各种导数的计算（4）

各种函数导数的计算

数学第四天

隐函数求导

\[[例题]：由方程e^y+xy-e=0所确定的隐函数的导数\frac {dy}{dx}. \]

\[解：e^{y(x)}+xy(x)-e=0 \]

\[对两边求导：e^{y(x)}\cdot y'(x)+y(x)+y'(x)\cdot x=0 \]

\[\Rightarrow [x+e^{y(x)}]y'=-y \]

\[\Rightarrow y'=\frac{-y}{x+e^y} \]

\[[例题]求方程e^y+xy-e=0所确定的隐函数的导数\frac {dy}{dx} \]

证：微分形式的不变性

\[在e^y+xy-e=0 两边同时微分 \]

\[d(e^y+xy-e)=0 \]

\[\Rightarrow de^y+d(xy)=0 \]

\[\Rightarrow e^y \cdot dy+ydx+xdy=0 \]

\[ \]

\[\Rightarrow (e^y+x)dy=-dy \]

\[\Rightarrow \frac {dy}{dx}=-\frac{y}{x+e^y} \]

反函数求导

\[\frac {dx}{dy}=\frac {1}{\frac {dy}{dx}}=\frac 1 {f'(x)} \]

\[求y=\arcsin x的导数 \]

\[y=\arcsin x \Rightarrow x=\arcsin y \]

\[\frac{dx}{dy}=\frac 1{\frac{dx}{dy}}=\frac1{\cos y} =\frac1{\sqrt{1-\sin ^2y}} \]

\[\Rightarrow (\arcsin x)'=\frac1{\sqrt {1-x^2}} \]

反函数的二阶导数

导数是微商‘

\[\frac {d^2x}{dy^2}=d\Big(\frac {\frac{dx}{dy}}{dy}\Big) =d\frac{d(\frac{dx}{dy})}{dx}\cdot \frac{dx}{dy} \]

\[\Rightarrow \Big[\frac1 {f'(x)}\Big]'\cdot \frac{dx}{dy} \]

\[\Rightarrow \frac{-f''(x)}{[f'(x)]^2}\cdot \frac1{f'(x)} =\frac {-f''(x)}{[f'(x)]^3} \]

分段函数求导

\[[例题]已知f(x)=\begin{cases} x^2\sin \frac1{x},\quad x \neq 0,\\ 0,\quad x=0 \end{cases} 求f'(x) \]

\[解：当x \neq 0时，f'(x)=2x\cdot\sin\frac1x+\cos \frac1x\cdot x^2 \cdot (-\frac1x^2) \]

\[\Rightarrow 2x\sin \frac1x-cos\frac1x \]

\[当x=0时，（一点导数考虑导数定义） \]

\[f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0} =\lim_{x\rightarrow 0}\frac{x^2\sin \frac1x-0}{x} \]

\[\Rightarrow \lim_{x\rightarrow0}x\sin \frac1x (有界函数乘无穷小仍是无穷小)=0 \]

\[\therefore f'(x)=\begin{cases} 2x\sin\frac1x-\cos\frac1x,\quad x\neq 0\\ 0,\quad x=0 \end{cases} \]

幂指函数求导

\[y=u(x)^{v(x)} \]

方法一

\[y=u(x)^{v(x)}=e^{\ln u(x)^{v(x)}} \]

\[=e^{v(x)\ln u(x)} \]

\[[例题]求 y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} (x>4) 的导数 \]

\[\ln y=\ln \Big[\frac{(x-1)(x-2)}{(x-3)(x-4)}\Big]^\frac12 \]

\[=\frac1y\cdot y'=\frac12(\frac1{x-1}+\frac1{x-2}-\frac1{x-3 }-\frac1{x-4}) \]

\[\Rightarrow y'=\frac y2(\frac1{x-1}-\frac1{x-2}-\frac1{x-3} -\frac1{x-4}) \]

由参数方程确定的函数的导数

\[设函数y=f(x)由\begin{cases} x=\varphi (t)\\ y=\Psi(t) \end{cases} 确定，\varphi'(t),\Psi'(t)存在，且\varphi'(t)\neq0,则 \]

\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{\Psi(x)}{\varphi(x)} \]

\[\frac {d^2y}{dx^2}=\frac {d(\frac {dy}{dx})}{dx} =\frac{d\frac{dy}{dx}}{dt}/\frac{dx}{dt} \]

\[= \frac {\frac {\Psi''(t)\varphi'(t)-\Psi'(t)\varphi''(t)} {[\varphi'(t)]^2}}{\varphi'(t)} \]

\[\frac{\Psi''(t)\varphi'(t)-\Psi'(t)\varphi''(t)}{[\varphi'(t)]^3} \]

\[[例题]已知椭圆的参数方程为\begin{cases} x=a\cos t\\ y=b\sin t \end{cases} 求椭圆在t=\frac\pi4 相应点处的切线方程。 \]

\[(\frac{\sqrt 2}{2}a,\frac{\sqrt 2}{2}b) \]

\[解：\frac{dy}{dx}\Big|_{t=\frac \pi4} =\frac{\frac {dy}{dt}}{\frac {dx}{dt}}\Big|_{t=\frac\pi4} \]

\[=\frac {b\cdot \cos t}{-a\sin t}\Big|_{t=\frac\pi4} =-\frac ba \]

\[切线方程为： y-\frac {\sqrt 2}2b=-\frac ba(x-\frac{\sqrt 2}2) \]

\[[例题]:计算由摆线的参数方程\begin{cases}x=a(t-\sin t)\\y=a(1- \cos t)\end{cases}所确定的y=y(x)二阶导数 \]

\[解：\frac {dy}{dx}=\frac {\frac {dy}{dt}}{\frac{dx}{dt}}= \frac {a\sin t}{a(1-\cos t)}=\frac{\sin t}{1-\cos t} \]

\[\frac{d^2y}{dx^2}=\frac {d(\frac {dy}{dx})}{dt}\Big/\frac {dx}{dt}=\frac {\frac{\cos t(1-\cos t)-\sin^2t}{(1-\cos t )^2}}{a(1-\cos t)} \]

\[=\frac {\cos t-\cos^2t-\sin^2t}{a(1-\cos t)^3} \]

\[=\frac{\cos t-1}{a(1-\cos t)^3}=-\frac 1 {a(1-\cos t)^2} \]