导数例题编练
First
\[\begin{align}
设f( x ) = x^{3} + 2\cos{x} + ln3,\quad求f ( x )' 和f( \frac { π } { 2 } ) '
\\ \\
f( x ) ' = ( x^{3} ) ' + (2\cos{x})' + ( ln3)'
\\ \\
( x^{3} ) ' = \lim_ { Δx \to0 } \frac { ( x +Δx) ^ 3 - x^{3} }{ Δx }
\\ \\
得:\quad
\lim_ { Δx \to 0 }\frac { ( 3x^{2}Δx+3xΔx ^ {2} +Δx^{3})Δx}{Δx}
\\ \\
=\lim_ { Δx \to0} 3x^{2}+3xΔx+Δx^{2}
\\ \\
\because \lim_ {Δx\to0} Δx = 0
\\ \\
\therefore \lim_{ Δx\to0} 3x^{2}
\\ \\
\because ( \cos{x}) ' = - \sin{x},
\quad
\therefore ( 2\cos{x}) '= - 2\sin{x}
\\ \\
( ln3) ' = \lim_ { Δx \to 0} \frac { \ln{3} - \ln{3} } { Δx}=0
\\ \\
f'(x)= 3x^{2} -2\sin{x}
\\ \\
f'(\frac{ π } { 2})=3( \frac { π } { 2} ) ^ { 2 }-2 \sin{\frac{ π }{2}}
\\ \\
=\frac { 3π ^ { 2} }{ 4} - 2\sin{\frac{ π } { 2 }}
\\ \\
\because \sin{\frac { π } {2}}=1
\\ \\
\therefore f'(\frac{ π } { 2})= \frac { 3π ^ { 2} }{ 4} -2
\\ \\
f'(x)= 3x^{2} -2\sin{x}
\\
f'(\frac{ π } { 2})= \frac { 3π ^ { 2} }{ 4 } -2
\end{align}
\]
Second
\[已知: y=\ln (1+e^{x})-x, \quad 求 y'
\]
\[\\
\]
\[y'=[\ln (1+e^{x})]'-(x)'
\]
\[\\
\]
\[设 u=\ln (1+e^{x})
\]
\[\\
\]
根据复合函数求导原则: $$\ y'=(\ln u)' \cdot [(1)'+(e^{x})'] -(x)' $$
\[\\
\]
\[首先: (x)'=1
\]
\[ \\
根据常见求导公式得出:
\]
\[\\
(\ln u)'=\frac{1}{u}=\frac{1}{1+e^{x}}
\]
\[\\
[(1)'+(e^{x})']=0+e^{x}
\]
\[\\ \\
\]
\[综合: (\ln u)' \cdot [(1)'+(e^{x})'] +(x)'
\iff \frac{1}{1+e^{x}} \cdot e^{x} - 1
\\
\]
\[y' = \frac{e^{x}}{1+e^{x}} - 1
\]
Third
\[y=\sqrt{x^{2}+x}, \quad y'=?
\]
\[\\ \\
\]
\[设u=x^{2}+x, 则y=\sqrt{u}
\]
\[\\ \\
\]
根据复合函数求导法则: $$\ $$
\[y'=(\sqrt{u})' \cdot (u)'
\]
\[\\ \\
\]
\[(u)' = (x^{2})'+(x)'=2x+1
\]
\[\\ \\
\]
\[(\sqrt{u})' = \frac{1}{2\sqrt{u}} = \frac{1}{2\sqrt{x^{2}+x}}
\]
\[\\ \\
\]
\[y'=\frac{2x+1}{2\sqrt{x^{2}+x}}
\]
Parametric Equation
\[参数方程:
\begin{cases}
\text{} x= \cos{t} \\
\text{} y= \sin{t} + t
\end{cases}
\quad, 求 \frac{dy}{dx}
\]
\[\\ \\
\]
\[\frac{dx}{dt} = (\cos{t})'=-\sin{t}
\]
\[\\ \\
\]
\[\frac{dy}{dt} = (\sin{t}+t)'=\cos{t}+t'
\]
\[\\ \\
\]
\[ 注: t 在此不是常数,而是函数: (t)=t_{0}
\\
\frac{d}{dt}(t)=\lim_{ h \to 0 } \frac{t_{0}+h-t_{0}}{h} = 1
\]
\[\\ \\
\]
\[\therefore \frac{dy}{dt}=\cos{t}+1
\]
\[\\ \\
\]
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]
\[\\ \\
\]
\[\therefore \frac{dy}{dx}=\frac{\cos{t}+1}{-\sin{t}}
\]
\[\\ \\
\]
\[\frac{dy}{dx}=\frac{\cos{t}}{-\sin{t}}+\frac{1}{-\sin{t}}
\]
\[\\ \\
\]
\[\frac{dy}{dx}=-\cot{t}-\csc{t}
\]
Fifth
\[y=2^{x}+x^{x}, \quad y'?
\]
\[\\ \\
\]
\[y'=(2^{x})'+(x^{x})'
\\
根据常见导数公式: \quad (2^{x})'=2^{x}\ln{2}\]
\[\\ \\
\]
\[x^{x}转为e为底的指数: \quad e^{x\ln{x}}, \quad e^{x\ln{x}}为复合函数
\]
\[\\ \\
\]
\[由链式法则: \quad (x^{x})'=(e^{x\ln{x}})'=(e^{x\ln{x}})' \cdot (x\ln{x})'
\]
\[\\ \\
\]
\[据导数乘法公式: \quad (x\ln{x})'=(x)'\ln{x}+x(\ln{x})'
\]
\[\\ \\
\]
\[\Rightarrow \ln{x}+x \cdot \frac{1}{x} = \ln{x}+1
\]
\[\\ \\
\]
\[\therefore (x^{x})'=e^{x\ln{x}}(\ln{x}+1)=x^{x}(\ln{x}+1)
\]
\[\\ \\
\]
\[y'= 2^{x}\ln{2} + x^{x}(\ln{x}+1)
\]
Sixth
\[\begin{align}
求下列分段函数的导: \\
f(x)=\begin{cases}
\text{} x^{2}\sin{\frac{1}{x}}+2x, \quad x \ne 0
\\
\text{} 0, \quad x = 0
\end{cases}
\\ \\
第一,当x=0时,用定义法(x=0为分段函数之分段点): \\
f'(0)=\lim_{ x \to 0}\frac{f(x)-f(0)}{x-0}
= \lim_{x \to 0}\frac{x^{2}\sin{\frac{1}{x}}+2x-0}{x}
\\
\lim_{x \to 0}\frac{x^{2}\sin{\frac{1}{x}}+2x}{x}
=\lim_{x \to 0}\frac{x(x\sin{\frac{1}{x}}+2)}{x}
=\lim_{x \to 0}(x\sin{\frac{1}{x}}+2)
\\
\lim_{x \to 0}x\sin{\frac{1}{x}}+\lim_{x \to 0}2=0+2=2
\\ \\
第二,当 x\ne0 时用公式法: \\
f'(x)=(x^{2}\sin{\frac{1}{x}}+2x)' \\
依导数乘法公式: \quad (x^{2})'\sin\frac{1}{x}+x^{2}(\sin\frac{1}{x})'+(2x)'
\\ \\
其中 \sin\frac{1}{x} 符合复合函数
\\
\therefore (\sin\frac{1}{x})'=(\sin\frac{1}{x})' \cdot (\frac{1}{x})'
=\cos{\frac{1}{x}} \cdot (-x^{-2})
\\ \\
f'(x)=2x\sin\frac{1}{x}+x^{2}\cos\frac{1}{x}(-x^{-2})+2
\\ \\
f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}+2
\\ \\
综上: \quad
f'(x)=\begin{cases}
\text{} 2x\sin\frac{1}{x}-\cos\frac{1}{x}+2, \quad x \ne 0
\\
\text{} 2, \quad x = 0
\end{cases}
\end{align}
\]
Seventh
\[\begin{align}
求导数: y=\sqrt{x+\cos{x}} \ ,\quad 并求其在x=0处的微分
\\ \\
据链式法则: \ y'=(\sqrt{x+\cos{x}})' \cdot (x+\cos{x})'
\\ \\
\frac{1}{2\sqrt{x+\cos{x}}} \cdot (1-\sin{x})
=\frac{1-\sin{x}}{2\sqrt{x+\cos{x}}}
\\ \\
求其在x=0处的微分, \ 依公式: dy=f'(x_{0})dx 将x=0代入:
\\ \\
y'=\frac{1-\sin{0}}{2\sqrt{0+\cos{0}}} \cdot dx
=\frac{1-0}{2\sqrt{0+1}}\cdot dx=\frac{1}{2}dx
\\ \\
综上:
\ y'=\frac{1-\sin{x}}{2\sqrt{x+\cos{x}}} \ , \ 其在x=0处的微分为\frac{1}{2}dx
\end{align}
\]
Ninth
\[
\begin{align}
已知\lim_{x \to 0} \frac{f(5x)-f(0)}{2x}=1, \ f'(0)=?
\\ \\
注: 利用导数定义求函数某一点之导数
\\ \\
思路: 令x_{0}=0, 则\Delta x=5x, 因此2x须转化为5x, 使其符合f'(0)的导数定义之形式
\\ \\
\lim_{x \to 0} f(0+5x)-f(0)=2x \cdot \frac{5}{2} \cdot \frac{2}{5}
\\ \\
\lim_{x \to 0} f(0+5x)-f(0)=5x \cdot \frac{2}{5}
\\ \\
\lim_{x \to 0} \frac{f(0+5x)-f(0)}{5x}=\frac{2}{5}
\\ \\
\because \lim_{x \to 0} \frac{f(0+5x)-f(0)}{5x}符合f'(0)的导数定义之形式
\\ \\
\therefore f'(0)=\frac{2}{5}
\end{align}
\]
Tenth
\[
\begin{align}
参数方程: \
\begin{cases}
\text{ } x = \frac{1}{2} \cos^{3}{t} \\
\text{ } y = \frac{1}{2} \sin^{3}{t}
\end{cases}
\ , \ 求其导
\\ \\
据参数方程求导公式: \ { \frac{dy}{dt} \div \frac{dx}{dt}} = \frac{dy}{dx}
\\ \\
\frac{dx}{dt}=\frac{1}{2} \cdot [(\cos{t})^{3}]' \cdot (\cos{t})'
=-\frac{3}{2} \cos^{2}{t}\sin{t}
\\ \\
\frac{dy}{dt}=\frac{1}{2} \cdot [(\sin{t})^{3}]' \cdot (\sin{t})'
=\frac{3}{2}\sin^{2}{t}\cos{t}
\\ \\
\frac{dy}{dx}=
\frac{\frac{3}{2}\sin^{2}{t}\cos{t}}{-\frac{3}{2} \cos^{2}{t}\sin{t}}
\\ \\
\because \frac{\sin{t}}{\cos{t}}=\tan{t} \ ,
\ \frac{\cos{t}}{\sin{t}}=\cot{t}
\\ \\
\therefore \frac{dy}{dx}=-\tan^{2}{t} \cdot \cot{t}
=-\tan^{2}{t} \cdot \frac{1}{\tan{t}}=-\tan{t}
\\ \\
参数方程的导数是: \quad \frac{dy}{dx}=-\tan{t}
\end{align}
\]
Eleven
\[\begin{align}
y=2x+\frac{1}{x+1}, \ 其二阶导数y''=?
\\ \\
首先求y': \ y'=2+\frac{0-(x+1)'}{(x+1)^{2}}
\\
y'=2-\frac{1}{(x+1)^{2}}
\\ \\
y''=[2-\frac{1}{(x+1)^{2}}]'
\\ \\
0-\frac{0-[(x+1)^{2}]'}{(x+1)^{4}}
\\ \\
-\frac{0-(2x+2+0)}{(x+1)^{4}}=-\frac{-2(x+1)}{(x+1)^{4}}
\\ \\
y''=\frac{2}{(x+1)^{3}}
\end{align}
\]
