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函数和差积商的求导法则

## 函数和差积商的求导法则
若 $u(x) 、 v(x)$ 在点 $x$ 处的导数均存在，则它们的和、差、积、商的导数 也都存在，且有
(1) $[u(x) \pm v(x)]^{\prime}=u^{\prime}(x) \pm v^{\prime}(x)$ ；
(2) $[u(x) \cdot v(x)]^{\prime}=u^{\prime}(x) v(x)+u(x) v^{\prime}(x)$ ；
(3)$\left[\frac{u(x)}{v(x)}\right]^{\prime}=\frac{u^{\prime}(x) v(x)-u(x) v^{\prime}(x)}{v^2(x)}(v(x) \neq 0)$.

证明 我们仅证明 (2) $[u(x) \cdot v(x)]^{\prime}=u^{\prime}(x) v(x)+u(x) v^{\prime}(x)$

$$
\begin{aligned}
{[u(x) \cdot v(x)]^{\prime} } & =\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x) \cdot v(x+\Delta x)-u(x) \cdot v(x)}{\Delta x} \\
& =\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x) v(x+\Delta x)-u(x) v(x+\Delta x)+u(x) v(x+\Delta x)-u(x) v(x)}{\Delta x} \\
& =\lim _{\Delta x \rightarrow 0}\left[\frac{u(x+\Delta x)-u(x)}{\Delta x} v(x+\Delta x)+u(x) \frac{v(x+\Delta x)-v(x)}{\Delta x}\right] \\
& =\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x)-u(x)}{\Delta x} v(x+\Delta x)+\lim _{\Delta x \rightarrow 0} u(x) \frac{v(x+\Delta x)-v(x)}{\Delta x} \\
& =u^{\prime}(x) v(x)+u(x) v^{\prime}(x) .
\end{aligned}
$$

注
(1) $[C u(x)]^{\prime}=C u^{\prime}(x)$ ， 由此可得 $[\alpha u(x)+\beta v(x)]^{\prime}=\alpha u^{\prime}(x)+\beta v^{\prime}(x)$.
(2) 若 $u^{\prime}(x) ， v^{\prime}(x) ， w^{\prime}(x)$ 均存在，则 $[u(x) \cdot v(x) \cdot w(x)]^{\prime}$ 存在，且 $[u(x) \cdot v(x) \cdot w(x)]^{\prime}=u^{\prime}(x) v(x) w(x)+u(x) v^{\prime}(x) w(x)+u(x) v(x) w^{\prime}(x)$

## 求导法则

$$ \boxed{
(u \pm v)^{\prime}=u^{\prime} \pm v^{\prime} ; 
\quad(u \cdot v)^{\prime}=u^{\prime} \cdot v+u \cdot v^{\prime} ；\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} \cdot v-u \cdot v^{\prime}}{v^2}(v \neq 0)  
}
$$

利用商的导数公式可以得到另外四个三角函数的计算公式.

$$
\begin{aligned}
& (\tan x)^{\prime}=\left(\frac{\sin x}{\cos x}\right)^{\prime}=\frac{\cos x \cdot \cos x-\sin x \cdot(-\sin x)}{\cos ^2 x}=\sec ^2 x ; \\
& (\cot x)^{\prime}=\left(\frac{\cos x}{\sin x}\right)^{\prime}=\frac{-\sin x \cdot \sin x-\cos x \cdot \cos x}{\sin ^2 x}=\frac{-1}{\sin ^2 x}=-\csc ^2 x ; \\
& (\sec x)^{\prime}=\left(\frac{1}{\cos x}\right)^{\prime}=\frac{0-(-\sin x)}{\cos ^2 x}=\sec x \tan x ; \\
& (\csc x)^{\prime}=\left(\frac{1}{\sin x}\right)^{\prime}=\frac{-\cos x}{\sin ^2 x}=-\csc x \cot x .
\end{aligned}
$$

求下列导数
(1) $y=3^x-\ln x+\sin x+\mathrm{e}$

$$
y^{\prime}=\left(3^x-\ln x+\sin x+\mathrm{e}\right)^{\prime}=\left(3^x\right)^{\prime}-(\ln x)^{\prime}+(\sin x)^{\prime}+(\mathrm{e})^{\prime}=3^x \ln 3-\frac{1}{x}+\cos x
$$

(2) $f(x)=x^3+4 \cos x-\sin \frac{\pi}{7}$

$$
f^{\prime}(x)=\left(x^3+\cos x-\sin \frac{\pi}{7}\right)^{\prime}=\left(x^3\right)^{\prime}+(4 \cos x)^{\prime}-\left(\sin \frac{\pi}{7}\right)^{\prime}=3 x^2-4 \sin x
$$

(3) $y=\left(x^2+3 a^x\right)(\sin x-1)$

$$
\begin{aligned}
y^{\prime}=\left[\left(x^2+3 a^x\right)(\sin x-1)\right]^{\prime} & =\left(x^2+3 a^x\right)^{\prime}(\sin x-1)+\left(x^2+3 a^x\right)(\sin x-1)^{\prime} \\
& =\left(2 x+3 a^x \ln a\right)(\sin x-1)+\left(x^2+3 a^x\right) \cos x
\end{aligned}
$$

(4) $y=(2 x+3)(1-x)(x+2)$

$$
\begin{aligned}
y^{\prime} & =(2 x+3)^{\prime}(1-x)(x+2)+(2 x+3)(1-x)^{\prime}(x+2)+(2 x+3)(1-x)(x+2)^{\prime} \\
& =2(1-x)(x+2)+(2 x+3)(-1)(x+2)+(2 x+3)(1-x) \\
& =-6 x^2-10 x+1
\end{aligned}
$$

(5) $y=\frac{x+1}{\ln x}$

$$
y^{\prime}=\frac{(x+1)^{\prime} \ln x-(x+1)(\ln x)^{\prime}}{(\ln x)^2}=\frac{\ln x-(x+1) \frac{1}{x}}{(\ln x)^2}=\frac{x \ln x-(x+1)}{x(\ln x)^2} \text {. }
$$

(6) $f(x)=\frac{2 \sqrt{x} \tan x}{1+x^2}$

$$
\begin{aligned}
f^{\prime}(x) & =2 \frac{(\sqrt{x} \tan x)^{\prime}\left(1+x^2\right)-\sqrt{x} \tan x\left(1+x^2\right)^{\prime}}{\left(1+x^2\right)^2} \\
& =2 \frac{\left(\frac{1}{2 \sqrt{x}} \tan x+\sqrt{x} \sec ^2 x\right)\left(1+x^2\right)-\sqrt{x} \tan x \cdot 2 x}{\left(1+x^2\right)^2}=\frac{2 x\left(1+x^2\right) \sec ^2 x+\left(1-3 x^2\right) \tan x}{\sqrt{x}\left(1+x^2\right)^2}
\end{aligned}

上一篇：函数的可导性与连续性的关系

下一篇：反函数的求导法则

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