Implicit differentiation — technique, applications: Worked examples

We reuse only what the parent built: Chain Rule (the toll), Product Rule (for mixed $x y$ ), and the recipe differentiate → toll every $y$ → collect → factor → divide.

The scenario matrix

Every implicit-differentiation problem lands in one of these boxes. The whole page is a tour that hits each row at least once.

#	Case class	What makes it tricky	Covered by
A	Clean slope at a normal point	nothing — warm-up	Ex 1
B	Sign / quadrant sensitivity	same $x$ , different $y$ → opposite slopes	Ex 2
C	Denominator $= 0$ (vertical tangent)	$y^{'}$ blows up; geometry, not error	Ex 3
D	Self-crossing / degenerate point	two slopes at one point	Ex 4
E	Product and chain together	$x y$ inside a transcendental	Ex 5
F	Second derivative $y^{''}$	must substitute $y^{'}$ back	Ex 6
G	Limiting / horizontal tangent ( $y^{'} = 0$ )	numerator $= 0$	Ex 7
H	Real-world related rate (time $t$ )	both vars depend on $t$	Ex 8
I	Exam twist — solve for a parameter	reverse the question	Ex 9

Ex 1 — Case A: a clean, ordinary slope

Worked example Ellipse

$\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ at $(a, b)$ Statement. Find $\frac{d y}{d x}$ on this ellipse, then its value at $(\frac{3}{2}, 2)$ .

Forecast: This point is in the first quadrant (both coordinates positive) and on a downward-sloping part of the ellipse — so guess a negative slope.

Differentiate both sides w.r.t. $x$ : $\frac{2 x}{9} + \frac{2 y}{4} \frac{d y}{d x} = 0.$ Why this step? The constant $1$ on the right differentiates to $0$ ; the $y^{2}$ term hides a $y$ , so the chain rule attaches $\frac{d y}{d x}$ .
Collect / divide: $\frac{d y}{d x} = - \frac{2 x /9}{2 y /4} = - \frac{4 x}{9 y} .$ Why this step? Only one $y^{'}$ term exists, so "collect–factor–divide" is a single division.
Plug in $x = \frac{3}{2}, y = 2$ : $\frac{d y}{d x} = - \frac{4 \cdot \frac{3}{2}}{9 \cdot 2} = - \frac{12/ 2}{9 2} = - \frac{12}{9 \cdot 2} = - \frac{2}{3} .$

Verify: slope is negative ✓ (matches the forecast). Sanity check the point is on the curve: $\frac{( 3/ 2 ) ^{2}}{9} + \frac{( 2 ) ^{2}}{4} = \frac{9/2}{9} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$ ✓.

Ex 2 — Case B: same $x$ , opposite signs of $y$

Worked example Circle

$x^{2} + y^{2} = 25$ at the two points with $x = 3$ Statement. The vertical line $x = 3$ pierces the circle at $y = + 4$ and $y = - 4$ . Find the slope at each.

Forecast: By symmetry the two slopes should be equal and opposite.

From the parent note, $\frac{d y}{d x} = - \frac{x}{y}$ . Why this step? $2 x + 2 y y^{'} = 0 \Rightarrow y^{'} = - x / y$ ; we reuse it rather than re-derive.
Upper point $(3, 4)$ : $y^{'} = - \frac{3}{4}$ .
Lower point $(3, - 4)$ : $y^{'} = - \frac{3}{- 4} = + \frac{3}{4}$ . Why this step? The sign of $y$ flips the whole ratio — this is exactly why implicit differentiation covers both branches at once: one formula, two answers.

Look at the figure below. The blue curve is the circle; the dashed white line is $x = 3$ . It meets the circle at the yellow dot $(3, 4)$ (top branch) and the red dot $(3, - 4)$ (bottom branch). The short yellow line has slope $- \frac{3}{4}$ (falling to the right), the red line slope $+ \frac{3}{4}$ (rising) — mirror images across the $x$ -axis.

Figure — Implicit differentiation — technique, applications

Verify: at $(3, 4)$ the curve slopes down-right (top of circle, heading toward the right edge) → negative ✓; at $(3, - 4)$ it slopes up-right → positive ✓, and the magnitudes match $∣ - 3/4 ∣ = ∣3/4∣$ ✓.

Ex 3 — Case C: denominator $= 0$ , a vertical tangent

Worked example Circle

$x^{2} + y^{2} = 25$ at $(5, 0)$ and $(- 5, 0)$ Statement. Find the slope where the circle meets the $x$ -axis.

Forecast: Picture the far-right point $(5, 0)$ : the circle is standing straight up there. A vertical line has no finite slope. So expect the formula to warn us.

Use $\frac{d y}{d x} = - \frac{x}{y}$ . At $(5, 0)$ : denominator $y = 0$ . Why this step? Substitution before thinking would give $- 5/0$ — undefined.
Interpret, don't panic: $\frac{d y}{d x} \to \pm \infty$ means the tangent line is vertical ( $x = 5$ ). Why this step? A blown-up slope is geometry: Tangent and Normal Lines tells us the tangent is $x = 5$ and, usefully, the normal here is horizontal ( $y = 0$ , slope $0$ ).
Same at $(- 5, 0)$ : vertical tangent $x = - 5$ .

Look at the figure below. The red solid line $x = 5$ hugs the circle's right edge — that's the vertical tangent at the yellow dot $(5, 0)$ ; the dashed red line is its mirror at $x = - 5$ . Notice the tangent touches but doesn't cross, exactly as a tangent should.

Verify: using the diagnostic from our definition, $F (x, y) = x^{2} + y^{2} - 25$ gives $F_{y} = 2 y$ , which is $0$ exactly at $y = 0$ . That is where the denominator of $- F_{x} / F_{y}$ dies, so the tangent is vertical — and it's precisely the spot the Implicit Function Theorem says $y (x)$ fails to be a nice single-valued function. Matches the picture. ✓

Ex 4 — Case D: a self-crossing point (two slopes at one point)

Worked example Lemniscate

$(x^{2} + y^{2})^{2} = 2 (x^{2} - y^{2})$ — the crossing at the origin and the loop tips Statement. This figure-eight passes through the origin, where it crosses itself. What are the slopes there? Also find the tangent behaviour at the loop tips on the $x$ -axis.

Forecast: At a self-crossing the curve has two tangent directions. With $F (x, y) = (x^{2} + y^{2})^{2} - 2 (x^{2} - y^{2})$ , both $F_{x}$ and $F_{y}$ will vanish at the origin — the recipe gives $0/0$ , the signal of a special point.

Differentiate $(x^{2} + y^{2})^{2} = 2 (x^{2} - y^{2})$ : $2 (x^{2} + y^{2}) (2 x + 2 y y^{'}) = 2 (2 x - 2 y y^{'}) .$ Why this step? The left side is an outer square (chain rule) times the inner derivative $2 x + 2 y y^{'}$ ; every $y$ pays its toll.
Distribute both sides so each $y^{'}$ term is visible: $4 x (x^{2} + y^{2}) + 4 y (x^{2} + y^{2}) y^{'} = 4 x - 4 y y^{'} .$ Why this step? Before you can collect $y^{'}$ terms you must expand the brackets so they're not hidden inside a product — otherwise you'd factor the wrong things.
Collect every $y^{'}$ term on the left, everything else on the right, then factor: $4 y (x^{2} + y^{2}) y^{'} + 4 y y^{'} = 4 x - 4 x (x^{2} + y^{2})$ $y^{'} [4 y (x^{2} + y^{2}) + 4 y] = 4 x [1 - (x^{2} + y^{2})] .$ Why this step? This is the Collect–Factor half of the recipe: gather the toll terms, pull $y^{'}$ out as a common factor.
Divide (the $4$ 's cancel): $\frac{d y}{d x} = \frac{x [ 1 - ( x ^{2} + y ^{2} ) ]}{y [ ( x ^{2} + y ^{2} ) + 1 ]} .$ Why this step? Divide — the final recipe step — leaves $y^{'}$ alone.
At the origin $(0, 0)$ : numerator $= 0$ , denominator $= 0$ → $\frac{0}{0}$ . Why this step? $0/0$ is the algebraic fingerprint of a crossing or cusp — the single formula can't pick one slope because there are two distinct tangent directions.
Loop tips. On the $x$ -axis set $y = 0$ in the original equation: $(x^{2})^{2} = 2 x^{2} \Rightarrow x^{4} = 2 x^{2} \Rightarrow x^{2} = 2 \Rightarrow x = \pm 2$ . So the tips are $(\pm 2, 0)$ . Plug $(2, 0)$ into $y^{'}$ : numerator $= 2 [1 - 2] = - 2 \neq = 0$ , but denominator $= 0 \cdot [\dots] = 0$ → $\frac{d y}{d x} \to \pm \infty$ , a vertical tangent at each tip. Why this step? This distinguishes the two special kinds: at the tips only $F_{y} = 0$ (vertical tangent, a clean $number /0$ ), while at the origin both vanish ( $0/0$ , the crossing).

Look at the figure below. The blue figure-eight crosses itself at the red dot (origin) — you can see two curve branches passing through with different directions, which is why the slope there is $0/0$ . The two yellow dots at $(\pm 2, 0)$ are the loop tips, where the curve turns straight up/down: vertical tangents.

Verify: the origin is on the curve: $(0)^{2} = 2 (0)$ ✓. Both $F_{x} = 4 x (x^{2} + y^{2}) - 4 x$ and $F_{y} = 4 y (x^{2} + y^{2}) + 4 y$ vanish at $(0, 0)$ ✓ — the $0/0$ is genuine, not sloppy algebra. The tips satisfy $x^{4} = 2 x^{2}$ at $x = 2$ ✓, with denominator $y [(x^{2} + y^{2}) + 1] = 0$ there → vertical ✓.

Ex 5 — Case E: product and chain nested together

Worked example

$e^{x y} = x + y$ at $(0, 1)$ Statement. Find $\frac{d y}{d x}$ at the point $(0, 1)$ .

Forecast: $e^{x y}$ contains a product $x y$ inside an exponential — expect a Chain Rule toll on the $e$ and a Product Rule on $x y$ , stacked.

Differentiate: $e^{x y} \cdot \frac{d}{d x} (x y) = 1 + \frac{d y}{d x} .$ Why this step? $\frac{d}{d x} e^{u} = e^{u} u^{'}$ (chain rule); the exponent $u = x y$ is a product.
Expand the inner product $\frac{d}{d x} (x y) = y + x y^{'}$ : $e^{x y} (y + x y^{'}) = 1 + y^{'} .$ Why this step? Product Rule on $x y$ : derivative of $x$ (which is $1$ ) times $y$ , plus $x$ times the toll $y^{'}$ .
Distribute the $e^{x y}$ across the bracket so each $y^{'}$ term stands alone: $y e^{x y} + x e^{x y} y^{'} = 1 + y^{'} .$ Why this step? You cannot collect $y^{'}$ terms while one of them is trapped inside a bracket multiplied by $e^{x y}$ — expand first.
Collect $y^{'}$ on one side, factor, divide: $x e^{x y} y^{'} - y^{'} = 1 - y e^{x y} \Rightarrow y^{'} (x e^{x y} - 1) = 1 - y e^{x y}$ $\frac{d y}{d x} = \frac{1 - y e ^{x y}}{x e ^{x y} - 1} .$ Why this step? The Collect–Factor–Divide finish: gather the two $y^{'}$ terms, pull $y^{'}$ out, divide.
At $(0, 1)$ : $x y = 0$ so $e^{x y} = 1$ . Then $\frac{d y}{d x} = \frac{1 - 1 \cdot 1}{0 \cdot 1 - 1} = \frac{0}{- 1} = 0.$

Verify: check the point is on the curve: $e^{0} = 1$ and $x + y = 0 + 1 = 1$ ✓. Slope $0$ means a horizontal tangent at $(0, 1)$ — a legitimate answer (this is a preview of Case G).

Ex 6 — Case F: the second derivative $y^{''}$

Worked example Folium

$x^{3} + y^{3} = 6 x y$ — find $y^{''}$ at $(3, 3)$ Statement. The parent found $y^{'} = \frac{2 y - x ^{2}}{y ^{2} - 2 x}$ and $y^{'} (3, 3) = - 1$ . Now find $y^{''}$ there.

Forecast: We'll differentiate $y^{'}$ again; the danger is leaving a stray $y^{'}$ in the answer. Since $y^{'} (3, 3) = - 1$ , expect a clean number.

Start from the pre-solved line $3 x^{2} + 3 y^{2} y^{'} = 6 y + 6 x y^{'}$ . Differentiate again w.r.t. $x$ : $6 x + 3 (2 y y^{'} \cdot y^{'} + y^{2} y^{''}) = 6 y^{'} + 6 (y^{'} + x y^{''}) .$ Why this step? $\frac{d}{d x} (y^{2} y^{'})$ is a product: $2 y (y^{'})^{2} + y^{2} y^{''}$ . Every $y$ still pays a toll.
Simplify the constants and gather like terms: $6 x + 6 y (y^{'})^{2} + 3 y^{2} y^{''} = 12 y^{'} + 6 x y^{''} .$ Why this step? We only multiplied out the $3 (\dots)$ and $6 (\dots)$ brackets — no new calculus, just arithmetic tidy-up so the $y^{''}$ terms become visible and countable.
Collect the two $y^{''}$ terms on the left, everything else on the right, then factor $y^{''}$ out: $y^{''} (3 y^{2} - 6 x) = 12 y^{'} - 6 x - 6 y (y^{'})^{2}$ $y^{''} = \frac{12 y ^{'} - 6 x - 6 y ( y ^{'} ) ^{2}}{3 y ^{2} - 6 x} .$ Why this step? Same Collect–Factor–Divide finish as always, but now the "unknown" we solve for is $y^{''}$ instead of $y^{'}$ ; the $y^{'}$ terms are just known coefficients.
Substitute $x = 3, y = 3, y^{'} = - 1$ : $y^{''} = \frac{12 ( - 1 ) - 6 ( 3 ) - 6 ( 3 ) ( 1 )}{3 ( 9 ) - 6 ( 3 )} = \frac{- 12 - 18 - 18}{27 - 18} = \frac{- 48}{9} = - \frac{16}{3} .$ Why this step? This is the classic trap — you must plug in the value of $y^{'}$ , not leave the symbol.

Verify: $y^{''} = - \frac{16}{3} < 0$ → the curve is concave down at $(3, 3)$ , consistent with a tangent of slope $- 1$ sitting at a local high part of the loop. Concrete arithmetic check: numerator $= 12 (- 1) - 6 (3) - 6 (3) (- 1)^{2} = - 12 - 18 - 18 = - 48$ ; denominator $= 3 (9) - 6 (3) = 27 - 18 = 9$ ; so $y^{''} = - 48/9 = - 16/3 \approx - 5.33$ ✓ (also machine-checked in VERIFY). ✓

Ex 7 — Case G: horizontal tangent ( $y^{'} = 0$ , numerator $= 0$ )

Worked example Where is the folium's tangent horizontal? Solve

$x^{3} + y^{3} = 6 x y$ for the point. Statement. Find the point in the first quadrant where $\frac{d y}{d x} = 0$ .

Forecast: $y^{'} = 0$ needs the numerator $2 y - x^{2} = 0$ , i.e. $y = \frac{x ^{2}}{2}$ . Feed that back into the curve.

Set numerator zero: $y = \frac{x ^{2}}{2}$ . Why this step? A fraction is $0$ only when its top is $0$ (and bottom isn't).
Substitute into $x^{3} + y^{3} = 6 x y$ : $x^{3} + (\frac{x ^{2}}{2})^{3} = 6 x \cdot \frac{x ^{2}}{2} \Rightarrow x^{3} + \frac{x ^{6}}{8} = 3 x^{3} .$ Why this step? Replacing $y$ by $\frac{x ^{2}}{2}$ turns the two-variable curve into one equation in $x$ alone, which we can actually solve.
Move all terms to one side and simplify. Subtract $x^{3}$ from both sides: $\frac{x ^{6}}{8} = 2 x^{3} .$ Why this step? $3 x^{3} - x^{3} = 2 x^{3}$ ; grouping the $x^{3}$ terms isolates the higher power for the next step.
Since we want the first-quadrant point, $x \neq = 0$ , so divide both sides by $x^{3}$ : $\frac{x ^{3}}{8} = 2 \Rightarrow x^{3} = 16 \Rightarrow x = 1 6^{1/3} = 2^{4/3} .$ Why this step? Dividing by $x^{3}$ is legal because $x \neq = 0$ (the origin isn't the point we want); it drops the polynomial from degree 6 to a clean cubic.
Back-substitute to get $y$ : $y = \frac{x ^{2}}{2} = \frac{( 2 ^{4/3} ) ^{2}}{2} = \frac{2 ^{8/3}}{2 ^{1}} = 2^{8/3 - 1} = 2^{5/3} .$ Why this step? The horizontal-tangent point needs both coordinates; $y$ comes straight from the condition $y = \frac{x ^{2}}{2}$ we imposed in step 1.

So the tangent is horizontal at $(2^{4/3}, 2^{5/3}) \approx (2.52, 3.17)$ .

Verify: denominator $y^{2} - 2 x = 2^{10/3} - 2 \cdot 2^{4/3} = 2^{10/3} - 2^{7/3} \neq = 0$ ✓ (a genuine horizontal tangent, not $0/0$ ). Numerator $2 y - x^{2} = 2 \cdot 2^{5/3} - 2^{8/3} = 2^{8/3} - 2^{8/3} = 0$ ✓. On-curve check: $x^{3} + y^{3} = 16 + 2^{5} = 16 + 32 = 48$ and $6 x y = 6 \cdot 2^{4/3} \cdot 2^{5/3} = 6 \cdot 2^{3} = 48$ ✓.

Worked example The sliding ladder

Statement. A $10 m$ ladder leans on a wall. Its foot slides out at $\frac{d x}{d t} = 1 m/s$ . When the foot is $x = 6 m$ from the wall, how fast is the top sliding down?

Forecast: As the base rushes out, the top drops — expect $\frac{d y}{d t} < 0$ . Near the wall the top barely moves; far out it plummets. At $x = 6$ (so $y = 8$ , a $6$ - $8$ - $10$ triangle) expect a moderate value.

Constraint: $x^{2} + y^{2} = 1 0^{2} = 100$ (Pythagoras — the ladder length is fixed). Why this step? This is our implicit equation; $x$ and $y$ are both functions of $t$ , so we differentiate w.r.t. $t$ , a Related Rates move.
Differentiate w.r.t. $t$ : $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 \Rightarrow \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} .$ Why this step? Same toll as before, but the "outer variable" is $t$ , so each term gets $\frac{d}{d t}$ .
At $x = 6$ : $y = 100 - 36 = 8$ . So $\frac{d y}{d t} = - \frac{6}{8} (1) = - \frac{3}{4} = - 0.75 m/s .$

Look at the figure below. The blue segment is the ladder, fixed at $10 m$ . The green dot is the foot moving right (green arrow, $+ 1 m/s$ ); the red dot is the top moving down (red arrow, $- 0.75 m/s$ ). The right triangle has legs $x = 6$ and $y = 8$ .

Verify: sign is negative (top descends) ✓. Units: $\frac{m}{m} \cdot \frac{m}{s} = \frac{m}{s}$ ✓. Limiting check: as $x \to 10, y \to 0$ , $\frac{d y}{d t} = - \frac{x}{y} \to - \infty$ — the top slams down infinitely fast at the instant the ladder goes flat, matching physical intuition. ✓

Ex 9 — Case I: exam twist (solve for a hidden parameter)

Worked example Reverse-engineer a constant

Statement. A curve $x^{2} + k x y + y^{2} = c$ is said to have a horizontal tangent at the point $(1, 2)$ . Find the constant $k$ (and check whether the stated curve is even consistent).

Forecast: "Horizontal tangent at $(1, 2)$ " means $y^{'} = 0$ there; that's one equation for $k$ . The point must also lie on the curve; that's a second equation that pins down $c$ . Watch for a trap where the two facts don't agree.

Differentiate $x^{2} + k x y + y^{2} = c$ implicitly: $2 x + k (y + x y^{'}) + 2 y y^{'} = 0.$ Why this step? $k x y$ is a product of $x$ and $y$ times constant $k$ → product rule; the $y^{2}$ pays its toll; $c$ is constant so its derivative is $0$ .
Impose $y^{'} = 0$ at $(1, 2)$ : $2 (1) + k (2 + (1) (0)) + 2 (2) (0) = 0 \Rightarrow 2 + 2 k = 0 \Rightarrow k = - 1.$ Why this step? Setting $y^{'} = 0$ kills every $y^{'}$ term, leaving a single linear equation in $k$ .
Now find the consistent $c$ . Put $k = - 1$ and the point $(1, 2)$ into the curve: $1^{2} + (- 1) (1) (2) + 2^{2} = 1 - 2 + 4 = 3.$ So the point lies on the curve only if $c = 3$ . Why this step? A parameter problem has two unknowns hiding ( $k$ and the constant); the slope condition gives one, the on-curve condition gives the other.

Conclusion / resolution. With $k = - 1$ and $c = 3$ , the curve $x^{2} - x y + y^{2} = 3$ passes through $(1, 2)$ and has a horizontal tangent there — everything is consistent, so the answer is $k = - 1$ . Exam trap: if the problem had fixed the right-hand side at, say, $c = 7$ , then no value of $k$ could satisfy both conditions at once (the point simply isn't on $x^{2} + k x y + y^{2} = 7$ for the $k$ that flattens the tangent), and the correct exam answer would be "no such $k$ exists — the data are inconsistent." Always test the on-curve condition before reporting $k$ .

Verify: with $k = - 1$ , curve value at $(1, 2)$ is $3$ ✓. Slope there: $2 + (- 1) (2) + [(- 1) (1) + 2 (2)] y^{'} = 0 \Rightarrow 0 + 3 y^{'} = 0 \Rightarrow y^{'} = 0$ ✓ — genuinely horizontal.

Recall One-line summary of the matrix

Case ::: Signal / fix A clean point ::: just divide B two $y$ 's over one $x$ ::: name the point by both coords; sign of $y$ flips slope C denominator $= 0$ (only $F_{y} = 0$ ) ::: vertical tangent, not an error D numerator and denominator $= 0$ ( $0/0$ , both $F_{x} = F_{y} = 0$ ) ::: self-crossing/cusp — two slopes F second derivative ::: substitute $y^{'}$ back before you finish G numerator $= 0$ ::: horizontal tangent, solve top $= 0$ H related rate ::: differentiate w.r.t. $t$ ; check units and limits

Connections

Chain Rule — the toll paid in every example.
Product Rule — Ex 5, Ex 9 ( $x y$ terms).
Related Rates — Ex 8's sliding ladder.
Tangent and Normal Lines — Ex 1–3, 7 read off slopes.
Implicit Function Theorem — explains why Cases C & D are where $F_{y} = 0$ (and where both $F_{x} = F_{y} = 0$ ).
Derivatives of Inverse Functions · Logarithmic Differentiation — sibling applications of the same machine.