Section03_求导类型

型一 定义求导

• 例1 $f(x)=\begin{cases}\ln(1-2x),&x<0;\\\sin ax, & x\ge 0;\end{cases}$ 已知$\exists\ f'(0)$，求 $a$
  • 解 $$\begin{array}{ll} \because & \exists\ f'(0) \\ \therefore & \exists\ \displaystyle{\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}} \\ \therefore & \displaystyle{\lim_{x\to 0^{-}}\frac{f(x)-0}{x-0} = \lim_{x\to 0^{+}}\frac{f(x)-0}{x-0}} \\ & \displaystyle{\lim_{x\to0^{-}}\frac{\ln(1-2x)}{x} = \lim_{x\to 0^{+}}\frac{\sin ax}{x}} \\ \therefore & a=-2 \\ \end{array}$$
• 例2 $f(x)$ 连续，$\displaystyle{\lim_{x\to 1}\frac{f(x)-1}{x-1}}=2$ 求 $\displaystyle{\lim_{x\to1}\frac{f^{2}(x)-1}{x^{2}-1}}$
  • 解 $$\begin{array}{ll} \because & \displaystyle{\lim_{x\to 1}}\frac{f(x)-1}{x-1} = 2 \\ \therefore & f(1) = 1 \text{ and } f'(1) = 2\\ \therefore & \displaystyle{\lim_{x\to 1}\frac{f^{2}(x)-1}{x^{2}-1}} \\ & = \displaystyle{\lim_{x\to 1}\frac{f(x)+1}{x+1}\frac{f(x)-1}{x-1}} \\ & = \displaystyle{\lim_{x\to 1}\frac{f(x)-1}{x-1}=2}\\ \end{array}$$

型二 显函数求导

• 例1 $\displaystyle{y =e^{\sin \frac{1}{x}} + \frac{1+x}{1-x}e^{\sqrt{x}}}$ 求 $y'$
  • 解 $$\begin{split} y' = e^{\sin \frac{1}{x}}\cos \frac{1}{x}\bigg(-\frac{1}{x^{2}}\bigg) + \frac{2}{(1-x)^{2}}e^{\sqrt{x}}+\frac{1+x}{1-x}e^{\sqrt{x}}\frac{1}{2\sqrt{x}} \end{split}$$
• 例2 $y=(\sin x)^{x}$ 求 $y'$
  • 解 $$\begin{split} y &= (\sin x)^{x} = e^{x\ln(\sin x)}\\ y' &= e^{x\sin x}\bigg(\ln(\sin x)+ x\frac{\cos x}{\sin x}\bigg) \\ &= (\sin x)^{x}(\ln(\sin x) + x\cot x)\\ \end{split}$$

型三 隐函数求导

$F(x,y)=0$ 被称为隐函数，$F(x,y)=0\Rightarrow y=f(x)$ 被称为隐函数显式化的过程

• 例1 $\displaystyle{e^{xy}=x^{2}+y+1}$ 求 $\displaystyle{\frac{dy}{dx}}$
  • 解 $$\begin{array}{ll} \because & e^{xy} = x^{2} +y + 1 \\ \therefore & e^{xy}(y+x\frac{dy}{dx}) = 2x+\frac{dy}{dx} \\ \therefore & (xe^{xy}-1)\frac{dy}{dx} = 2x-ye^{xy} \\ & \frac{dy}{dx} = \frac{2x-ye^{xy}}{xe^{xy}-1} \end{array}$$
• 例2 $e^{xy}+\tan(xy) = y$ 求 $\left.\frac{dy}{dx}\right\vert_{x=0}$
  • 解 $$\begin{array}{ll} \because & x=0 \\ \therefore & y = 1\\ \because & e^{xy}+\tan(xy)=y \\ \therefore & e^{xy}(y+x\frac{dy}{dx}) + \sec^{2}(xy)(y+x\frac{dy}{dx}) = \frac{dy}{dx} \\ \because & x= 0, y=1 \\ \therefore & \left.\frac{dy}{dx}\right\vert_{x=0} = 2 \\ \end{array}$$
• 例3 $2^{xy}+x=y$ 求 $\left.dy\right\vert_{x=0}$
  • 解 $$\begin{array}{ll} \because & x=0 \\ \therefore & y=1 \\ \because & 2^{xy}+x =y \\ \therefore & 2^{xy}\ln{2}\cdot(y+x\frac{dy}{dx}) + 1=\frac{dy}{dx} \\ \therefore & \left.\frac{dy}{dx}\right\vert_{x=0} = \ln 2 + 1\\ & \left.dy\right\vert_{x=0} = (\ln{2}+1)\cdot dx \end{array}$$

型四 参数方程确定的函数 （不要求）

定理 $\phi(t),\psi(t)$ 可导且 $\phi'(t)\ne 0$ 由 $\displaystyle{\begin{cases}x= \phi(t)\\y=\psi(t)\end{cases}}$ 可得 $y=f(x)$ 且 $\displaystyle{\frac{dy}{dx} = \frac{dy/dt}{dx/dt}} = \frac{\phi'(t)}{\psi'(t)}$

• 例1 设$\displaystyle{\begin{cases}x= t-\ln(1+t)\\y=t^{2}\end{cases}}$ 求$\displaystyle{\frac{dy}{dx},\ \frac{d^{2}y}{dx^{2}}}$
  • 解 $$\begin{split} & \frac{dy}{dx} = \frac{(t^{2})'}{[t-\ln(1+t)]'} = \frac{2t}{1-\frac{1}{1+t}} = 2(t+1) \\ & \frac{d^{2}y}{dx^{2}} = \frac{[2(t+1)]'}{[t-\ln(1+t)]'} = \frac{2}{1-\frac{1}{1+t}} = \frac{2(1+t)}{t} \end{split}$$
• 例2 $\displaystyle{\begin{cases}x=\arctan x\\y=\ln(1+t^{2})\end{cases}}$ 求$\displaystyle{\frac{dy}{dx},\ \frac{d^{2}y}{dx^{2}}}$
  • 解 $$\begin{split} & \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2t}{1+t^{2}}}{\frac{1}{1+t^{2}}} = 2t \\ & \frac{d^{2}y}{dx^{2}} = \frac{dy/dt}{dx/dt} = \frac{2}{\frac{1}{1+t^2}} = 2(1+t^{2}) \end{split}$$

型五 分段函数的求导

• 例1 $\displaystyle{f(x)=\begin{cases}e^{2x},&x<0\\ax+b,& x\ge0\end{cases} \text{ and } \exists\ f'(0)}$ 求 $a,b$
  • 解 $$\begin{array}{ll} \because & \exists\ f'(0) \\ \therefore & \displaystyle{\lim_{x\to 0^{-}}f(x) = f(0)} \text{ and } f_{-}'(x) = f_{+}'(x) \\ \therefore & \begin{cases} \displaystyle{\lim_{x\to 0^{-}}e^{2x} = b} \\ \displaystyle{\lim_{x\to 0^{-}}\frac{e^{2x}-e^{0}}{x-0} = \lim_{x\to 0^{-}}\frac{ax+b-b}{x-0}\Rightarrow a= 2} \end{cases}\\ \therefore & a=2 \text{ and } b=1 \\ \end{array}$$
• 例2 $\displaystyle{f(x)=\begin{cases}\frac{\sin x}{x},& x\ne 0\\1, & x=0\end{cases}}$
  1. 求 $f'(x)$ $$\begin{array}{ll} & \text{When } x\ne 0 \\ & f'(x) = (\frac{\sin x}{x})' = \frac{x\cos x-\sin x}{x^{2}} \\ & \text{When } x = 0 \\ & f'(x) = \displaystyle{\lim_{x\to 0}\frac{f(x) -f(0)}{x-0}} = \displaystyle{\lim_{x\to 0}\frac{\sin x - x}{x^{2}}} = 0 \\ \therefore & f'(x) = \begin{cases} \frac{x\cos x - \sin x}{x^{2}}, & x\neq 0 \\ 0, & x=0 \end{cases} \\ \end{array}$$
  2. 讨论 $f'(x)$ 在 $x=0$ 处的连续性 $$\begin{array}{ll} \because & \\ & f'(x) = \begin{cases} \frac{x\cos x - \sin x}{x^{2}}, & x\neq 0 \\ 0, & x=0 \end{cases} \\ \therefore & \displaystyle{\lim_{x\to 0} f'(x) = \lim_{x\to 0} \frac{x \cos x- \sin x}{x^{2}} =\lim_{x\to 0}\frac{-x\sin x}{2x} = 0 }=f'(0)\\ \therefore & f'(x) \text{在}x=0 \text{处连续} \\ \end{array}$$

型六 高阶导数

常用解题方法

1. 归纳法

例题

• 例1 $\displaystyle{y=\frac{1}{2x+3}}$ 求 $y^{(n)}$
  • 解 $$\begin{array}{ll} \because & y = (2x+3)^{-1} \\ \therefore & y' = (-1)\cdot(2x+3)^{-2}\cdot2 \\ & y'' = (-1)(-2)\cdot(2x+3)^{-3}\cdot2^{2} \\ & \cdots \\ & y^{(n)} = (-1)(-2)\cdots(-n)\cdot(2x+3)^{-(n+1)}\cdot 2^{n} \\ & = (-1)^{n}n!\cdot(2x+3)^{-(n+1)}\cdot 2^{n} \\ & = \frac{(-1)^{n} \cdot n!\cdot 2^{n}}{(2+3)^{n+1}} \end{array}$$
• 例2 $y=\frac{1}{x^{2}-x-2}$ 求 $y^{(n)}$
  • 解 $$\begin{array}{ll} \because & y = \frac{1}{x^{2}-x-2} = \frac{1}{(x+1)(x-2)} = \frac{1}{3}(\frac{1}{x-2}-\frac{1}{x+1}) \\ \therefore & y^{(n)} = \frac{1}{3}\big[\frac{(-1)^{n}\cdot n!}{(x-2)^{n+1}}-\frac{(-1)^{n}\cdot n!}{(x+1)^{n+1}}\big] \\ \end{array}$$
• 例3 $y = e^{x}\cdot \sin x$ 求 $y^{(n)}$
  • 解 $$\begin{array}{ll} \because & y = e^{x}\cdot \sin x \\ \therefore & y' = e^{x}\sin x + e^{x}\cos x \\ \because & \sin x + \cos x = \sqrt{2}\cdot\sin (x + \frac{\pi}{4}) \\ \therefore & y' = \sqrt{2}\cdot e^{x}\sin (x+\frac{\pi}{4}) \\ \therefore & y^{(n)} = (\sqrt{2})^{n}\cdot e^{x}\sin (x+\frac{n\pi}{4}) \\ \end{array}$$