初等函数求导

匿名 (未验证) 提交于 2019-12-03 00:39:02

remoon

  • 求C的导数
    \[f(x) = C\]
    \[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
    \[=\lim_{h \to 0} \frac{C-C}{h}\]
    \[=\lim_{h \to 0} \frac{0}{h}\]
    \[=0\]

  • 幂函数导数
    \[f(x) = x^n(n \in R)\]
    \[\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]
    \[=\lim_{h \to 0} \frac{(x+h)^n - x^n}{h}\]
    \[=\lim_{h \to 0} x^{n-1} \frac{(1+\frac{h}{x})^n - 1}{\frac{h}{x}}\]
    \[=\lim_{h \to 0} x^{n-1} \frac{n\frac{h}{x}}{\frac{h}{x}}\]
    \[=\lim_{h \to 0} nx^{n-1}\]
    \[=nx^{n-1}\]

  • 正弦函数导数
    \[f(x) = \sin x\]
    \[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
    \[=\lim_{h \to 0} \frac{sin(x+h) - sin(x)}{h}\]
    \[=\lim_{h \to 0} \frac{2cos(x+\frac{h}{2})sin(\frac{h}{2})}{h}\]
    \[=\lim_{h \to 0} \frac{cos(x+\frac{h}{2})sin(\frac{h}{2})}{\frac{h}{2}}\]
    \[=\lim_{h \to 0} cos(x+\frac{h}{2})\]
    \[=cos(x)\]

  • 余弦函数导数
    \[f(x) = \cos x\]
    \[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
    \[=\lim_{h \to 0} \frac{cos(x+h) - cos(x)}{h}\]
    \[=\lim_{h \to 0} \frac{-2sin(x+\frac{h}{2})sin(\frac{h}{2})}{h}\]
    \[=\lim_{h \to 0} -\frac{sin(x+\frac{h}{2})sin(\frac{h}{2})}{\frac{h}{2}}\]
    \[=\lim_{h \to 0} -sin(x+\frac{h}{2})\]
    \[=-sin(x)\]

  • 指数函数导数
    \[f(x) = a^x\]
    \[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
    \[=\lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\]
    \[=\lim_{h \to 0} \frac{a^x(a^h - 1)}{h}\]
    \[=\lim_{h \to 0} a^x \frac{a^h - 1}{h}\]
    \[t = a^h - 1\]
    那么有\[h = \log_a t+1\]
    \(h\to0\)ʱ\(t\to0\)
    \[=a^x\lim_{t\to0} \frac{t}{\log_a t+1}\]
    \[=a^x \lim_{t\to0} \frac{1}{\frac{log_a t+1}{t}}\]
    \[=a^x \lim_{t \to 0} \frac{1}{log_a (t+1)^{\frac{1}{t}}}\]
    \[=a^x \lim_{t\to0} \frac{1}{log_a e}\]
    \[=a^x \lim_{t\to0} log_e a\]
    \[=a^x In a\]

  • 对数函数导数
    \[f(x) = log_a x\]
    \[\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]
    \[=\lim_{h \to 0} \frac{log_a (x+h) - log_a x}{h}\]
    \[=\lim_{h \to 0} \frac{1}{h} * log_a \frac{x+h}{x}\]
    \[=\lim_{h \to 0} \frac{1}{x} * \frac{x}{h} * log_a(1+\frac{h}{x})\]
    \[=\lim_{h \to 0} \frac{1}{x} * log_a (1+\frac{h}{x})^{\frac{x}{h}}\]
    \[=\lim_{h \to 0} \frac{1}{x} * log_a e\]
    \[=\lim_{h \to 0} \frac{1}{x} * \frac{1}{In a}\]
    \[= \frac{1}{x In a}\]

原文:https://www.cnblogs.com/reverymoon/p/9255702.html

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