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初等函数导数证明

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
  \[f(x) = a^x\]
  \[\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
  \[=\lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\]
  \[=\lim_{h \to 0} \frac{a^x(a^h - 1)}{h}\]
  \[=\lim_{h \to 0} a^x \frac{a^h - 1}{h}\]
  令\[t = a^h - 1\]
  那么有\[h = \log_a t+1\]
  当\(h\to0\)时\(t\to0\)
  \[=a^x\lim_{t\to0} \frac{t}{\log_a t+1}\]
  \[=a^x \lim_{t\to0} \frac{1}{\frac{log_a t+1}{t}}\]
  \[=a^x \lim_{t \to 0} \frac{1}{log_a (t+1)^{\frac{1}{t}}}\]
  \[=a^x \lim_{t\to0} \frac{1}{log_a e}\]
  \[=a^x \lim_{t\to0} log_e a\]
  \[=a^x In a\]

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

初等函数求导

标签：求导   spl   play   导数   证明   strong   splay   正弦   isp   

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