导数

\[\begin{aligned} & [u(x)\cdot v(x)]'\\ = & \lim_{\Delta x \to 0} \frac {u(x+\Delta x)v(x+\Delta x) - u(x)v(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {u(x+\Delta x)v(x+\Delta x) - u(x+\Delta x)v(x)+u(x+\Delta x)v(x) - u(x)v(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {u(x+\Delta x)[v(x+\Delta x) - v(x)]} {\Delta x} + \frac {v(x)[u(x+\Delta x) - v(x)]}{\Delta x}\\ = & \lim_{\Delta x \to 0} v(x)\frac {u(x+\Delta x) - u(x)}{\Delta x} + u(x)\frac {v(x+\Delta x) - v(x)}{\Delta x}\\ = & v(x)u'(x) + u(x)v'(x)\\ = & u'(x)v(x) + v'(x)u(x) \end{aligned} \]


\[\begin{aligned} & [u(x)+v(x)]'\\ = & \lim_{\Delta x \to 0} \frac {u(x+\Delta x) + v(x+\Delta x) - u(x)-v(x)}{\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {u(x+\Delta x) - u(x)} {\Delta x} + \frac {v(x+\Delta x)-v(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {u(x+\Delta x) - u(x)} {\Delta x} + \lim_{\Delta x \to 0} \frac {v(x+\Delta x)-v(x)} {\Delta x}\\ = & u'(x) + v'(x) \end{aligned} \]


\(f(x) = C\)\(C\) 为常数)的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(x+\Delta x) - f(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {C - C} {\Delta x}\\ = & 0 \end{aligned} \]


\(f(x) = x^n(n\in \mathbb{N_*})\) 的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(x+\Delta x)-f(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {(x+\Delta x)^n-x^n} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {nx^{x-1}\Delta x + \frac {n(n-1)}{2}x^{n-2}{\Delta x}^2+ \cdots + {\Delta x} ^ n} {\Delta x}\\ = & \lim_{\Delta x \to 0} {nx^{x-1} + \frac {n(n-1)}{2}x^{n-2}\Delta x + \cdots + {\Delta x} ^ {n-1}}\\ = & nx^{n-1} \end{aligned} \]


\(f(x) = x^\mu(\mu\in \mathbb R)\) 的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(x+\Delta x) - f(x)}{\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {(x+\Delta x)^\mu - x^\mu} {\Delta x}\\ = & \lim_{\Delta x \to 0} x^{\mu-1}\frac {(1+\frac {\Delta x} x)^\mu - 1} {\frac {\Delta x} x}\\ & \because (1+x)^{\alpha}-1\sim \alpha x~~~(x \to 0)&第三个等价无穷小公式\\ = & \lim_{\Delta x \to 0} \mu x^{\mu-1}\\ = & \mu x^{\mu-1} \end{aligned} \]


\(f(x) = \sin x\) 的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(x+\Delta x) - f(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {\sin(x+\Delta x) - \sin x}{\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {\sin x\cos {\Delta x} + \cos x\sin \Delta x - \sin x} {\Delta x} \\ = & \lim_{\Delta x \to 0} \frac {\cos x\sin{\Delta x}}{\Delta x}\\ & \because \sin x\sim x~~(x\to 0)\\ = & \cos x \end{aligned} \]


\(f(x) = a^x(a > 0, a \neq 1)\) 的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(x+\Delta x)-f(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {a^{x+\Delta x} - a^x} {\Delta x}\\ = & a^x\lim_{\Delta x \to 0} \frac {a^{\Delta x}-1} {\Delta x}\\ & \because \lim_{x \to 0} \frac {a^x -1} x = \ln a\\ = & a^x\ln a \end{aligned} \]

特殊地,当 \(a = e\) 时,因为 \(\ln e = 1\),所以:

\[(e^x)' = e^x \]


\(f(x) = \log_a x(a > 0,a\neq 1)\) 的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(x+\Delta x)-f(x)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {\log_a(x + \Delta x) - \log_a x} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {\log_a \frac {x+\Delta x} x }{\Delta x}\\ = & \lim_{\Delta x \to 0} \frac 1 x \cdot \frac x {\Delta x} \log_a(1+\frac {\Delta x} x)\\ = & \frac 1 x \lim_{\Delta x \to 0} \frac {\log_a(1+\frac {\Delta x} x)}{\frac x {\Delta x}}\\ & \because \lim_{\Delta x \to 0} \frac {\log_a(1+x)} x = \frac 1 {\ln a}\\ = & \frac 1 {x\ln a} \end{aligned} \]

特殊地,当 \(a = e\) 时,我们可以得到自然对数函数的导数公式:

\[(\ln x)' = \frac 1 x \]


\(f(x) = |x|\)\(x = 0\) 处的导数

\[\begin{aligned} & f'(x)\\ = & \lim_{\Delta x \to 0} \frac {f(0 + \Delta x) - f(0)} {\Delta x}\\ = & \lim_{\Delta x \to 0} \frac {|\Delta x|} {\Delta x}\\ \end{aligned} \]

\[\therefore f'(x) = \begin{cases} -1 & h < 0\\ 1 & h > 0 \end{cases} \]

由此可得,\(f(x)\)\(x = 0\) 处不可导

posted @ 2024-03-06 10:59  ricky_lin  阅读(8)  评论(0编辑  收藏  举报