Second Derivatives of Implicit Functions

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Textbook Reference

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Try This First

For the circle $x^2 + y^2 = r^2$, you know $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Predict: what should $\dfrac{d^2y}{dx^2}$ look like? Is the upper arc of the circle concave up or concave down?

The upper arc is a “hill” shape -- concave down. So you expect $\dfrac{d^2y}{dx^2} < 0$ on the upper arc. The formula derivation below will confirm this.

Prerequisite Check

• Find $dy/dx$ for a given implicit equation
• Apply the quotient rule to differentiate $dy/dx$ with respect to $x$
• Substitute $dy/dx$ back into an expression

Quick Reference

To find $d^2y/dx^2$ implicitly:

1. Find $dy/dx$ by implicit differentiation.
2. Differentiate $dy/dx$ with respect to $x$ (implicit differentiation again -- treat $dy/dx$ as a function of $x$, and $y$ as a function of $x$).
3. Substitute the expression for $dy/dx$ from Step 1 wherever $dy/dx$ appears in Step 2.
4. Simplify, expressing the result in terms of $x$ and $y$.

Key Concepts

1. The Procedure for a Circle

Example 1. Find $\dfrac{d^2y}{dx^2}$ for $x^2 + y^2 = r^2$.

Step 1. $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Step 2. Differentiate $-x/y$ with respect to $x$ (quotient rule, treating $y = y(x)$): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = -\frac{y - x\frac{dy}{dx}}{y^2}. \]

Step 3. Substitute $dy/dx = -x/y$: \[ \frac{d^2y}{dx^2} = -\frac{y - x(-x/y)}{y^2} = -\frac{y + x^2/y}{y^2} = -\frac{y^2 + x^2}{y^3} = -\frac{r^2}{y^3}. \]

(Used $x^2 + y^2 = r^2$ in the last step.)

Interpretation. For the upper arc, $y > 0$, so $d^2y/dx^2 = -r^2/y^3 < 0$. Concave down -- matching the “hill” picture. For the lower arc, $y < 0$: $d^2y/dx^2 > 0$. Concave up -- matching a “valley.”

2. Why Substituting $dy/dx$ Matters

After differentiating again in Step 2, the result contains $dy/dx$. This is still an implicit expression. To get a formula purely in $x$ and $y$, you substitute the Step 1 expression for $dy/dx$.

If you do NOT substitute, the formula is: \[ \frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2}. \]

This is correct but contains $dy/dx$ -- it is valid, but less useful for determining concavity at a specific point. After substitution, you get: \[ \frac{d^2y}{dx^2} = -\frac{r^2}{y^3}. \]

Evaluating concavity at $(3, 4)$ on $x^2 + y^2 = 25$: $d^2y/dx^2 = -25/64 < 0$. Concave down at this point.

3. General Example

Example 2. Find $\dfrac{d^2y}{dx^2}$ for $x^2 + y^3 = 5$.

Step 1. $2x + 3y^2\,dy/dx = 0 \Rightarrow dy/dx = -2x/(3y^2)$.

Step 2. Differentiate: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{-2x}{3y^2}\right] = -\frac{2 \cdot 3y^2 - 2x \cdot 6y\,dy/dx}{9y^4} = -\frac{6y^2 - 12xy\,dy/dx}{9y^4}. \]

Step 3. Substitute $dy/dx = -2x/(3y^2)$: \[ = -\frac{6y^2 - 12xy \cdot (-2x/(3y^2))}{9y^4} = -\frac{6y^2 + 8x^2/y}{9y^4} = -\frac{6y^3 + 8x^2}{9y^5}. \]

4. Multiple Representations: Concavity and Graph

For the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$:

$dy/dx = -4x/(9y)$.

$d^2y/dx^2 = -4/(9y^3) \cdot (x^2/9 + y^2/4)^{-1} \cdot 1 = -4 \cdot (1/9)/y^3 \cdot (9) = -4/(9y^3)$ (after using the ellipse equation $x^2/9 + y^2/4 = 1$ to simplify).

Wait, let me redo: differentiate $dy/dx = -4x/(9y)$:

$\frac{d^2y}{dx^2} = -\frac{4 \cdot 9y - 4x \cdot 9\,dy/dx}{81y^2} = -\frac{4}{9} \cdot \frac{y - x\,dy/dx}{y^2}$.

Substitute $dy/dx = -4x/(9y)$:

$= -\frac{4}{9} \cdot \frac{y + 4x^2/(9y)}{y^2} = -\frac{4}{9} \cdot \frac{9y^2 + 4x^2}{9y^3}$.

Using $4x^2/9 + y^2 = 4$ (rearranging the ellipse equation): $4x^2 + 9y^2 = 36$.

$d^2y/dx^2 = -\frac{4}{9} \cdot \frac{36}{9y^3} = -\frac{16}{9y^3}$.

Upper ellipse ($y > 0$): $d^2y/dx^2 < 0$ (concave down). Lower ellipse ($y < 0$): $d^2y/dx^2 > 0$ (concave up).

5. Ask Why: Why Two Applications of Implicit Differentiation?

The second derivative $d^2y/dx^2$ is $\dfrac{d}{dx}\left[\dfrac{dy}{dx}\right]$. Since $dy/dx$ is a function of $x$ (involving $y(x)$), differentiating it with respect to $x$ requires the same chain rule and implicit-differentiation technique used in the first step.

The two-step structure is unavoidable: each derivative step uses the previous derivative result.

Named Misconception: Stopping After One Differentiation

A common error: computing $dy/dx$ correctly, then reporting that as the answer to “find $d^2y/dx^2$.” These are different quantities. $dy/dx$ is the slope; $d^2y/dx^2$ is the rate of change of the slope, which governs concavity.

Common Errors

Common Misconceptions

Leveled Practice

Level 1 -- Computing $d^2y/dx^2$

Problem 1. Find $\dfrac{d^2y}{dx^2}$ for $x^2 + y^2 = 9$ at $(0, 3)$.

Problem 2. Find $\dfrac{d^2y}{dx^2}$ for $y^2 = 4x$ (parabola) and evaluate at $(1, 2)$.

Level 2 -- Applying to Concavity

Problem 3. For $x^2 - y^2 = 1$ (hyperbola), find $d^2y/dx^2$ and determine the concavity of the upper branch.

Level 3 -- Low-Floor-High-Ceiling Extension

Problem 4 (Extension).

(a) (Floor) For $y^3 = x$, find $dy/dx$ and $d^2y/dx^2$ explicitly (solve for $y = x^{1/3}$) and confirm they match the implicit differentiation result.

(b) (Mid) Show that for any conic $ax^2 + bxy + cy^2 = d$, the second derivative $d^2y/dx^2$ can be expressed in terms of $a$, $b$, $c$, $d$, and the original conic equation.

(c) (Ceiling) The curvature of a curve at a point is $\kappa = |y''|/(1+(y')^2)^{3/2}$. For the circle $x^2+y^2=r^2$, compute $\kappa$ at any point and verify it equals $1/r$ (the reciprocal of the radius).

Mastery Checklist

• Find $dy/dx$ by implicit differentiation
• Differentiate $dy/dx$ with respect to $x$ a second time using the chain and quotient rules
• Substitute the expression for $dy/dx$ to eliminate it from the result
• Use the original curve equation to simplify the final expression
• Determine concavity of an implicit curve at a specific point

Mental Model

The second implicit derivative requires two complete rounds of implicit differentiation, each building on the last. The first round gives a slope formula $y' = f(x,y)$. The second round differentiates $f(x,y)$ with respect to $x$, treating $y$ as a function and applying the chain rule wherever $y$ appears. The result still contains $y'$, which you then replace using the first-round formula.

The circle result $d^2y/dx^2 = -r^2/y^3$ is a memorable test case: the sign confirms concavity (negative for upper arc), and the magnitude captures curvature.

Connections

Within MATH161

• Concavity and inflection points: The sign of $d^2y/dx^2$ tells you concavity on an implicit curve, exactly as for explicit functions.
• Curvature: In more advanced work, the curvature formula $\kappa = |y''|/(1+(y')^2)^{3/2}$ requires both first and second implicit derivatives.

Back to Calculus I Skills | Previous: Tangent Lines to Implicit Curves