【笔记】一元函数微分学

导数

可导性判定

1. $f(x)$在$x_0$可导的充分条件为：$f(x)$在$x_0$左右可导并有 $f'_-(x)=f'_+(x)$
2. 可导一定连续，连续不一定可导

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链式法则求导

设$y=f(x)$,则$f'(t)=frac{{ m d}y}{{ m d}x}=frac{{ m d}y}{{ m d}t}frac{{ m d}t}{{ m d}x}$

常用于复杂的复合函数求导

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$eg.1求(a^x)'$
$a^x=e{xlna} ightarrow 令u=xlna Rightarrow$
$$
frac{{ m d}a^x}{{ m d}x}=frac{{ m d}e^u}{{ m d}u}frac{{ m d}u}{{ m d}x}=e^ulna=axlna
$$

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$eg.2设y=sinsqrt{x^2+x+1},求y'$
$设u=x^2+x+1,v=sqrt{u}，由链式法则Rightarrow $
$$frac{{ m d}y}{{ m d}x}=frac{{ m d}y}{{ m d}v}frac{{ m d}v}{{ m d}u}frac{{ m d}u}{{ m d}x}=cosvfrac{1}{2sqrt{u}}(2x+1)=frac{2x+1}{2sqrt{x^{2+x+1}}cossqrt{x}2+x+1}$$

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对数求导

$y=f(x) ightarrow lny=lnf(x) ightarrow (lny)'=(lnf(x))'Rightarrow$
$y'frac{1}{y}=(lnf(x))'Rightarrow y'=y(lnf(x))'$

$eg.3设y=sqrt[3]{frac{x^{2(x-1)(x-2)}{(x+1)(x+3)}2}},求y'$
$egin{aligned}
y' & =y(lnfrac{x^{2(x-1)(x-2)}{(x+1)(x+3)}2})' \
& = y(frac{2}{3}lnx+frac{1}{3}ln(x-1)+frac{1}{3}ln(x+2)-frac{1}{3}ln(x+1)-frac{2}{3}ln(x+3))' \
& = sqrt[3]{frac{x^{2(x-1)(x-2)}{(x+1)(x+3)}2}}(frac{2}{3(x^2+1)}+frac{2}{x(x+3)}+frac{1}{3(x+2)}) \
end{aligned}
$

对数求导用于函数中同时包含幂函数、乘法、除法的复杂复合函数

反函数求导

$由链式法则可得：frac{{ m d}y}{{ m d}x}=frac{1}{frac{{ m d}x}{{ m d}y}}$
$eg.4设x=arcsiny,求x'$
$x=arcsiny ightarrow y=sinxRightarrow $
$$
frac{{ m d}x}{{ m d}y} = frac{1}{frac{{ m d}y}{{ m d}x}} = frac{1}{frac{{ m d}sinx}{{ m d}x}} = frac{1}{cosx} = frac{1}{sqrt{1-sin^{2x}}=frac{1}{sqrt{1-y}2}}
$$

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幂指函数求导

先将底化为常数，再对指数进行求导

$eg.5设y=x^x求y'$
$y'=(e^{{xlnx})'=(xlnx)'e}{xlnx}=(lnx+1)x^x$

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高阶导数

$(sinx)^{(n)}=sin(frac{npi}{2}+x)$
$(cosx)^{(n)}=cos(frac{npi}{2}+x)$
$ln(1+x)^{{(n)}=frac{(-1)}{n-1}(n-1)!}{(1+x)^n}$
$sin(ax+b)^{(n)}=ansin(ax+b+frac{npi}{2})$
$(e^{ax+b}){(n)}=a^ne{ax+b}$
$[u(x)v(x)]^{{(n)}=sum_{k=0}}{n}C_{n}^{k}u{(k)}v^{(n-k)}$

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