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Higher Derivatives and Motion · Tutorial 232 of 999

Third Derivatives of Implicit Curves

Differentiate an implicit equation three times while keeping track of every derivative of \(y\), then evaluate and verify \(y'''\) at a specified point.

Intermediate 9 min read

What You'll Learn

  • Identify when an implicit curve can be treated locally as \(y\) depending on \(x\)
  • Differentiate first- and second-derivative relations to obtain a third-derivative equation
  • Track how both \(y'\) and \(y''\) change during another differentiation
  • Evaluate \(y'''\) at a point using exact intermediate values
  • Check a third-derivative result in the differentiated relation
  • Avoid omitting terms when differentiating products involving \(y\), \(y'\), and \(y''\)

Why a Third Implicit Derivative Needs Care

An equation such as \(x^2+y^2=25\) can describe a curve without giving \(y\) explicitly as a formula in \(x\). At a point where the curve is locally a graph \(y=y(x)\), implicit differentiation lets us find its derivatives directly from the equation.

Finding a third derivative requires more than differentiating the original equation three times mechanically. After the first differentiation, \(y'\) is itself a quantity that changes with \(x\); after the second, so is \(y''\). In particular, the product rule can create several terms involving \(y'\) and \(y''\). Keeping them all is the central task.

Definition: If \(y\) is a differentiable function of \(x\), then \(y'\), \(y''\), and \(y'''\) denote its first, second, and third derivatives with respect to \(x\). For an implicit equation \(F(x,y)=0\), these derivatives describe a local graph on the curve wherever \(y\) can be treated as a differentiable function of \(x\).

For a polynomial equation \(F(x,y)=0\), the usual local condition that permits this treatment at a point is \(F_y\ne 0\), where \(F_y\) is the partial derivative with respect to \(y\). This is the implicit function theorem’s graph condition. If \(F_y=0\), the point may still lie on a curve, but this calculation does not establish that the curve is locally a differentiable graph \(y(x)\) there.

The Differentiation Pattern

When differentiating along a curve \(y=y(x)\), the chain rule gives \(\frac{d}{dx}F(x,y(x))=F_x+F_y y'\). The equation \(F(x,y(x))=0\) therefore gives a first-derivative relation. Differentiating that relation again produces a relation involving \(y''\); differentiating once more produces one involving \(y'''\).

The formula below records the third differentiation for a sufficiently differentiable function \(F\). Subscripts denote partial derivatives, and \(p=y'\), \(q=y''\), and \(r=y'''\) are used just to make the expression easier to read.

Theorem: Along a differentiable graph \(y=y(x)\) satisfying \(F(x,y)=0\), differentiating three times gives
$$ F_{xxx}+3F_{xxy}y'+3F_{xyy}(y')^2+F_{yyy}(y')^3 +3(F_{xy}+F_{yy}y')y''+F_y y'''=0. $$

This formula is valid when the indicated derivatives exist and the curve is being differentiated as a function \(y(x)\). In examples, it is often less error-prone to differentiate the first- or second-derivative relation directly than to memorize the full formula.

Why the Third-Derivative Formula Has These Terms

Here is a derivation that also explains why terms involving \(y'y''\) must not be lost. Start with the second-derivative relation obtained by differentiating \(F_x+F_y y'=0\):

$$ F_{xx}+2F_{xy}y'+F_{yy}(y')^2+F_y y''=0. $$

Differentiate each term with respect to \(x\), using the chain rule for each partial derivative. For instance, \(\frac{d}{dx}F_{xx}=F_{xxx}+F_{xxy}y'\). The four terms give:

$$ \begin{aligned} \frac{d}{dx}F_{xx} &=F_{xxx}+F_{xxy}y',\\ \frac{d}{dx}\bigl(2F_{xy}y'\bigr) &=2F_{xxy}y'+2F_{xyy}(y')^2+2F_{xy}y'',\\ \frac{d}{dx}\bigl(F_{yy}(y')^2\bigr) &=F_{xyy}(y')^2+F_{yyy}(y')^3+2F_{yy}y'y'',\\ \frac{d}{dx}\bigl(F_y y''\bigr) &=F_{xy}y''+F_{yy}y'y''+F_y y'''. \end{aligned} $$

Adding the expressions gives three copies of \(F_{xxy}y'\), three of \(F_{xyy}(y')^2\), three of \(F_{xy}y''\), and three of \(F_{yy}y'y''\). Grouping those terms yields the displayed third-derivative relation. The \(y'y''\) contribution comes from differentiating products; it cannot be discarded just because the original equation is polynomial.

Worked Examples

Worked Example: A Third Derivative on a Circle

Find \(y'''\) at \((3,4)\) on the circle \(x^2+y^2=25\), treating \(y\) locally as a function of \(x\).

First check that the point lies on the curve: \(3^2+4^2=9+16=25\). Differentiate the equation once:

$$ 2x+2yy'=0 \quad\Longrightarrow\quad x+yy'=0. $$

At \((3,4)\), \(3+4y'=0\), so \(y'=-\frac34\). Differentiate the simplified relation \(x+yy'=0\) again. The product rule gives:

$$ 1+(y')^2+yy''=0. $$

Substitute \(y=4\) and \(y'=-\frac34\):

$$ 1+\frac{9}{16}+4y''=0 \quad\Longrightarrow\quad \frac{25}{16}+4y''=0 \quad\Longrightarrow\quad y''=-\frac{25}{64}. $$

Differentiate \(1+(y')^2+yy''=0\). The derivative of \((y')^2\) is \(2y'y''\), and the derivative of \(yy''\) is \(y'y''+yy'''\). Therefore:

$$ 3y'y''+yy'''=0. $$

At the point, \(y'y''=(-\frac34)(-\frac{25}{64})=\frac{75}{256}\). Thus \(3y'y''=\frac{225}{256}\), and

$$ \frac{225}{256}+4y'''=0 \quad\Longrightarrow\quad y'''=-\frac{225}{1024}. $$

Check: Substitution into the third-derivative relation gives \(3(\frac{75}{256})+4(-\frac{225}{1024})=\frac{225}{256}-\frac{900}{1024}=0\), since \(\frac{900}{1024}=\frac{225}{256}\).

Worked Example: A Mixed Polynomial Equation

Find \(y'''\) at \((1,2)\) on \(x^2+xy+y^2=7\).

The point satisfies the equation because \(1^2+(1)(2)+2^2=1+2+4=7\). Differentiate once, remembering that both occurrences of \(y\) depend on \(x\):

$$ 2x+y+xy'+2yy'=0 \quad\Longrightarrow\quad 2x+y+(x+2y)y'=0. $$

At \((1,2)\), \(4+5y'=0\), so \(y'=-\frac45\). Differentiate the last relation term by term:

$$ 2+y'+(1+2y')y'+(x+2y)y''=0, $$

which simplifies to \(2+2y'+2(y')^2+(x+2y)y''=0\). At the point, the terms not involving \(y''\) sum to \(2-\frac85+\frac{32}{25}=\frac{50-40+32}{25}=\frac{42}{25}\). Hence

$$ \frac{42}{25}+5y''=0 \quad\Longrightarrow\quad y''=-\frac{42}{125}. $$

Differentiate the simplified second-derivative relation. The derivative of \(2(y')^2\) is \(4y'y''\), while the product \((x+2y)y''\) differentiates to \((1+2y')y''+(x+2y)y'''\). Combining the \(y''\)-terms gives:

$$ (3+6y')y''+(x+2y)y'''=0. $$

At \((1,2)\), \(3+6y'=3-\frac{24}{5}=-\frac95\), and \(x+2y=5\). Therefore

$$ \left(-\frac95\right)\left(-\frac{42}{125}\right)+5y'''=0 \quad\Longrightarrow\quad \frac{378}{625}+5y'''=0 \quad\Longrightarrow\quad y'''=-\frac{378}{3125}. $$

Check: Substituting this value into the differentiated relation gives \(\frac{378}{625}+5(-\frac{378}{3125})=\frac{378}{625}-\frac{1890}{3125}=0\), because \(\frac{1890}{3125}=\frac{378}{625}\).

Worked Example: Tracking a Product of \(y\) and \(y'\)

Find \(y'''\) at \((1,1)\) on the curve \(x^2y+y^3=2\). This example illustrates why differentiating a term involving \(y(y')^2\) requires care.

The point is on the curve because \(1^2(1)+1^3=2\). Differentiate the equation once:

$$ 2xy+x^2y'+3y^2y'=0. $$

At \((1,1)\), \(2+4y'=0\), so \(y'=-\frac12\). Differentiating the first-derivative relation gives:

$$ 2y+4xy'+x^2y''+6y(y')^2+3y^2y''=0. $$

At the point this becomes \(2-2+y''+6(\frac14)+3y''=0\), or \(\frac32+4y''=0\). Thus \(y''=-\frac38\).

Differentiate the second-derivative relation. In particular, the derivative of \(6y(y')^2\) uses the product rule twice:

$$ \frac{d}{dx}\bigl(6y(y')^2\bigr) =6(y')^3+12yy'y''. $$

Including the derivatives of all the other terms gives:

$$ 6y'+(6x+18yy')y''+6(y')^3+(x^2+3y^2)y'''=0. $$

At \((1,1)\), the coefficient \(6x+18yy'\) is \(6+18(-\frac12)=-3\), and \(x^2+3y^2=4\). Substitution yields:

$$ 6\left(-\frac12\right) +(-3)\left(-\frac38\right) +6\left(-\frac12\right)^3 +4y'''=0, $$

so \(-3+\frac98-\frac34+4y'''=0\). The constant terms combine to \(-\frac{24}{8}+\frac98-\frac68=-\frac{21}{8}\). Therefore

$$ -\frac{21}{8}+4y'''=0 \quad\Longrightarrow\quad y'''=\frac{21}{32}. $$

Check: Substituting \(y'''=\frac{21}{32}\) gives \(-\frac{21}{8}+4(\frac{21}{32})=-\frac{21}{8}+\frac{21}{8}=0\), as required.

A Reliable Process

For a specific implicit equation, it is usually clearest to write down each differentiated relation before substituting the point’s coordinates. This keeps the algebraic derivatives separate from the numerical evaluation.

1
Check the point and graph condition.
Verify that the point satisfies \(F(x,y)=0\). If using the local graph condition, check that \(F_y\ne 0\) there.
2
Find \(y'\).
Differentiate \(F(x,y)=0\) with respect to \(x\), treating \(y\) as a function of \(x\), then solve the resulting relation at the point.
3
Find \(y''\).
Differentiate the first-derivative relation, applying the product and chain rules to every term, and then evaluate at the point.
4
Find \(y'''\) and check.
Differentiate the second-derivative relation, substitute the known values of \(x\), \(y\), \(y'\), and \(y''\), solve, and substitute the result back into the third-derivative relation.

Common Mistake: Dropping a Product-Rule Term

A frequent error is to differentiate \(6y(y')^2\) as if \(y\) were constant, writing only \(12y y'y''\), or as if \((y')^2\) were constant, writing only \(6(y')^3\). Neither is complete: both factors depend on \(x\).

$$ \frac{d}{dx}\bigl(6y(y')^2\bigr) =6y'(y')^2+6y\cdot 2y'y'' =6(y')^3+12yy'y''. $$

In the third worked example, omitting \(12yy'y''\) would also omit \(12yy'\) from the coefficient multiplying \(y''\) after the entire second-derivative relation is differentiated. At \((1,1)\), that missing contribution changes the equation and gives the wrong value of \(y'''\). The correction is to differentiate each factor explicitly before collecting like terms.

Why Third Implicit Derivatives Matter

The value \(y'''\) measures how the second derivative changes as \(x\) changes along the curve. It can distinguish curves whose positions, slopes, and concavity agree at a point but whose higher-order behavior differs. In calculation, its main lesson is practical: every derivative in an implicit relation is itself a function of \(x\), so another differentiation must account for its change.

Before accepting an answer, check that the point lies on the original curve, that the values of \(y'\) and \(y''\) satisfy their differentiated relations, and that the proposed \(y'''\) satisfies the third-derivative relation. These checks catch sign errors and missing product-rule terms.

Check Your Understanding

For each question, show the differentiation steps and keep exact values until the final answer.

  1. On \(x^2+y^2=25\), find \(y'\), \(y''\), and \(y'''\) at \((4,3)\).
  2. For \(x^2+xy+y^2=7\), explain why differentiating \((x+2y)y'\) produces both \((1+2y')y'\) and \((x+2y)y''\).
  3. Differentiate \(5y(y')^2\) with respect to \(x\).
  4. For an equation \(F(x,y)=0\), what local condition allows the usual implicit function theorem to treat the curve as a differentiable graph \(y(x)\)?
  5. Why should \(y'\) and \(y''\) be found before solving the third-derivative relation at a point?

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