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有人要我解一下隐式微分

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tan(x/y)=x+y

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我们已经讲过几个隐式微分的视频

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但这个对于第一年学微积分的同学来说

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还是很让人头疼的

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因此 我想再举个例子

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多看几题没有坏处

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因此 我们看看这个

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因此 要用隐式微分法做这个题

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我们只要在方程两边加上

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微分算子d/dx

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这个对x求导

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左边对x求导 相当于

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右边对x求导

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右边非常简单

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左边有点复杂

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因此 我们在这边看一下

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把左边稍微换个写法

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换个颜色

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假设 a=tanb

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假设b=x/y

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然后 很明显 a就是这个

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我的意思是把b带到这里

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可以把这整个写成a

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因此 如果对x求导

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这是我们要做的

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两边都对它求导

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这个是da/dx

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等于 x对x求导

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这个非常简单 等于1

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加上y对x求导

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我这么写

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导数算子

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dy/dx

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这就是我们做的

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运用y的导数算子

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我们不知道它等于多少 我们要求解它

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但是 很明显 我不能这么放着

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不能把da/dx就这么放着

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我们刚刚算出a a等于这个 对吧？

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a=tanb b=y/x

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我这么写的原因是 我想让你们看看

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当求这个的导数时

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可以通过链式法则求解

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这个不是什么新的知识

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因此 导数…我把链式法则写在这儿

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da/dx 等于

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da/db 乘以

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db/dx

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这是链式法则 很容易记住

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因为 这两个db抵消

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就只剩下da/dx

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如果把这个看成普通的分数的话

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因此 da/db等于多少?

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等于1除以(cosb)^2

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如果你们不记得了

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实际上不难证明

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只要把这个写成sinb/cosb

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但是这个是很多人都知道的

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一个三角函数的导数

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我们已经做过一个视频 证明过这个了

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有的书也把这个写成（secb）^2

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但是我们知道sec方相当于

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（1/cos）^2

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我喜欢把它化成基本三角函数

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或正割函数 比如sec和csc

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然后 db/dx等于多少？

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这个很有意思

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我把b重写一下 实际上

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写成b=（xy）^-1

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因此 db/dx

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我们可以利用链式法则

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可以说 我写一下

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db/dx等于

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（xy）^-1的导数

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因此 x的导数等于1

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乘以y^(-1) 加上 导数…

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因此我写一下

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加上d（y^-1)/dx

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乘以第一项 乘以x

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因此 这里

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很显然 我还没完全化简

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我还没算出这个等于多少

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我只是简单地利用了乘积法则

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把第一项求导

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x的导数等于1 乘以第二项 加上

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第二项的导数 乘以第一项

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这是我们做的

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因此 db/dx

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就是这个

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因此 它等于…用黄色 因此 乘以

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我还是用蓝色吧 因为前面就是用的蓝色

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这是蓝的

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db/dx等于y^(-1)

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或者1/y 加上d（1/y）dx 乘以x

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我把它写下来

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因此 我们刚刚算出 或者说差不多算出

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da/dx等于多少

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我们把它代进去

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但是还没做完

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1/y的对x求导等于多少

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再利用链式法则

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我讲清楚一点

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我知道这个看起来有点复杂

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但我想这是说的通的

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我们令c=1/y

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因此 dc/dx 利用链式法则

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等于 dc/dy

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乘以dy/dx

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dc/dy等于多少？

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它相当于

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可以把这个写成y^(-1)

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因此 等于-y^（-2）

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就是这个

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这个是这个

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我们不知道dy/dx等于多少

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这是我们要求解的

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因此 等于它乘以dy/dx

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这是从链式法则得到的

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因此 这个

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这是 这个对x的导数

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也就相当于dc/dx

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因此 可以把这个写在这儿

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可以把这个写成-y^（-2）dy/dx

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然后 当然要乘以x

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然后加上（1/y）

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这整个乘以1/（cosb）^2

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现在我们化简得差不多了

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希望链式法则不会把你们弄糊涂了

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因为我真的想说 所有的

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隐式微分 这些dy/dx不需要…

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这个不需要记住

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它们是从链式法则推导来的

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因此我们解出da/dx 等于这个式子

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我写一下 它等于1/（cosb）^2

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b等于多少？它等于x/y

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(cos(x/y))^2 乘以这些东西

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乘以这整个

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1/y加上 或者应该说减去

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把这个整理一下 等于x/y^2 乘以dy/dx

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然后 它等于右边

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等于1加dy/dx

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现在 我们要做的就是求解dy/dx

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因此 我们回顾一下 怎么得到的这个

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看看链式法则的每一步

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但是 如果你们掌握了

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你们可以直接到这一步

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你们可以这么考虑…

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右边你们知道

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x的导数等于1 y对x求导

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即dy/dx

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左边

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对这整个对x/y求导

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因此 tan的导数等于1/cos^2

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因此 等于1/(cos(x/y))^2

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把它乘以x/y

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对x的导数

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x/y对x求导 就是…

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有点复杂

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这就是为什么我们要写在边上

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x的导数 等于1 乘以1/y

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也就是这一项 加上1/y对x的导数

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即 -1/（y^2）乘以dy/dx

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这是根据链式法则得来的 乘以dx

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这就是为什么我们要写在边上

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这样就不容易粗心 犯错

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如果你们熟练之后

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你们可以直接在脑子里面做 当然

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它等于右边

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因此 从这里开始只是单纯的代数了

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只要解出dy/dx

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因此 先把等式两边乘以

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(cos（x/y）)^2

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因此 很显然 这边变成1

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左边变成1/y减去

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x/(y^2) 乘以dy/dx 等于…

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把方程两边都同时乘以

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这里的分母

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等于（cosx/y）^2 加上

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(cosx/y)^2 乘以dy/dx

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现在要怎么做

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可以把方程两边

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都减去（cosx/y）^2

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得到1/y 减去（cosx/y）^2

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我只是把方程两边都减去这个

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从本质上说是把它移到了左边

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我要做的是…

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把dy/dx项从非dy/dx项中分离出来

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因此 我要把dy/dx项移到右边

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因此 把两边都加上x/(y^2)

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然后这个等于x/y

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把它用原来的颜色写

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稍微有点不同的颜色

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因此 等于x/(y^2) dy/dx用橙色

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dy/dx 然后这一项

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加上(cosx/y)^2 乘以dy/dx

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我想差不多能算出来了

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我们把右边的dy/dx分离出来

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因此 这个等于dy/dx 乘以x/(y^2) 加上

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(cosx/y)^2

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它等于这里这个

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等于1/y 减去（cosx/y）^2

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现在解dy/dx 只要把方程两边都除以

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这个式子

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然后得到什么？

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得到 如果两边都除以那个式子

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得到1/y减去（cosx/y）^2 除以

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这整个式子

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x/(y^2) 加上(cosx/y)^2 等于dy/dx

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然后就做完了

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我们只要多次利用链式法则 就能

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解出隐式微分 tan(y/x)等于x+y

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实际上难的部分是得到这一步

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这一步之后就只是单纯的代数了

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只要解出dy/dx

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然后 得到结果在这儿

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不管怎样 希望这对你们有帮助