Visual walkthrough — Implicit differentiation — technique, applications

Step 0 — The words we will use (built from scratch)

Before any symbol appears, let's fix a picture in words.

That's the whole cast. Everything below is about finding $\frac{dy}{dx}$ when $y$ won't come out of the equation.

Step 1 — Look at the curve and the number we want

WHAT. We draw the circle $x^2+y^2=25$ and pick one point on it, say $P=(3,4)$ (check: $9+16=25$ ✓). We want the slope of the tangent line at $P$.

WHY. Seeing the target first stops the algebra from feeling like magic. The slope is a real, visible thing — the tilt of the red line touching the circle at $P$.

PICTURE. The red line in the figure touches the circle at $P$ and nowhere else nearby. It clearly tilts downhill (goes down as we move right), so we already expect $\frac{dy}{dx}$ to come out negative at this point.

Step 2 — Why we can't just solve for $y$ first

WHAT. Try to free $y$: from $x^2+y^2=25$ we get $y^2=25-x^2$, so $y=\pm \sqrt{25-x^2}$.

WHY. That $\pm$ is the whole problem. One equation secretly holds two functions — the top half ($+$) and the bottom half ($-$). To differentiate the ordinary way we'd pick a branch and drag a square root through the chain rule. Messy, and we'd have to redo it for the other half.

PICTURE. The circle splits at the two red dots where it's vertical ($x=\pm 5$). Above the dashed line is the $+\sqrt{}$ branch; below is the $-\sqrt{}$ branch. No single formula $y=f(x)$ covers both at once.

Step 3 — The one gear that makes it work: the toll

WHAT. We recall the Chain Rule in its most useful shape here. If something is built out of $y$, and $y$ itself rides along with $x$, then

$\frac{d}{dx}\left(\underset{\text{outer square}}{\underbrace{y^2}}\right)=\underset{\begin{array}{c}\text{derivative of}\\ \text{the square}\end{array}}{\underbrace{2y}}\cdot \underset{\begin{array}{c}\text{rate }y\\ \text{itself moves}\end{array}}{\underbrace{\frac{dy}{dx}}}.$

WHY. When we nudge $x$ by a hair, $x$ changes at rate $1$ — so $\frac{d}{dx}(x^2)=2x\cdot 1=2x$, no toll. But $y$ does not change at rate $1$; it changes at rate $\frac{dy}{dx}$ (it's tied to $x$ by the equation). The chain rule charges us that rate as a "toll" every time we pass through $y$.

PICTURE. Two side-by-side ramps. On the left, nudging $x$ moves the height of $x^2$ at the plain rate $2x$. On the right, to change $y^2$ we must first move $y$ (rate $\frac{dy}{dx}$, the red arrow), then the square responds ($2y$). The red toll-arrow is the only new idea on this whole page.

Step 4 — Differentiate the whole equation, term by term

WHAT. Apply $\frac{d}{dx}$ to both sides of $x^2+y^2=25$:

$\underset{=2x\text{(toll-free)}}{\underbrace{\frac{d}{dx}(x^2)}}+\underset{=2y\frac{dy}{dx}\text{(pays toll)}}{\underbrace{\frac{d}{dx}(y^2)}}=\underset{=0\text{(constant, flat)}}{\underbrace{\frac{d}{dx}(25)}}$

so

$\underset{\begin{array}{c}\text{from the}\\ x^2\text{ term}\end{array}}{\underbrace{2x}}+\underset{\begin{array}{c}\text{from the}\\ y^2\text{ term}\end{array}}{\underbrace{2y\frac{dy}{dx}}}=\underset{\begin{array}{c}25\text{ never}\\ \text{changes}\end{array}}{\underbrace{0}}.$

WHY. Differentiating both sides keeps a true equation true — if two quantities are always equal, they change at the same rate. The $25$ is a fixed number: it never moves, so its rate of change is $0$. The $x^2$ term is ordinary. The $y^2$ term pays the toll from Step 3.

PICTURE. Each term of the equation is labelled with its "response to a tiny push in $x$": the $x^2$ block responds by $2x$, the $y^2$ block by $2y\cdot \frac{dy}{dx}$ (red, because of the toll), and the constant block does not budge.

Step 5 — Solve for the slope (Collect, Factor, Divide)

WHAT. We now have one equation with one unknown, $\frac{dy}{dx}$. Isolate it:

$2y\frac{dy}{dx}=-2x⟹\frac{dy}{dx}=-\frac{2x}{2y}=\boxed{-\frac{x}{y}}.$

Symbol by symbol: $-2x$ is the $x^2$ term moved to the other side (Collect); $\frac{dy}{dx}$ already stands alone so there is nothing to Factor here; dividing by $2y$ is the Divide; the $2$'s cancel.

WHY. This is plain algebra now — the calculus finished in Step 4. Note we never needed the formula for $y(x)$; the answer is allowed to mention $y$ because $y$ is a real number once we stand at a point.

PICTURE. At $P=(3,4)$ the formula gives $\frac{dy}{dx}=-\frac{3}{4}=-0.75$. The red tangent's tilt in the figure is exactly $3$ down for every $4$ across — a downhill line, matching our Step 1 prediction.

Step 6 — Every case: all four quadrants and the sign of the slope

WHAT. The one formula $-\frac{x}{y}$ silently handles all four regions of the circle. Let's read its sign everywhere:

WHY. A single explicit branch would have forced separate square-root formulas per half; here one expression covers both signs of $y$ and both signs of $x$ automatically. This is the payoff the parent note promised: "covers both branches at once."

PICTURE. Four sample points, one per quadrant, each with its red tangent. Watch the tilt flip uphill/downhill exactly as the sign table says.

Step 7 — The degenerate cases: where the slope blows up or vanishes

WHAT. Two special situations:

• Top and bottom $(0,\pm 5)$: here $x=0$, so $\frac{dy}{dx}=-\frac{0}{\pm 5}=0$. Flat tangent.
• Left and right $(\pm 5,0)$: here $y=0$, so $\frac{dy}{dx}=-\frac{\pm 5}{0}$ — undefined (division by zero). Vertical tangent.

WHY. The formula doesn't crash randomly; it fails exactly where the picture says a slope shouldn't exist as a number. A vertical line has "infinite" rise per run — no finite slope. This is also precisely where $F_y=\frac{\partial }{\partial y}(x^2+y^2)=2y=0$, so the Implicit Function Theorem warns us $y$ isn't a function of $x$ there (the $\pm$ branches meet). The maths and the picture agree perfectly.

PICTURE. At the red dots $(\pm 5,0)$ the tangent is a vertical red line ($y^{\prime }$ undefined); at $(0,\pm 5)$ the tangent is a flat red line ($y^{\prime }=0$). The two failure points are the same two dots from Step 2.

The one-picture summary

WHAT. One figure compresses the entire journey: the equation, the term-by-term toll, the solved slope, and the four-quadrant sign pattern, all on the same circle.

PICTURE. The circle carries its four sample tangents (colour-keyed by slope sign), the two degenerate red dots, and the boxed result $\frac{dy}{dx}=-\frac{x}{y}$ in the middle. This single image is the derivation.

Connections

• Implicit Differentiation — the parent recipe this page draws out.
• Chain Rule — the "toll" gear in Step 3 is exactly this.
• Product Rule — needed the moment an $xy$ term appears (see parent's folium example).
• Tangent and Normal Lines — what the slope $-x/y$ is for.
• Implicit Function Theorem — justifies the "$y$ is some function of $x$" promise and predicts the degenerate points.
• Related Rates — the same term-by-term toll, but the hidden variable is time $t$.

Flashcards