常见函数之导数
Constant
\[\begin{align}
f(x)=C, \quad 求f'(x)
\\ \\
f'(x)=\lim_{Δ\to 0}\frac{f(x+\Delta{x} )-f(x)}{\Delta{x} }
\\ \\
\lim_{\Delta{x} \to 0}\frac{C-C}{\Delta{x} }=0
\\ \\
\therefore (C)'=0
\end{align}
\]
\(\log_{a}{x}\)
\[\begin{align}
求f'(\log_{a}{x})
\\ \\
\lim_{\Delta{x} \to 0} \frac{f(x+\Delta{x} )-f(x)}{\Delta{x} }
\\ \\
\Rightarrow
\lim_{\Delta{x} \to 0}\frac{\log_{a}{(x+\Delta{x} )}-\log_{a}{x}}{\Delta{x} }
\\ \\
=\lim_{\Delta{x} \to 0} \frac{1}{\Delta{x} } \cdot
\log_{a}{\frac{x+\Delta{x} }{x}}
\\ \\
=\lim_{\Delta{x} \to 0} \frac{1}{\Delta{x} } \cdot
\log_{a}{(\frac{x}{x}+\frac{\Delta{x} }{x})}
\\ \\
=\lim_{\Delta{x} \to 0} \frac{1}{\Delta{x} } \cdot
\log_{a}{(1+\frac{\Delta{x} }{x})}
\\ \\
设u=\frac{\Delta{x} }{x}, \quad 则 \frac{1}{u} =\frac{x}{\Delta{x} },
\quad
\frac{1}{\Delta{x} }=\frac{1}{x} \cdot \frac{1}{u}
\\ \\
=\lim_{\Delta{x} \to 0} \frac{1}{x}
\cdot \frac{1}{u}
\cdot \log_{a}{(1+u)}
\\ \\
=\frac{1}{x} \cdot \lim_{\Delta{x} \to 0}
\log_{a}{(1+u)} ^{\frac{1}{u}}
\\ \\
据第二重要极限, 且 u \to 0( \because \Delta{x} \to 0)
\\ \\
\Rightarrow
\frac{1}{x} \times \lim_{\Delta{x} \to 0} \log_{a}{e}
\\ \\
=\frac{1}{x} \times \lim_{\Delta{x} \to 0} \frac{\ln{e}}{\ln{a}}
\\ \\
=\lim_{\Delta{x} \to 0} \frac{1}{x} \cdot
\frac{\log_{e}{e}}{\ln{a}}
\\ \\
=\lim_{\Delta{x} \to 0} \frac{1}{x} \cdot
\frac{1}{\ln{a}}=\lim_{\Delta{x} \to 0}\frac{1}{x\ln_{a}}
\\ \\
\therefore
(\log_{a}{x})'=\frac{1}{x\ln{a}}
\end{align}
\]
\(\sqrt x\)
\[\begin{align}
求y'(\sqrt[]{x})
\\ \\
y = \lim_{ \Delta{x} \to 0 } \frac{\sqrt[]{x+\Delta{x} }-\sqrt[]{x}}{ \Delta{x} }
\\ \\
\lim_{ \Delta{x} \to 0 } \frac{(\sqrt[]{x+\Delta{x} }-\sqrt[]{x})(\sqrt[]{x+\Delta{x} }+\sqrt[]{x})}{\Delta{x} (\sqrt[]{x+\Delta{x} }+\sqrt[]{x})}
\\ \\
\lim_{ \Delta{x} \to 0 }\frac{(\sqrt[]{x+\Delta{x} })^{2}-(\sqrt[]{x})^{2}} { \Delta{x} (\sqrt[]{x+\Delta{x} }+\sqrt[]{x})}
\\ \\
\lim_{ \Delta{x} \to 0 }\frac{x+\Delta{x} -x}{\Delta{x} (\sqrt[]{x+\Delta{x} }+\sqrt[]{x})}
\\ \\
\lim_{ \Delta{x} \to 0 }\frac{1}{\sqrt[]{x+\Delta{x} }+\sqrt[]{x}}
\\ \\
\because \lim_{ \Delta{x} \to 0 }\Delta{x} =0
\\ \\
\therefore\lim_{ \Delta{x} \to 0 }\frac{1}{\sqrt[]{x+0}+\sqrt[]{x}}=\frac{1}{2\sqrt[]{x}}
\\ \\
(\sqrt[]{x})'=\frac{1}{2\sqrt[]{x}}
\end{align}
\]
\(\tan{x}\)
\[\begin{align}
求(\tan{x})'
\\ \\
\because \tan{x} =\frac{\sin{x}}{\cos{x}}
\\ \\
\therefore (\tan{x})'=(\frac{\sin{x}}{\cos{x}})'
\\ \\
=\lim_{\Delta{x} \to 0}\frac{(\sin{x})' \cos{x}-\sin{x}(\cos{x})'}{\cos^{2}{x}}
\\ \\
\lim_{\Delta{x} \to 0}\frac{\cos{x}\cos{x}-\sin{x}\cdot -\sin{x}}{\cos^{2}{x}}
\\ \\
\lim_{\Delta{x} \to 0}\frac{\cos^{2}{x}+\sin^{2}{x}}{\cos^{2}{x}}
\\ \\
\because \cos^{2}{x}+\sin^{2}{x}=1
\\ \\
\therefore \lim_{\Delta{x} \to 0}\frac{\cos^{2}{x}+\sin^{2}{x}}{\cos^{2}{x}}=\frac{1}{\cos^{2}{x}}
\\ \\
\because \cos{x}=\frac{1}{\sec{x}}
\\ \\
\therefore \lim_{\Delta{x} \to 0}\frac{1}{\cos^{2}{x}}=\sec^{2}{x}
\\ \\
(\tan{x})'=\sec^{2}{x}
\end{align}
\]
\(\sin{x}\)
\[\begin{align}
(\sin{x})'=\lim_{\Delta{x} \to 0}\frac{\sin (x+\Delta{x})-\sin{x}}{\Delta{x}}
\\ \\
由和差化积公式: \\
\sin{(x+\Delta{x})}-\sin{x}=2\cos{\frac{x+\Delta{x}+x}{2}} \sin{\frac{x+\Delta{x}-x}{2}}
\\ \\
\lim_{\Delta{x} \to 0}2\cdot \cos{\frac{2x+\Delta{x}}{2}} \cdot \sin{\frac{\Delta{x}}{2}} \cdot\frac{1}{\Delta{x}}
\\ \\
\lim_{\Delta{x} \to 0} \cos{[\frac{2(x+\frac{\Delta x}{2})}{2}]} \cdot \lim_{\Delta{x} \to 0} \sin{\frac{\Delta{x}}{2}} \cdot 2\frac{1}{\Delta{x}}
\\ \\ \\ \\
\lim_{\Delta{x} \to 0} \cos{[\frac{2(x+\frac{\Delta x}{2})}{2}]}
=\lim_{\Delta{x} \to 0} \cos{(x+\frac{\Delta x}{2})}
\\ \\
\because \lim_{\Delta{x} \to 0} \frac{\Delta x}{2} =0
\\ \\
\therefore \lim_{\Delta{x} \to 0} \cos{(x+\frac{\Delta x}{2})} =\cos x
\\ \\ \\ \\
\enspace 设\frac{\Delta{x}}{2}=a,
\enspace
则\frac{2}{\Delta{x}}=\frac{1}{a},
\enspace (\Delta{x} \to 0)=(a\to 0)
\\ \\
据第一重要极限: \\
\lim_{\Delta{x} \to 0} \sin{\frac{\Delta{x}}{2}} \cdot 2\frac{1}{\Delta{x}}
=\lim_{a \to 0}\sin{a} \cdot \frac{1}{a}=1
\\ \\ \\ \\
\lim_{\Delta{x} \to 0}{\cos (x)} \cdot
\lim_{a \to 0}\frac{\sin{a}}{a}
\\ \\
\Rightarrow \cos{x} \cdot 1=\cos{x}
\\ \\
\therefore (\sin{x})'=\cos{x}
\end{align}
\]
\(x^{u}\)
\[\begin{align}
y=x^{u},y'=?
\\ \\
y=x^{u}=e^{u\ln{x}}
\\ \\
设k=u\ln{x},则y=x^{u}=e^{k}=e^{u\ln{x}},
\quad y'可用复合函数求导法则来求导
\\ \\
y'=(e^{k})'\cdot (u\ln{x})'
\\ \\
(e^{k})'=e^{k}
\\ \\
(e^{u\ln{x}})'=\frac{u}{x\ln{e}}=ux^{-1}
\\ \\
e^{k}\cdot ux^{-1}=e^{u\ln{x}}\cdot e^{-\ln{x}}\cdot u
\\ \\
u\cdot e^{\ln{x}(u-1)}
\\ \\
\because \frac{1}{x}=e^{-\ln{x}}
\\ \\
\therefore e^{\ln{x}}=x^{-1}\cdot x^{2}=x
\\ \\
u\cdot e^{\ln{x}(u-1)}=ux^{u-1}
\\ \\
\therefore y'=ux^{u-1}
\end{align}
\]
\(a^{x}\)
PS: \(a^{x}\)与\(x^{u}\) 的区别在于\(x\)是底数还是指数
\[\begin{align}
y=a^{x}, \enspace y'=? ,\enspace a 为常数
\\ \\
\lim_{x \to 0} \frac {a^{(x+\Delta{x})} -a^{x} }{\Delta{x}}
\\ \\
\lim_{x \to 0} \frac { a^{x} \cdot a^{\Delta{x}} - a^{x} }{\Delta{x}}
\\ \\
\lim_{x \to 0} \frac { a^{x}(a^{\Delta{x}} - 1) }{\Delta{x}}
\\ \\
设 t = a^{\Delta{x}} - 1, 当\Delta{x} \to 0 时, t \to 0,二者等价
\\ \\
a^{\Delta{x}}=t+1, \Delta{x} = \log_{a}{(t+1)}
\\ \\
得到: \lim_{t \to 0} a^{x} \cdot \frac{t}{\log_{a}{(t+1)}}
\\ \\
\lim_{t \to 0} a^{x} \cdot \frac{1}{\log_{a}{(t+1)}} \cdot \frac{1}{\frac{1}{t}}
\\ \\
\lim_{t \to 0} a^{x} \cdot \frac{1}{ \frac{1}{t} \log_{a}{(t+1)}}
\\ \\
根据对数运算公式进行变形: \\
\lim_{t \to 0} a^{x} \cdot \frac{1}{ \log_{a}{(t+1)^{\frac{1}{t}} }}
\\ \\
根据第二重要极限公式得出: \\
a^{x} \frac{1}{\log_{a}{e}}
\\ \\
根据对数的倒数关系式得到如下: \\
\frac{1}{\log_{a}{e}} = \ln{a}
\\ \\
\therefore y' = a^{x}\ln{a}
\end{align}
\]
\(ax\)
\[\begin{align}
f(x)=ax, \enspace 求f'(x), \enspace a为常数
\\ \\
f'(x)=\lim_{\Delta{x} \to 0} \frac{a(x+\Delta{x})-ax}{\Delta{x}}
\\ \\
\lim_{\Delta{x} \to 0} \frac{ax+a\Delta{x}-ax}{\Delta{x}}
\\ \\
\lim_{\Delta{x} \to 0} \frac{a\Delta{x}}{\Delta{x}}=a
\\ \\
\therefore f'(x)=(ax)'=a
\end{align}
\]
\(ax^{n}\)
\[\begin{align}
f(x)=ax^{n}, \enspace a为常数, \enspace f'(x)=?
\\ \\
设x^{n}=u, \enspace 则ax^{n}=au
\\ \\
由链式法则: \enspace f'(x)=(au)' \cdot (x^{n})'
\\ \\
(au)'=a, \enspace \enspace (x^{n})'=nx^{n-1}
\\ \\
\therefore f'(x)=a \cdot nx^{n-1}
\end{align}
\]
\(\cos{x}\)
\[\begin{align}
由三角函数和差化积公式:\\
\cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}
\\ \\
因此:\\
\cos (x+\Delta x)-\cos x \Rightarrow -2 \sin \frac{x+\Delta x+x}{2} \sin \frac{x+\Delta x-x}{2}
\\ \\
转化: \\
\lim _{\Delta x \rightarrow 0} \frac{\cos (x+\Delta x)-\cos x}{\Delta x}=
\lim _{\Delta x \rightarrow 0} -2 \sin \frac{2 x + \Delta x}{2} \sin \frac{\Delta x}{2} \cdot \frac{1}{\Delta x}
\\ \\
-\lim _{\Delta x \rightarrow 0}
\sin \left[\frac{2\left(x+\frac{\Delta x}{2}\right)}{2}\right]
\cdot \sin \frac{\Delta x}{2}
\cdot \frac{2}{\Delta x}
\\ \\ \\
设\frac{\Delta x}{2}=t,
\quad
则\frac{2}{\Delta x}=\frac{1}{t},
\quad
(\Delta x \rightarrow 0) \Rightarrow(t \rightarrow 0)
\\ \\
因此得: \enspace -1 \cdot \lim _{t \rightarrow 0} \frac{\sin t}{t} \cdot \sin \left(x+\frac{\Delta x}{2}\right)
\\ \\ \\ \\
\because \lim_{\Delta x \to 0} \frac{\Delta x}{2}=0
\\ \\
\therefore \lim_{\Delta x \to 0} \sin \left(x+\frac{\Delta x}{2}\right)=0
\\ \\ \\ \\
根据第一重要极限: \enspace \lim _{t \rightarrow 0} \frac{\sin t}{t}=1
\\ \\
\Rightarrow -1 \cdot(1 \cdot \sin x)=-\sin x
\\ \\
\therefore (\cos x)'=-\sin x
\end{align}
\]
\(\cot x\)
\[\begin{align}
(\cot x)^{\prime}=?
\\ \\
(\cot x)^{\prime}=(\frac{1}{\tan x})^{\prime}
\\ \\
(\frac{1}{\tan x})^{\prime}=
\frac{(1)^{\prime}\tan x-(\tan x)^{\prime}}{\tan^{2}x}=
\frac{-\frac{1}{\cos^{2}x}}{\tan^{2}x}
\\ \\
\Rightarrow
-\frac{1}{\cos x}\cdot\frac{1}{\tan^{2}x}=
-\frac{1}{\cos^{2}x}\cdot\frac{1}{\frac{\sin^{2}x}{\cos^{2}x}}
=-\frac{1}{\sin^{2}x}
\\ \\
\because \frac{1}{\sin x}=\csc x
\\ \\
\therefore(\cot x)^{\prime}=-\frac{1}{\sin^{2}x}=-\csc^{2}x
\end{align}
\]
\(\sec x\)
\[\begin{eqnarray}
\left(\sec x\right)^{\prime}=?
\\ \\
(\sec x)^{\prime}=(\frac{1}{\cos x})
\\ \\
据导数乘法法则:
\\
\Rightarrow\frac{0-\left(-\sin x\right)}{\cos^{2}x}
\\ \\
\Rightarrow\frac{\sin x}{\cos^{2}x}=\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}
\\ \\
=\tan x\sec x
\end{eqnarray}
\]
\(\arccos x\)
\[\begin{align}
\text { 若: } y=\arccos x, \quad y^{\prime}=?
\\ \\
y=\arccos x \Leftrightarrow \cos y=x
\\ \\
(\arccos x)^{\prime}=(\cos y)^{\prime}=(x)^{\prime}
\\ \\
\cos y 为复合函数,基于链式法则:
\\ \\
(\cos y)^{\prime} \Rightarrow(\cos y)^{\prime} \cdot y^{\prime}=(x)^{\prime}
\\ \\
-\sin y \cdot y^{\prime}=1
\\ \\
\because -\sin y=-\sqrt{1-\cos ^{2} y}
\\ \\
\because \cos y=x \Rightarrow \cos ^{2} y=x^{2}
\\ \\
\therefore\left(-\sqrt{1-x^{2}}\right) \cdot y^{\prime}=1
\\ \\
y^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}
\end{align}
\]
\(\arctan (x)\)
\[\begin{align}
y=\arctan(x), \quad y^{\prime}=?
\\ \\
y=\arctan(x)\Rightarrow\tan(y)=x
\\ \\
\arctan^{\prime}(x)=\tan^{\prime}(y)=(x)^{\prime}
\\ \\
据链式法则:
\\ \\
\tan^{\prime}(y)=\tan^{\prime}(y)\cdot y^{\prime}
\\ \\
\Rightarrow\tan^{\prime}(y)\cdot y^{\prime}=(x)^{\prime}
\\ \\
\Rightarrow\sec^{2}(y)\cdot y^{\prime}=1
\\ \\
y^{\prime}=\frac{1}{\sec^{2}(y)}\Rightarrow\frac{1}{1+\tan^{2}(y)}
\\ \\
\because\tan(y)=x\Rightarrow\tan^{2}(y)=x^{2}
\\ \\
\therefore y^{\prime}=\frac{1}{\sec^{2}(y)}=\frac{1}{1+\tan^{2}(y)}=\frac{1}{1+x^{2}}
\\ \\
\arctan^{\prime}(x)=\frac{1}{1+x^{2}}
\end{align}
\]
