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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.

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

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.

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.