常用极限计算方式
\begin{align}
\lim_{x \to 0} \frac{\log_{a}(x+1)}{x} & = \lim_{x \to 0} \log_{a}(x+1)^{\frac{1}{x}} \\
& = \log_{a}e \\
& = \frac{1}{\ln(a)}
\end{align}
\begin{align}
\lim_{x \to 0} \frac{a^x – 1}{x} & = \lim_{x \to 0} \frac{a^x – 1}{\log_a(a^x – 1 + 1)} \\
& = \ln(a)
\end{align}
幂函数导数
\begin{align}
f(x)&=x^\mu (\mu \in R) \\
f^{‘}(x) & = \lim_{h \to 0} \frac{(x+h)^\mu -x^\mu}{h} \\
\end{align}
当 \(x \neq 0\)
\begin{align}
f^{‘}(x) & = \lim_{h \to 0} x^{\mu-1}\frac{(1+\frac{h}{x})^\mu -1}{\frac{h}{x}} \\
& = \frac{1}{\mu}x^{\mu -1}
\end{align}
当 \(x = 0\)
\begin{align}
f^{‘}(x) & = \lim_{h \to 0} \frac{(h)^\mu}{h} \\
& = \frac{1}{\mu}x^{\mu -1}
\end{align}
进一步考虑，当 $\mu \geq 1$ 时
\begin{align}
f^{‘}(x) = 0 = \frac{1}{\mu}x^{\mu -1}
\end{align}
当 \(\mu < 1\) 时 \(f^{‘}(x)\) 不存在
综上，当 \(x \neq 0\) 或 \(\mu \geq 1\) 时 \(f^{‘}(x) = \frac{1}{\mu}x^{\mu – 1}\) 。否则导数不存在
正弦函数导数
\begin{align}
f(x)&=\sin(x) \\
f^{‘}(x) & =\lim_{h \to 0}\frac{\sin(x+h) – \sin(x)}{h} \\
& = \lim_{h \to 0}\frac{1}{h}(\sin(x+\frac{h}{2}+\frac{h}{2}) – \sin(x + \frac{h}{2} – \frac{h}{2})) \\
& = \lim_{h \to 0}\frac{1}{h}(2*\cos(x)*\sin(\frac{h}{2})) \\
& = \cos(x)
\end{align}
指数函数
\begin{align}
f(x) & = a^x (a>0,a \neq 1) \\
f^{‘}(x) & = \lim_{h \to 0}\frac{a^{x+h}-a^x}{h} \\
& = \lim_{h \to 0} a^x \frac{a^h – 1}{h} \\
& = a^x \ln(a)
\end{align}