)=\frac{2}{5} \end{align}$$

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Tenth

$$\begin{align} 参数方程: \ \begin{cases} \text{ } x = \frac{1}{2} \cos^{3}{t} \\ \text{ } y = \frac{1}{2} \sin^{3}{t} \end{cases} \ , \ 求其导 \\ \\ 据参数方程求导公式: \ { \frac{dy}{dt} \div \frac{dx}{dt}} = \frac{dy}{dx} \\ \\ \frac{dx}{dt}=\frac{1}{2} \cdot [(\cos{t})^{3}]' \cdot (\cos{t})' =-\frac{3}{2} \cos^{2}{t}\sin{t} \\ \\ \frac{dy}{dt}=\frac{1}{2} \cdot [(\sin{t})^{3}]' \cdot (\sin{t})' =\frac{3}{2}\sin^{2}{t}\cos{t} \\ \\ \frac{dy}{dx}= \frac{\frac{3}{2}\sin^{2}{t}\cos{t}}{-\frac{3}{2} \cos^{2}{t}\sin{t}} \\ \\ \because \frac{\sin{t}}{\cos{t}}=\tan{t} \ , \ \frac{\cos{t}}{\sin{t}}=\cot{t} \\ \\ \therefore \frac{dy}{dx}=-\tan^{2}{t} \cdot \cot{t} =-\tan^{2}{t} \cdot \frac{1}{\tan{t}}=-\tan{t} \\ \\ 参数方程的导数是: \quad \frac{dy}{dx}=-\tan{t} \end{align}$$

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Eleven

$$\begin{align} y=2x+\frac{1}{x+1}, \ 其二阶导数y''=? \\ \\ 首先求y': \ y'=2+\frac{0-(x+1)'}{(x+1)^{2}} \\ y'=2-\frac{1}{(x+1)^{2}} \\ \\ y''=[2-\frac{1}{(x+1)^{2}}]' \\ \\ 0-\frac{0-[(x+1)^{2}]'}{(x+1)^{4}} \\ \\ -\frac{0-(2x+2+0)}{(x+1)^{4}}=-\frac{-2(x+1)}{(x+1)^{4}} \\ \\ y''=\frac{2}{(x+1)^{3}} \end{align}$$

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