WEBVTT

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-可微函数x和y由以下方程关联

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Sin(x)+Cos(y)

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将等于2的平方根

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它也告诉我们dx/dt=5

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他也告诉我们要在当y=π/4且x大于0小于π/2时

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求出基于t的y的导数

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所以他告诉了我们基于t的x的导数

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并且我们要找到

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基于t的y的导数

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认为x和y都是基于t的函数的假设是合理的

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所以你甚至可以在这将函数重新列一遍

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你可以将其重新写为sin(x)，其中

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x是t的函数

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加上

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cos(y)，y也是t的函数

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整个式子等于2的平方根

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现在你可能有些困惑

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你未曾将x设为具有3个未知量的函数

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或你未曾将y当做一个除了x外另有变量的函数

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但请记住，x和y仅是自变量

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这可以是f(t),那可以是g(t)

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而不是x(t)或y(t)

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这也许会让你觉得自然些

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不必说如果我们要求出dt

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我们要做的是对这个方程两边的 t 求导

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所以让我们处理它吧

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所以我们将从等式左侧入手

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所以我们将把它与 t 相关联，对 t 取它的导数。

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我们将对t取它的导数

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接着我们将取等式右侧这一常数项

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对t的导数

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所以我们逐一思考这些式子

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所以这是什么呢

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让我换种颜色

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我正在用水笔解的式子

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我该如何写它呢

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所以我取对t的函数

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我有某数的sin值，其自身为关于t的函数

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所以我将在这里应用链式法则

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首先，我将求sinx的导数

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我将之写作sinx(t))

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但为了简化，我将其恢复为原式

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接着我将其与x的导数相乘

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你可以说，对于t的导数

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乘上dx/dt

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这和你之前处理链式法则相比，

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可能有些反常

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之前我们只处理与x或y相关的

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但这就是现在发生的，我将取sin外的

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对于某值的某值的导数，

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在这种情况下，即为x

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接着我将求某值的导数

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在这种情况下，即为x对于t的导数

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我们可以对第二个多项式

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用同样的方式处理

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所以我将求d/dy乘上，

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我猜你将说

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cos(y)

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接着我将之相乘：

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乘上对于t的y的导数

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接着它们整体将等于多少呢

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对于t的常数的导数

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根号2是个常数

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将不会随t的变化而变化

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所以它的导数，其变化率即为0

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好的，现在我们求到了

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所有的值

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所以第一步，sin(x)的导数为

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cos(x)乘x对于t的导数

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我将在这写下

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x对于t导数

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接着我们将有，这里是加上

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y对于t的导数

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所以加上y对于t的导数

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我在这把顺序调换了下

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所以它将提前

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现在，对于y的cos(y)的导数是多少呢？

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其值为-sin(y)

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接着，让我先把sin(y)写在这

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再加个负号

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将这（符号）擦除并换为减号

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而这依旧将等于0

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因此我们可以求出什么呢

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它告诉我们x对于t的导数的值为5

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他就在这告诉了我们

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所以这将等于5

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我们想求出对于t的y的导数的值

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他告诉了我们y的值，π/4

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这里，y是π/4，所以写下π/4

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所以我们看一下，我们得求出它

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我们仍有两个未知量

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我们不知道x的值，我们也不知道

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y对于t的导数值

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那就是我们所需求的值

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所以x的值是多少呢？

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当y等于π/4时，x的值是多少呢

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好的，为了求出它，

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我们将回到原式

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所以当y等于π/4时，你会得到

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让我把它写下

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sin(x)，加上

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cos(π/4),

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（cos(π/4)）的值为二分之根号二

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cos(π/4)

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我们可以转化为我们（熟悉的角度）或用单位圆解决

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我们正处于第一象限中，

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如果我们以角度制思考，它的值为45°

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它将等于二分之根号二

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我们可从等式两侧同时减去二分之根号二，

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我们将的值

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sin(x)的值为，好的，如果你从根号二中减去

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二分之根号二，

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你将减去它的一半

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所以你仍留有它的一半

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所以就是二分之根号二

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所以，x的值是多少呢，我将取它的sin值

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记住，这是在角度制中

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如果我们用单位圆思考，它将在第一象限。

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在这情况下x将是一个角，

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就在这

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所以那将是又一次的π/4

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所以x的值为π/4

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当y等于π/4

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