# 导数的四则运算

$\begin{array}{l}[f(x)\pm g(x)]’=f'(x)\pm g'(x)\\ \left[f(x)\cdot g(x)\right]’=f'(x)g(x)+f(x)g'(x)\\ \left[\frac{f(x)}{g(x)}\right]’=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}\end{array}$

\begin{align*}[f(x)\pm g(x)]’&=\lim_{h\to0}\frac{[f(x+h)\pm g(x+h)]-[f(x)\pm g(x)]}{h}\\ &=\lim_{h\to 0}\frac{(f(x+h)-f(x))\pm(g(x+h)-g(x))}{h}\\ &=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\pm \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=f'(x)\pm g'(x)\end{align*}

\begin{align*}[f(x)g(x)]’&=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}+\lim_{h\to0}\frac{f(x)g(x+h)-f(x)g(x)}{h}\\ &= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\cdot g(x+h)+\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\cdot f(x)\\ &=f'(x)g(x)+f(x)g'(x)\end{align*}

\begin{align*}\left(\frac{f(x)}{g(x)}\right)’&=\lim_{h\to 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}=\lim_{h\to0}\frac{\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}}{h}\\&= \lim_{h\to0}\frac{f(x+h)g(x)-f(x)g(x+h)}{h\cdot g(x+h)g(x)}\\&=\lim_{h\to0}\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{h\cdot g(x+h)g(x)}\\ &=\lim_{h\to0}\frac{\frac{f(x+h)g(x)-f(x)g(x)}{h}-\frac{f(x)g(x+h)-f(x)g(x)}{h}}{ g(x+h)g(x)}\\&=\lim_{h\to0}\frac{\frac{f(x+h)-f(x)}{h}\cdot g(x)-\frac{g(x+h)-g(x)}{h}\cdot f(x)}{ g(x+h)g(x)}\\ &=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} \end{align*}

(1) $$f(x)=5x^3+\cos x-e^x$$

(2) $$f(x)=x^3e^x$$

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

（2）由乘法的导数法则 $f'(x)=(x^3e^x)=(x^3)’e^x+x^3(e^x)’=3x^2e^x+x^3e^x$

（3）由商的求导法则 \begin{align*}f'(x)&=\frac{(4-3x)'(3x^2+x)-(4-3x)(3x^2+x)’}{(3x^2+x)^2}\\ &=\frac{-3(3x^2+x)-(4-3x)(6x+1)}{(3x^2+x)^2}\\&=\frac{9x^2-24x-4}{(3x^2+x)^2}\end{align*}

$(\tan x)’=\left(\frac{\sin x}{\cos x}\right)’=\frac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2x$

$(\cot x)’=\left(\frac{\cos x}{\sin x}\right)’=\frac{-\sin x\sin x-\cos x\cos x}{\sin^2x}=-\frac{1}{\sin^2x}=-\csc^2 x$

$(\sec x)’=\left(\frac{1}{\cos x}\right)’=\frac{0-(-\sin x)}{\cos^2x}=\frac{\sin x}{\cos^2x}=\tan x\sec x$

$(\csc x)’=\left(\frac{1}{\sin x}\right)’=\frac{0-\cos x}{\sin^2x}=-\cot x\csc x$

$\begin{array}{ll}(\tan x)’=\sec^2x& (\cot x)’=-\csc^2x\\ (\sec x)’=\tan x\sec x\quad& (\csc x)’=-\cot x\csc x\end{array}$