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