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【前言】
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【正文】
例2. 证明函数 y = \sqrt[3]{x},y = \sqrt {{x^2}} = \left| x \right|x = 0 点连续,但是在 x = 0 点不可导。

证:
y = \sqrt[3]{x} ,自变量在 x = 0 点有增量 \Delta x ,则 \Delta y = \sqrt[3]{{0 + \Delta x}} - \sqrt[3]{0} = \sqrt[3]{{\Delta x}}
因此 {(\Delta y)^3} = \Delta x
因为 \mathop {\lim }\limits_{\Delta x \to 0} {(\Delta y)^3} = {\left( {\mathop {\lim }\limits_{\Delta x \to 0} \Delta y} \right)^3} = \mathop {\lim }\limits_{\Delta x \to 0} \Delta x = 0
所以 \mathop {\lim }\limits_{\Delta x \to 0} \Delta y = 0
所以 y = \sqrt[3]{x}x = 0 点连续(注:由第18课的连续性定义可知)
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下面证明导数不存在。
第一个函数:
\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sqrt[3]{{\Delta x}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{{(\Delta x)}^{\frac{2}{3}}}}} = \infty
因此 y = \sqrt[3]{x}x = 0 点不可导。

第二个函数:
y = \sqrt {{x^2}} = \left| x \right| = \left\{ {\begin{array}{*{20}{c}}{x,x \ge 0}\\{ - x,x < 0}\end{array}} \right. ,易证 y = \left| x \right|x = 0 点连续(这里就不详细写了)
设自变量 xx = 0 点有增量 \Delta x ,则:

\Delta y = \left| {0 + \Delta x} \right| - \left| 0 \right| = \left| {\Delta x} \right| = \left\{ {\begin{array}{*{20}{c}}{\Delta x,\Delta x > 0}\\{ - \Delta x,\Delta x < 0}\end{array}} \right.
x = 0 处的右导数 {{f'}_ + }(0) = \mathop {\lim }\limits_{\Delta x \to {0^ + }} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to {0^ + }} \frac{{\Delta x}}{{\Delta x}} = 1
x = 0 处的左导数 {{f'}_ - }(0) = \mathop {\lim }\limits_{\Delta x \to {0^ - }} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to {0^ - }} \frac{{ - \Delta x}}{{\Delta x}} = - 1
因为 {{f'}_ + }(0) \ne {{f'}_ - }(0)
所以 y = f(x) = \left| x \right|x = 0 点不可导(注:由第23课开头的定义可知)
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从函数图形上很容易看出来:

对右图,在 x = 0 点处,切线垂直于 x 轴,斜率为无穷大,故不可导。
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五、几个基本初等函数的导数公式
1. 常数 Cf(x) \equiv C, - \infty < x < + \infty
下面推导其导数:
y = f(x) \equiv C,\;\forall x \in ( - \infty , + \infty )
\Delta y = f(x + \Delta x) - f(x) = C - C = 0
f'(x) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{0}{{\Delta x}} = 0
因此 {\left( C \right)^\prime } = 0
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2. 幂函数 y = f(x) = {x^\alpha }\alpha 为实常数)
下面推导其导数:
\alpha = n(n \in N) 时,有 \Delta y = f(x + \Delta x) - f(x) = {(x + \Delta x)^n} - {x^n}
按二项式定理展开前面的 {(x + \Delta x)^n} ,得:
\Delta y = \left[ {{x^n} + n{x^{n - 1}}\Delta x + \frac{{n(n - 1)}}{{2!}}{x^{n - 2}}{{(\Delta x)}^2} + \cdots + {{(\Delta x)}^n}} \right] - {x^n}
 = n{x^{n - 1}}\Delta x + \frac{{n(n - 1)}}{{2!}}{x^{n - 2}}{(\Delta x)^2} + \cdots + {(\Delta x)^n}
因此 \frac{{\Delta y}}{{\Delta x}} = n{x^{n - 1}} + \frac{{n(n - 1)}}{{2!}}{x^{n - 2}}\Delta x + \cdots + {(\Delta x)^{n - 1}}
因此 \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \left[ {n{x^{n - 1}} + \frac{{n(n - 1)}}{{2!}}{x^{n - 2}}\Delta x + \cdots + {{(\Delta x)}^{n - 1}}} \right] = n{x^{n - 1}}
(注:从第二项开始,每一项的极限均为0)
因此 ({x^n})' = n{x^{n - 1}}
\alpha 为任何实常数时, ({x^\alpha })' = \alpha {x^{\alpha - 1}} ,这个结论以后再证明。
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3. 正弦、余弦函数 y = f(x) = \sin x,f(x) = \cos x
先来推导正弦函数的导数:
y = \sin x,\; - \infty < x < + \infty
\forall x \in ( - \infty , + \infty ) ,自变量有增量 {\Delta x} ,函数 y = \sin x 的增量 \Delta y = \sin (x + \Delta x) - \sin x = 2\sin \frac{{\Delta x}}{2}\cos (x + \frac{{\Delta x}}{2})
(注:三角函数的和差化积公式)
因此 \frac{{\Delta y}}{{\Delta x}} = \frac{{2\sin \frac{{\Delta x}}{2}\cos (x + \frac{{\Delta x}}{2})}}{{\Delta x}}
因此 \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{2\sin \frac{{\Delta x}}{2}}}{{\Delta x}} \cdot \mathop {\lim }\limits_{\Delta x \to 0} \cos (x + \frac{{\Delta x}}{2}) = 1 \cdot \cos x = \cos x
(注: \mathop {\lim }\limits_{\Delta x \to 0} \frac{{2\sin \frac{{\Delta x}}{2}}}{{\Delta x}} = 1 是重要极限之一; y = \cos x 是连续函数,因此 \mathop {\lim }\limits_{\Delta x \to 0} \cos (x + \frac{{\Delta x}}{2}) 的极限号可以放进去)
因此 (\sin x)' = \cos x
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再来推导余弦函数的导数:
y = \cos x,\; - \infty < x < + \infty
\forall x \in ( - \infty , + \infty )\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos (x + \Delta x) - \cos x}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{ - 2\sin \frac{{\Delta x}}{2}\sin \left( {x + \frac{{\Delta x}}{2}} \right)}}{{\Delta x}}
 = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \frac{{\Delta x}}{2}}}{{\frac{{\Delta x}}{2}}} \cdot \mathop {\lim }\limits_{\Delta x \to 0} \sin \left( {x + \frac{{\Delta x}}{2}} \right) = - \sin x
因此 (\cos x)' = - \sin x
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4. 对数函数 y = f(x) = {\log _a}x\;(a > 0,a \ne 1)
y = {\log _a}x,\;0 < x < + \infty
\forall x \in (0, + \infty ) ,设自变量 x 有增量 {\Delta x} ,函数对应的增量:
\Delta y = {\log _a}(x + \Delta x) - {\log _a}x = {\log _a}\left( {\frac{{x + \Delta x}}{x}} \right) = {\log _a}\left( {1 + \frac{{\Delta x}}{x}} \right)
因此 \frac{{\Delta y}}{{\Delta x}} = \frac{1}{{\Delta x}}{\log _a}\left( {1 + \frac{{\Delta x}}{x}} \right) = \frac{1}{x} \cdot \frac{x}{{\Delta x}}{\log _a}\left( {1 + \frac{{\Delta x}}{x}} \right) = \frac{1}{x}{\log _a}{\left( {1 + \frac{{\Delta x}}{x}} \right)^{\frac{x}{{\Delta x}}}}
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因此 \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \left[ {\frac{1}{x}{{\log }_a}{{\left( {1 + \frac{{\Delta x}}{x}} \right)}^{\frac{x}{{\Delta x}}}}} \right] = \frac{1}{x} \cdot \mathop {\lim }\limits_{\Delta x \to 0} \left[ {{{\log }_a}{{\left( {1 + \frac{{\Delta x}}{x}} \right)}^{\frac{x}{{\Delta x}}}}} \right]
 = \frac{1}{x} \cdot {\log _a}\left[ {\mathop {\lim }\limits_{\Delta x \to 0} {{\left( {1 + \frac{{\Delta x}}{x}} \right)}^{\frac{x}{{\Delta x}}}}} \right] = \frac{1}{x} \cdot {\log _a}e = \frac{1}{x} \cdot \frac{1}{{\ln a}} = \frac{1}{{x\ln a}}
(注: \mathop {\lim }\limits_{\Delta x \to 0} {\left( {1 + \frac{{\Delta x}}{x}} \right)^{\frac{x}{{\Delta x}}}} = e 是重要极限之一,即 \mathop {\lim }\limits_{\alpha \to 0} {(1 + \alpha )^{\frac{1}{\alpha }}} = e
因此 {\left( {{{\log }_a}x} \right)^\prime } = \frac{1}{{x\ln a}}
(\ln x)' = \frac{1}{{x\ln e}} = \frac{1}{x}
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本课推导的常用的导数公式总结:
{\left( C \right)^\prime } = 0
{\left( {{x^\alpha }} \right)^\prime } = \alpha {x^{\alpha - 1}}
{\left( {\sin x} \right)^\prime } = \cos x
{\left( {\cos x} \right)^\prime } = - \sin x
{\left( {{{\log }_a}x} \right)^\prime } = \frac{1}{{x\ln a}}
{\left( {\ln x} \right)^\prime } = \frac{1}{x}

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(第24课完)
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