Implicit Differentiation
Introduced| English | Meaning |
|---|---|
| implicit differentiation/ɪmˈplɪsɪt ˌdɪfəˌrenʃɪˈeɪʃn/ | Differentiating an equation involving two variables while treating one variable as a function of the other. |
| implicit/ɪmˈplɪsɪt/ | Expressed by a relationship between variables rather than by isolating one variable as a function of another. |
When you can't solve for $y$
- The circle $x^2+y^2=25$ defines $y$ implicitly — 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.
An equation like $x^2+y^2=25$ defines $y$ ____, since it is not solved for $y$.
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That is why we use implicit differentiation.
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 $\tfrac{dy}{dx}$.
Implicit 隐式 differentiation on the circle x² + y² = 25: the tangent slope at (3, 4) is −3/4
Differentiating with respect to $x$, $\dfrac{d}{dx}[y^2]=$
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Chain rule: $2y$ times the inner derivative $\tfrac{dy}{dx}$.
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.
A tangent slope on a curve
y = √(25 − x²) (upper semicircle)
Even a curve you can't solve for $y$ has a tangent at each point — implicit differentiation finds that slope.
Put the implicit-differentiation steps in order.
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Differentiate, collect, solve, then evaluate.
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.
For $x^2+y^2=25$, $\dfrac{dy}{dx}=-\dfrac{x}{y}$. Find the slope at $(3,4)$ (a decimal).
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$-\tfrac{3}{4}=-0.75$.
The factor $\tfrac{dy}{dx}$ is added to every term, including pure $x$ terms like $x^2$.
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Only terms containing $y$ get the $\tfrac{dy}{dx}$ factor.
Differentiating $xy$ with respect to $x$ gives...
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Product rule: $\frac{d}{dx}[xy]=1\cdot y+x\cdot\tfrac{dy}{dx}$.
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: $\frac{d}{dx}[xy]=y+x\tfrac{dy}{dx}$.
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.
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 $y$.
Write from a circle slope of negative 0.75
- x squared + y squared = 25.
- The point (3, 4) is on the circle.
What is dy/dx at (3, 4)?
Negative 3 over 4 is the tangent slope.
On x squared + y squared = 25, the slope at (3, 4) is what?
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-3/4 = -0.75.