Single-Variable-Calculus · Unit 5 · Video 3 · Interactive Practice

Implicit Differentiation of a Quartic Curve

IKey Formulas

Formula	Name	What it takes
$y^4 + xy^2 - 2 = 0$	The curve	Quartic in $y$ , quadratic in $y^2$
$4y^3y' + y^2 + 2xy\,y' = 0$	Differentiated as it stands	Chain rule on $y^4$ , product rule on $xy^2$
$y' = \dfrac{-y^2}{4y^3 + 2xy}$	Slope formula	A point on the curve: both $x$ and $y$
$y = \pm\sqrt{\dfrac{-x + \sqrt{x^2 + 8}}{2}}\ \ (\star)$	Explicit solution	Quadratic formula with $y^2$ as the unknown

Key Insight: The implicit route never differentiates $(\star)$ — but $(\star)$ is still what supplies the $y$ that the slope formula demands.

IIVisualization 1 — Two Real Branches, Not Four

Four sign combinations in $(\star)$ , yet only two of them put a real point on the curve.

x 1

💡 The two lost roots are not missing, only imaginary: $y^2 < 0$ gives $y = \pm i\sqrt{|y^2|}$ , so the quartic still has four roots above every $x$ .

IIIVisualization 2 — The Implicit Route, Line by Line

Differentiating the equation as it stands takes five lines and never touches $(\star)$ .

Step 1

Step 1 — Chain rule on $y^4$

\frac{d}{dx}\,y^4 = 4y^3\,y'

Step 2 — Product rule on $xy^2$

\frac{d}{dx}\left(xy^2\right) = (1)\,y^2 + x\,(2y\,y')

Step 3 — The differentiated equation

4y^3y' + y^2 + 2xy\,y' = 0

Step 4 — Factor out $y'$ , then divide

\left(4y^3 + 2xy\right)y' = -y^2

y' = \frac{-y^2}{4y^3 + 2xy}

Step 5 — Substitute $(1, 1)$

y' = \frac{-(1)^2}{4(1)^3 + 2(1)(1)} = -\frac{1}{6}

IVVisualization 3 — The Slope at Any Point

The formula returns a slope at every point of the curve, and it needs both coordinates.

Point x 1 Branch

💡 This curve has neither a horizontal nor a vertical tangent: $-y^2 = 0$ forces $y = 0$ , and $4y^3 + 2xy = 2y(2y^2 + x) = 0$ forces $x = -2y^2$ , which turns the equation into $y^4 = -2$ .

VQuiz Questions

Problem 1 · Differentiating the Equation

Given: $y^4 + xy^2 - 2 = 0$ , where $y$ is a function of $x$ — find the equation that results from applying $\dfrac{d}{dx}$ to both sides.

✅ Correct! The chain rule turns

y^4

into

4y^3y'

, and the product rule splits

xy^2

into

(1)y^2 + x(2y\,y')

❌ Every $y$ term carries a $y'$ . Since

y

is a function of

x

, the chain rule gives

\dfrac{d}{dx}y^4 = 4y^3y'

, not

4y^3

❌ The product rule differentiates the $x$ factor too. That piece is

(1)(y^2) = y^2

, and it is the only term with no

y'

in it.

❌ The second factor needs the chain rule as well.

\dfrac{d}{dx}y^2 = 2y\,y'

, so that term is

x(2y\,y') = 2xy\,y'

Show solution

Differentiate term by term, remembering that $y$ depends on $x$ :

\frac{d}{dx}\,y^4 = 4y^3\,y' \qquad \text{(chain rule)}

\frac{d}{dx}\left(xy^2\right) = (1)\,y^2 + x\,(2y\,y') = y^2 + 2xy\,y' \qquad \text{(product rule)}

The constant $-2$ and the right-hand side $0$ both differentiate to $0$ , so

4y^3y' + y^2 + 2xy\,y' = 0

Exactly one term, $y^2$ , is free of $y'$ — that is the term which will cross the equals sign when the equation is solved for $y'$ .

Problem 2 · The Mirror Point

Given: the point $(1, -1)$ , which also lies on the curve since $(-1)^4 + (1)(-1)^2 - 2 = 0$ — find $y'$ there.

✅ Correct! The denominator is

4(-1)^3 + 2(1)(-1) = -6

while the numerator stays

-1

, so

y' = \dfrac{-1}{-6} = \dfrac{1}{6}

❌ That is the slope at $(1, 1)$ . Reflecting a curve across the

x

-axis negates every slope, so the two points cannot share one.

❌ The $2xy$ term does not drop out. At

(1, -1)

it equals

2(1)(-1) = -2

, making the denominator

-4 - 2 = -6

, not

-4

❌ Watch the cube: $(-1)^3 = -1$ . So

4y^3 = -4

, not

+4

, and the denominator is

-6

Show solution

Substitute $x = 1$ , $y = -1$ into the slope formula. The numerator is $-y^2$ , which never notices the sign of $y$ :

-y^2 = -(-1)^2 = -1

The denominator does notice it, in both terms:

4y^3 + 2xy = 4(-1)^3 + 2(1)(-1) = -4 - 2 = -6

y' = \frac{-1}{-6} = \frac{1}{6}

At $(1, 1)$ the slope was $-\dfrac{1}{6}$ . The lower branch is the upper one reflected across the $x$ -axis, and reflection flips the sign of every slope — which is exactly what the formula reports.

Problem 3 · The Curve at $x = 0$

Given: $x = 0$ on the curve $y^4 + xy^2 - 2 = 0$ — find the points there and the slope at the upper one.

Which $y$ satisfy the equation?

What is $y'$ at $\left(0,\ \sqrt[4]{2}\right)$ ?

✅ Correct! With

x = 0

the denominator collapses to

4y^3

, so

y' = \dfrac{-y^2}{4y^3} = \dfrac{-1}{4y} = \dfrac{-1}{4\sqrt[4]{2}} \approx -0.210

❌ Not quite. At

x = 0

the equation reduces to

y^4 = 2

, so the fourth root is what is wanted.

❌ One square root short.

y^4 = 2

gives

y^2 = \sqrt{2}

, and then

y = \pm\sqrt[4]{2} \approx \pm 1.189

❌ The line $x = 0$ crosses both branches.

y^4 = 2

has two real roots,

+\sqrt[4]{2}

and

-\sqrt[4]{2}

❌ Not quite. Put

x = 0

into

y' = \dfrac{-y^2}{4y^3 + 2xy}

and simplify the powers of

y

before evaluating.

❌ One factor of $y$ survives the cancellation.

\dfrac{-y^2}{4y^3} = \dfrac{-1}{4y}

, and

y = \sqrt[4]{2} \neq 1

❌ That is $-y^2/4$ . The denominator is

4y^3 = 4 \cdot 2^{3/4}

, not

4

❌ Only the term $2xy$ vanishes at $x = 0$ . The denominator is

4y^3 = 4 \cdot 2^{3/4} \approx 6.73 \neq 0

Show solution

Step 1 — the points. Setting $x = 0$ kills the middle term:

y^4 - 2 = 0 \quad\Longrightarrow\quad y^4 = 2 \quad\Longrightarrow\quad y^2 = \sqrt{2} \quad\Longrightarrow\quad y = \pm\sqrt[4]{2} \approx \pm 1.189

(Only the positive root of $y^2 = \pm\sqrt{2}$ survives: $y^2 = -\sqrt{2}$ has no real solution — the same reason $(\star)$ has only two real branches.)

Step 2 — the slope. With $x = 0$ the second term of the denominator vanishes, and the powers of $y$ cancel:

y' = \frac{-y^2}{4y^3 + 2(0)y} = \frac{-y^2}{4y^3} = \frac{-1}{4y}

y' = \frac{-1}{4\sqrt[4]{2}} = -\frac{2^{3/4}}{8} \approx -0.210

At the lower point $\left(0, -\sqrt[4]{2}\right)$ the same formula gives $+\dfrac{1}{4\sqrt[4]{2}}$ .

Problem 4 · A Different Implicit Curve

Given: the cubic curve $y^3 + xy - 2 = 0$ , which also passes through $(1, 1)$ — find $y'$ at that point.

✅ Correct!

3y^2y' + y + xy' = 0

gives

(3y^2 + x)y' = -y

, so

y' = \dfrac{-y}{3y^2 + x} = \dfrac{-1}{4}

❌ The chain rule is missing on $y^3$ .

\dfrac{d}{dx}y^3 = 3y^2y'

, not

3y^2

— the

y'

has to be there to be factored out.

❌ Check the sign. The term without

y'

+y

, so moving it across gives

(3y^2 + x)y' = -y

and a negative slope.

❌ The product rule on $xy$ leaves a term with no $y'$ .

\dfrac{d}{dx}(xy) = y + xy'

, and that lone

y

is what makes the slope nonzero.

Show solution

Same three moves as the quartic — chain rule, product rule, factor out $y'$ :

\frac{d}{dx}\,y^3 = 3y^2y', \qquad \frac{d}{dx}(xy) = y + xy', \qquad \frac{d}{dx}(2) = 0

3y^2y' + y + xy' = 0

\left(3y^2 + x\right)y' = -y \quad\Longrightarrow\quad y' = \frac{-y}{3y^2 + x}

At $(1, 1)$ , which is on the curve because $1 + 1 - 2 = 0$ :

y' = \frac{-1}{3(1)^2 + 1} = -\frac{1}{4}

As before, the answer is an algebraic formula in $x$ and $y$ : quick to evaluate once a point is known, and useless without one.

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