Implicit Differentiation · 隐函数求导
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| implicit differentiation/ɪmˈplɪsɪt ˌdɪfəˌrenʃɪˈeɪʃn/ | 隐函数求导 | yǐn hán shù qiú dǎo |
| implicit/ɪmˈplɪsɪt/ | 隐式 | yǐn shì |
When you can't solve for $y$
- The circle $x^2+y^2=25$ defines $y$ implicitly 隐式 — you can't write it as a single $y=f(x)$.
- Yet the curve still has a slope at each point. How do we get $\tfrac{dy}{dx}$ without solving for $y$?
- The trick: differentiate both sides with respect to $x$, treating $y$ as a hidden function of $x$.
- This is implicit differentiation 隐函数求导.
当你无法解出 $y$ 时
- 圆 $x^2+y^2=25$ 隐式地定义了 $y$——你无法把它写成单个 $y=f(x)$。
- 可是曲线在每点仍有斜率。不解出 $y$,我们怎么求 $\tfrac{dy}{dx}$?
- 诀窍:对两边关于 $x$ 求导,把 $y$ 当作 $x$ 的隐藏函数。
- 这就是隐函数求导。
An equation like $x^2+y^2=25$ defines $y$ ____, since it is not solved for $y$. · 像 $x^2+y^2=25$ 这样的方程定义了 $y$ ____，因为它未解出 $y$。
That is why we use implicit differentiation. · 这就是我们使用隐函数求导的原因。
Every $y$ term triggers the chain rule
- Because $y$ secretly depends on $x$, differentiating a $y$-term needs the chain rule.
- $\dfrac{d}{dx}[y^2]=2y\cdot\dfrac{dy}{dx}$ — the extra $\tfrac{dy}{dx}$ is the inner derivative.
- $\dfrac{d}{dx}[x^2]=2x$ as usual (no $y$, no extra factor).
- So each time you differentiate a term containing $y$, tack on a $\tfrac{dy}{dx}$.
每个 $y$ 项都触发链式法则
- 因为 $y$ 暗地里依赖 $x$,对 $y$ 项求导需要链式法则。
- $\dfrac{d}{dx}[y^2]=2y\cdot\dfrac{dy}{dx}$——那个多出的 $\tfrac{dy}{dx}$ 是内层导数。
- $\dfrac{d}{dx}[x^2]=2x$ 照旧(没有 $y$,没有额外因子)。
- 所以每次对含 $y$ 的项求导,都要补上一个 $\tfrac{dy}{dx}$。
Differentiating with respect to $x$, $\dfrac{d}{dx}[y^2]=$ · 对 $x$ 求导，$\dfrac{d}{dx}[y^2]=$
Chain rule: $2y$ times the inner derivative $\tfrac{dy}{dx}$. · 链式法则：$2y$ 乘以内层导数 $\tfrac{dy}{dx}$。
Solve for $\tfrac{dy}{dx}$
- After differentiating, you get an equation with $\tfrac{dy}{dx}$ scattered around.
- Collect all $\tfrac{dy}{dx}$ terms on one side, everything else on the other.
- Factor out $\tfrac{dy}{dx}$ and divide to isolate it.
- The result usually depends on both $x$ and $y$ — that's expected for a curve.
解出 $\tfrac{dy}{dx}$
- 求导后,你得到一个 $\tfrac{dy}{dx}$ 散落各处的方程。
- 把所有 $\tfrac{dy}{dx}$ 项收集到一边,其余的放另一边。
- 提取 $\tfrac{dy}{dx}$ 并相除,把它孤立出来。
- 结果通常同时依赖 $x$ 和 $y$——对一条曲线这是意料之中的。
A tangent slope on a curve · 曲线上的一条切线斜率
y = √(25 − x²) (upper semicircle) · y = √(25 − x²) （上半圆）
Even a curve you can't solve for $y$ has a tangent at each point — implicit differentiation finds that slope. · 即使无法解出 $y$ 的曲线，每一点也有切线——隐函数求导找到该斜率。
Put the implicit-differentiation steps in order. · 将隐函数求导步骤排序。
Differentiate, collect, solve, then evaluate. · 求导、移项、求解，然后代入求值。
Evaluate at a point
- To get an actual slope, plug a point $(x,y)$ on the curve into your $\tfrac{dy}{dx}$ formula.
- For $x^2+y^2=25$: differentiating gives $2x+2y\tfrac{dy}{dx}=0$, so $\tfrac{dy}{dx}=-\dfrac{x}{y}$.
- At $(3,4)$: slope $=-\tfrac34$. At $(3,-4)$: slope $=+\tfrac34$.
- The same $x$ can give different slopes — because the curve has two $y$-branches.
在某点求值
- 要得到实际斜率,把曲线上的一个点 $(x,y)$ 代入你的 $\tfrac{dy}{dx}$ 公式。
- 对 $x^2+y^2=25$:求导得 $2x+2y\tfrac{dy}{dx}=0$,所以 $\tfrac{dy}{dx}=-\dfrac{x}{y}$。
- 在 $(3,4)$:斜率 $=-\tfrac34$。在 $(3,-4)$:斜率 $=+\tfrac34$。
- 同一个 $x$ 可以给出不同斜率——因为曲线有两条 $y$ 分支。
For · 支持 $x^2+y^2=25$, $\dfrac{dy}{dx}=-\dfrac{x}{y}$. Find the slope at $(3,4)$ (a decimal). · 对于 $x^2+y^2=25$，$\dfrac{dy}{dx}=-\dfrac{x}{y}$。求 $(3,4)$ 处的斜率（小数形式）。
$-\tfrac{3}{4}=-0.75$.
The factor $\tfrac{dy}{dx}$ is added to every term, including pure $x$ terms like $x^2$. · 因子 $\tfrac{dy}{dx}$ 加到每一项上，包括纯 $x$ 项如 $x^2$。
Only terms containing $y$ get the $\tfrac{dy}{dx}$ factor. · 只有包含 $y$ 的项才获得 $\tfrac{dy}{dx}$ 因子。
Differentiating $xy$ with respect to $x$ gives... · 对 $xy$ 关于 $x$ 求导得到……
Product rule: $\frac{d}{dx}[xy]=1\cdot y+x\cdot\tfrac{dy}{dx}$. · 乘积法则：$\frac{d}{dx}[xy]=1\cdot y+x\cdot\tfrac{dy}{dx}$。
The chain-rule factor $\tfrac{dy}{dx}$ appears only on terms containing $y$. $\frac{d}{dx}[x^3]=3x^2$ (no factor), but $\frac{d}{dx}[y^3]=3y^2\tfrac{dy}{dx}$. Forgetting the $\tfrac{dy}{dx}$ on $y$-terms is the defining mistake of implicit differentiation. And a product like $xy$ needs the product rule: $\frac{d}{dx}[xy]=y+x\tfrac{dy}{dx}$.
链式法则因子 $\tfrac{dy}{dx}$ 只出现在含 $y$ 的项上。$\frac{d}{dx}[x^3]=3x^2$(无因子),但 $\frac{d}{dx}[y^3]=3y^2\tfrac{dy}{dx}$。在 $y$ 项上忘掉 $\tfrac{dy}{dx}$ 是隐函数求导的标志性错误。而像 $xy$ 这样的乘积需要乘积法则:$\frac{d}{dx}[xy]=y+x\tfrac{dy}{dx}$。
Find $\tfrac{dy}{dx}$ for $x^2+y^2=25$ at the point $(3,4)$.
- Differentiate: $2x+2y\dfrac{dy}{dx}=0$.
- Solve: $\dfrac{dy}{dx}=-\dfrac{2x}{2y}=-\dfrac{x}{y}$.
- At $(3,4)$: $\dfrac{dy}{dx}=-\dfrac{3}{4}$ — the tangent slope there.
求 $x^2+y^2=25$ 在点 $(3,4)$ 处的 $\tfrac{dy}{dx}$。
- 求导:$2x+2y\dfrac{dy}{dx}=0$。
- 求解:$\dfrac{dy}{dx}=-\dfrac{2x}{2y}=-\dfrac{x}{y}$。
- 在 $(3,4)$:$\dfrac{dy}{dx}=-\dfrac{3}{4}$——那里的切线斜率。
Implicit differentiation: differentiate both sides with respect to $x$, treating $y$ as a function of $x$ so every $y$-term picks up a $\tfrac{dy}{dx}$ (chain rule); then collect, factor, and solve for $\tfrac{dy}{dx}$. Plug in a point on the curve to get a slope. The answer usually depends on both $x$ and $y$.
隐函数求导:对两边关于 $x$ 求导,把 $y$ 当作 $x$ 的函数,于是每个 $y$ 项都带上一个 $\tfrac{dy}{dx}$(链式法则);然后收集、提取、解出 $\tfrac{dy}{dx}$。代入曲线上的一个点即得斜率。答案通常同时依赖 $x$ 和 $y$。