Implicit Differentiation
Comprehensive Worksheet

Intermediate–Advanced ⌛ Est. 3–4 Hours 75 Problems

A rigorous, publication-grade resource covering implicit differentiation from first principles through advanced multi-rule applications, curve analysis, and AP-level exam strategy. Suitable for AP Calculus AB/BC, university Calculus I and II, and students seeking genuine mastery.

Learning Objectives

Introduction

There is a quiet revolution that happens somewhere in every calculus course — the moment a student realizes that not every curve on the coordinate plane can be described by a neat formula of the form $y = f(x)$. A circle, for instance. Or a folium. Or the graph of $\sin(xy) = x^2 - y$. These curves exist. They have slopes at their points. Yet no amount of algebraic cleverness will isolate $y$ on one side of the equation in a usable closed form. The question, then, is not whether to differentiate these relations. It is how.

Implicit differentiation is the answer. Its intellectual roots trace back to Leibniz himself, who in the late seventeenth century understood that the relationship between two variables could be far richer than a simple functional assignment. When Newton and Leibniz built the calculus, they envisioned it as a tool for understanding rates of change in general — not just for functions in the modern sense. Implicit differentiation is, in a way, the most faithful expression of that original vision.

The technique rests on a single, elegant idea: we treat $y$ as a function of $x$ — even if we cannot write out what that function is explicitly — and we differentiate both sides of an equation with respect to $x$. The chain rule then requires that every derivative involving $y$ carries a factor of $dy/dx$, which is precisely the quantity we want to find. After differentiating both sides, we collect all $dy/dx$ terms on one side and solve algebraically. The result is a formula for $dy/dx$ that typically contains both $x$ and $y$ — not a defect, but a feature. The slope at any point on the curve can be found by substituting the coordinates of that point.

The technique is not merely a computational trick. It is a profound extension of the very concept of a derivative. In multivariable calculus, implicit differentiation evolves into the implicit function theorem — one of the cornerstones of advanced analysis. In differential equations, it underlies the method of exact equations. In physics, it is used whenever two dynamical quantities are constrained by a physical law and both are changing in time (the machinery known as related rates is implicit differentiation in disguise). The curve in a phase diagram. The constraint surface in a thermodynamic system. The level set in an optimization problem. All of these are implicitly defined relations, and all of them can be interrogated using the techniques you will practice here.

This worksheet is designed for students who have already encountered basic differentiation rules — the power rule, product and quotient rules, and the chain rule — and are now ready to see those tools applied in a setting that demands greater structural awareness. If you can differentiate $f(x) = \sin(x^3)$ and correctly account for the chain rule, you are prepared to begin. If you can also differentiate $f(x) = x^2 e^{\sin x}$ without hesitation, you will find most of this worksheet deeply satisfying rather than overwhelming.

Mastery of implicit differentiation looks like this: you encounter a relation between $x$ and $y$, your eye immediately identifies which terms require the chain rule and which require the product rule, you differentiate systematically without losing sign or coefficient, and you isolate $dy/dx$ with algebraic confidence. You can find slopes, write tangent line equations, classify horizontal and vertical tangencies, and compute second derivatives. That is the full picture, and this worksheet builds toward it in stages.

Conceptual Foundation

Formal Definition and Core Procedure

The Implicit Differentiation Procedure

Given a relation $F(x,y) = 0$ where $y$ is implicitly a differentiable function of $x$:

$$\frac{d}{dx}[F(x,y)] = 0$$

Differentiate every term with respect to $x$. Any term involving $y$ (or a function of $y$) requires the chain rule, producing a factor of $\dfrac{dy}{dx}$. Collect all $\dfrac{dy}{dx}$ terms, factor, and solve.

The formal justification comes from the chain rule applied to composite functions. If $y = y(x)$ is differentiable, then for any differentiable function $g$:

$$\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$$

This is the chain rule with outer function $g$ and inner function $y(x)$. Implicit differentiation is simply the disciplined application of this rule to every $y$-containing term in an equation, combined with the linearity of differentiation (which means we can differentiate an equation term-by-term).

When to Use Implicit Differentiation

SituationApproachReason
$y$ is isolated (e.g., $y = 3x^2 - x$)Direct differentiationSimpler; no benefit to implicit method
$y$ cannot be isolated (e.g., $x^3 + y^3 = 6xy$)Implicit differentiationExplicit form does not exist in closed form
$y$ can be isolated but result is messy (e.g., $\sqrt{x} + \sqrt{y} = 4$)Either; implicit is often cleanerAvoids differentiating a messy radical
Both $x$ and $y$ appear as functions of $t$ (related rates)Implicit differentiation w.r.t. $t$Relates $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Structural Pattern Recognition

Before applying the procedure, scan the equation for its structural features. Terms that are purely in $x$ (like $x^3$ or $\cos x$) differentiate normally. Terms purely in $y$ (like $y^4$ or $e^y$) produce a $dy/dx$ factor. The real complexity — and the most common source of errors — comes from mixed terms like $x^2 y$, $xy^3$, or $\frac{x}{y}$, which require the product or quotient rule on top of the chain rule.

Professor's Note: Before you write a single derivative symbol, underline every $y$ (or function of $y$) in the equation. This makes it visually clear which terms will generate a $dy/dx$ factor, reducing the chance of forgetting one in a hurry.

The Second Derivative

Finding $d^2y/dx^2$ implicitly requires differentiating the expression for $dy/dx$ again with respect to $x$. Since $dy/dx$ typically contains both $x$ and $y$, this second differentiation will itself require implicit differentiation — and will generate another $dy/dx$ term, into which you substitute the first derivative expression. The algebra intensifies, but the logic does not.

Comparison to Related Techniques

Students sometimes confuse implicit differentiation with logarithmic differentiation. The techniques overlap when the equation can be made implicit by taking a logarithm (useful for expressions like $y = x^x$), but they are distinct strategies. Logarithmic differentiation is typically chosen to simplify products and powers; implicit differentiation is chosen because $y$ cannot be isolated. For the chain rule, the connection is even closer: implicit differentiation simply is the chain rule, applied systematically across an entire equation rather than to a single composite expression.

Worked Examples

Example 1 — Foundational: Circle

Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$.

Step 1 — Differentiate both sides w.r.t. $x$

$$\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]$$

The left side: $x^2$ differentiates to $2x$. For $y^2$, the chain rule gives $2y \cdot \dfrac{dy}{dx}$. The right side is constant, so its derivative is $0$.

$$2x + 2y\frac{dy}{dx} = 0$$
Step 2 — Isolate $dy/dx$

Subtract $2x$ from both sides: $2y\dfrac{dy}{dx} = -2x$. Divide by $2y$ (assuming $y \neq 0$):

$$\frac{dy}{dx} = -\frac{x}{y}$$
Why this makes sense

At the point $(3, 4)$ on the circle, $\dfrac{dy}{dx} = -\dfrac{3}{4}$. This is indeed the slope of the tangent to the unit circle at that point — the tangent is perpendicular to the radius, which has slope $\dfrac{4}{3}$, and $-\dfrac{3}{4}$ is its negative reciprocal. ✓

Answer: $\dfrac{dy}{dx} = -\dfrac{x}{y}$
Example 2 — Mixed Term: Product Rule Required

Find $\dfrac{dy}{dx}$ for $x^2 y + y^3 = 7$.

Step 1 — Identify term types

$x^2 y$ is a mixed product — it requires the product rule. $y^3$ is a pure $y$-term — it requires the chain rule only. The right side is constant.

Step 2 — Differentiate term by term

For $x^2 y$: Product rule with $u = x^2$, $v = y$.

$$\frac{d}{dx}[x^2 y] = 2x \cdot y + x^2 \cdot \frac{dy}{dx}$$

For $y^3$: Chain rule.

$$\frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}$$

Putting it together:

$$2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$$
Step 3 — Collect and factor $dy/dx$
$$\frac{dy}{dx}(x^2 + 3y^2) = -2xy$$ $$\frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}$$
Answer: $\dfrac{dy}{dx} = \dfrac{-2xy}{x^2 + 3y^2}$
Example 3 — Tangent Line at a Point

Find the equation of the tangent line to $x^3 + y^3 = 6xy$ (the folium of Descartes) at the point $(3, 3)$.

Step 1 — Verify the point lies on the curve

$3^3 + 3^3 = 27 + 27 = 54$ and $6(3)(3) = 54$. ✓

Step 2 — Differentiate implicitly
$$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$
Step 3 — Solve for $dy/dx$
$$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$$ $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$ $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$
Step 4 — Evaluate at $(3,3)$
$$\frac{dy}{dx}\bigg|_{(3,3)} = \frac{2(3) - 9}{9 - 6} = \frac{-3}{3} = -1$$
Step 5 — Write tangent line equation

Using point-slope form: $y - 3 = -1(x - 3)$, so $y = -x + 6$.

Answer: Tangent line is $y = -x + 6$
Example 4 — Trigonometric Relation

Find $\dfrac{dy}{dx}$ for $\sin(x + y) = x \cos y$.

Step 1 — Differentiate left side

$\dfrac{d}{dx}[\sin(x+y)]$: outer function is $\sin$, inner function is $x+y$.

$$\cos(x+y) \cdot \frac{d}{dx}[x + y] = \cos(x+y)\left(1 + \frac{dy}{dx}\right)$$
Step 2 — Differentiate right side

$x\cos y$ is a product of $x$ and $\cos y$.

$$\frac{d}{dx}[x\cos y] = 1\cdot\cos y + x\cdot(-\sin y)\frac{dy}{dx} = \cos y - x\sin y\frac{dy}{dx}$$
Step 3 — Set equal and collect
$$\cos(x+y) + \cos(x+y)\frac{dy}{dx} = \cos y - x\sin y \frac{dy}{dx}$$ $$\frac{dy}{dx}\left[\cos(x+y) + x\sin y\right] = \cos y - \cos(x+y)$$ $$\frac{dy}{dx} = \frac{\cos y - \cos(x+y)}{\cos(x+y) + x\sin y}$$
Common Mistake Warning

Students often forget to differentiate the argument $(x+y)$ in $\sin(x+y)$, and write only $\cos(x+y)$ without the factor $\left(1 + \dfrac{dy}{dx}\right)$. This loses the $\dfrac{dy}{dx}$ term contributed by the left side.

Answer: $\dfrac{dy}{dx} = \dfrac{\cos y - \cos(x+y)}{\cos(x+y) + x\sin y}$
Example 5 — Exponential with Mixed Variables

Find $\dfrac{dy}{dx}$ for $e^{xy} = x + y$.

Step 1 — Differentiate left side

$\dfrac{d}{dx}[e^{xy}]$: outer function is $e^u$, inner function is $u = xy$.

$$e^{xy} \cdot \frac{d}{dx}[xy] = e^{xy}\left(y + x\frac{dy}{dx}\right)$$
Step 2 — Differentiate right side
$$1 + \frac{dy}{dx}$$
Step 3 — Expand and collect
$$ye^{xy} + xe^{xy}\frac{dy}{dx} = 1 + \frac{dy}{dx}$$ $$xe^{xy}\frac{dy}{dx} - \frac{dy}{dx} = 1 - ye^{xy}$$ $$\frac{dy}{dx}(xe^{xy} - 1) = 1 - ye^{xy}$$ $$\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} - 1}$$
Answer: $\dfrac{dy}{dx} = \dfrac{1 - ye^{xy}}{xe^{xy} - 1}$
Example 6 — Second Derivative Implicitly

Find $\dfrac{d^2y}{dx^2}$ for $x^2 + y^2 = r^2$ (circle of radius $r$).

Step 1 — First derivative (established earlier)
$$\frac{dy}{dx} = -\frac{x}{y}$$
Step 2 — Differentiate again w.r.t. $x$

Using the quotient rule on $-x/y$:

$$\frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2}$$
Step 3 — Simplify
$$= -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{\frac{y^2 + x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}$$

Since $x^2 + y^2 = r^2$:

$$\frac{d^2y}{dx^2} = -\frac{r^2}{y^3}$$
Geometric Interpretation

At the top of the circle ($y = r > 0$), $\dfrac{d^2y}{dx^2} = -\dfrac{r^2}{r^3} = -\dfrac{1}{r} < 0$, confirming the curve is concave down there — as expected for the upper semicircle.

Answer: $\dfrac{d^2y}{dx^2} = -\dfrac{r^2}{y^3}$
Example 7 — Conceptual Trap: Horizontal and Vertical Tangents

For the curve $y^2 - x^2 y = 4$, find all points where the tangent line is horizontal or vertical.

Step 1 — Find $dy/dx$

Differentiating: $2y\dfrac{dy}{dx} - \left(2xy + x^2\dfrac{dy}{dx}\right) = 0$

$$\frac{dy}{dx}(2y - x^2) = 2xy$$ $$\frac{dy}{dx} = \frac{2xy}{2y - x^2}$$
Step 2 — Horizontal tangents: numerator = 0 (denominator ≠ 0)

$2xy = 0 \Rightarrow x = 0$ or $y = 0$.

If $x = 0$: the original equation gives $y^2 = 4$, so $y = \pm 2$. Check denominator at $(0, 2)$: $2(2) - 0 = 4 \neq 0$. ✓ Similarly for $(0, -2)$. Horizontal tangents at $(0, 2)$ and $(0, -2)$.

If $y = 0$: the equation gives $0 - 0 = 4$, which is impossible. So $y = 0$ yields no points on the curve.

Step 3 — Vertical tangents: denominator = 0 (numerator ≠ 0)

$2y - x^2 = 0 \Rightarrow y = \dfrac{x^2}{2}$. Substitute into the original equation:

$$\left(\frac{x^2}{2}\right)^2 - x^2\cdot\frac{x^2}{2} = 4 \implies \frac{x^4}{4} - \frac{x^4}{2} = 4 \implies -\frac{x^4}{4} = 4 \implies x^4 = -16$$

No real solutions. Therefore there are no vertical tangents on this curve.

Trap to Avoid

Students often set numerator = 0 to find vertical tangents and denominator = 0 for horizontal ones. This is backwards. Horizontal tangent $\Leftrightarrow$ slope is zero $\Leftrightarrow$ numerator is zero. Vertical tangent $\Leftrightarrow$ slope is undefined $\Leftrightarrow$ denominator is zero.

Horizontal tangents at $(0, 2)$ and $(0, -2)$; no vertical tangents.
Example 8 — Multi-Rule: Logarithm with Mixed Variables

Find $\dfrac{dy}{dx}$ for $\ln(x^2 + y^2) = 2\arctan\!\left(\dfrac{y}{x}\right)$ (equation of a logarithmic spiral in Cartesian form).

Step 1 — Differentiate left side
$$\frac{d}{dx}\left[\ln(x^2+y^2)\right] = \frac{2x + 2y\frac{dy}{dx}}{x^2 + y^2}$$
Step 2 — Differentiate right side

$\dfrac{d}{dx}\left[\arctan\!\left(\dfrac{y}{x}\right)\right]$. Let $u = y/x$. Then $\dfrac{d}{dx}[\arctan u] = \dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}$.

$$\frac{du}{dx} = \frac{\frac{dy}{dx}\cdot x - y}{x^2}$$ $$\frac{d}{dx}\left[\arctan\!\left(\frac{y}{x}\right)\right] = \frac{1}{1 + y^2/x^2}\cdot\frac{x\frac{dy}{dx}-y}{x^2} = \frac{x^2}{x^2+y^2}\cdot\frac{x\frac{dy}{dx}-y}{x^2} = \frac{x\frac{dy}{dx}-y}{x^2+y^2}$$
Step 3 — Set equal
$$\frac{2x + 2y\frac{dy}{dx}}{x^2 + y^2} = \frac{2(x\frac{dy}{dx}-y)}{x^2+y^2}$$

Multiply both sides by $x^2 + y^2$:

$$2x + 2y\frac{dy}{dx} = 2x\frac{dy}{dx} - 2y$$ $$2x + 2y = 2x\frac{dy}{dx} - 2y\frac{dy}{dx} = 2\frac{dy}{dx}(x-y)$$ $$\frac{dy}{dx} = \frac{x+y}{x-y}$$
Alternate Method Note

This result is most cleanly verified by converting to polar coordinates, where the logarithmic spiral satisfies $r = e^\theta$ and $dr/d\theta = r$. The Cartesian slope can be recovered via $dy/dx = (r\cos\theta + r\sin\theta)/(r\cos\theta - r\sin\theta) = (x+y)/(x-y)$. Both routes agree — a satisfying confirmation.

Answer: $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$

Practice Problems

A. Core Skill Development

Foundational

Find $dy/dx$ using implicit differentiation. These problems involve single-rule applications and straightforward algebraic isolation.

  1. $x^2 + y^2 = 49$
  2. $3x + 5y = 15$
  3. $x^3 + y^3 = 8$
  4. $x^2 - y^2 = 1$ (unit hyperbola)
  5. $\sqrt{x} + \sqrt{y} = 4$
  6. $y^4 = x^3$
  7. $x^2 y = 10$
  8. $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ (ellipse)
  9. $x^{1/3} + y^{1/3} = 1$
  10. $y^2 + 2y = x$

B. Structural Recognition

Intermediate

Each problem contains a structural feature — product of $x$ and $y$, a quotient, or a nested composition — that must be identified before differentiating. Write out your rule identification before computing.

  1. $xy^2 + x^2 y = 6$
  2. $x^2 e^y = y + 1$
  3. $\ln(xy) = x - y$
  4. $y\sin x = x^2 - y^2$
  5. $\tan(y) = x^2 + 1$
  6. $e^y + ye^x = 3$
  7. $\cos(x - y) = y^2$
  8. $\sqrt{xy} = x - y$
  9. $x\ln y = y^2 - 1$
  10. $\dfrac{y}{x+y} = x^2$

C. Multi-Step Problems

Intermediate–Advanced

These problems involve multiple rules applied simultaneously, or require more careful algebraic manipulation to isolate $dy/dx$. Work systematically.

  1. $\sin(xy) + \cos(x+y) = 1$
  2. $x^2 y^3 - 3xy = 4$
  3. $e^{x+y} = x^2 + y^2$
  4. $\arctan(y/x) = \ln\sqrt{x^2+y^2}$
  5. $x^3 + y^3 = 3xy + 1$ (find $dy/dx$ and points of horizontal tangency)
  6. $(x^2 + y^2)^2 = 4xy$ (lemniscate-like curve; find slope at $(1,1)$)
  7. $y = x^y$ (use logarithmic + implicit differentiation)
  8. $\sin^2 x + \cos^2 y = 1$
  9. $\arcsin(y) = x^2 \arccos(y)$
  10. $\ln(x^2 + y) = xy^2$

D. Second Derivatives and Concavity

Advanced

Find $d^2y/dx^2$ and, where specified, interpret the concavity geometrically.

  1. $4x^2 + 9y^2 = 36$ (ellipse). Find $d^2y/dx^2$ and verify that it is negative on the upper half.
  2. $xy = 4$ (hyperbola). Find $d^2y/dx^2$.
  3. $x^3 - y^3 = 1$. Find $d^2y/dx^2$.
  4. $\sin y = x$. Find $d^2y/dx^2$ in terms of $y$.
  5. $y^2 = x^3$ (semi-cubical parabola). Find $d^2y/dx^2$ and determine where the curve is concave up.

E. Tangent Lines, Normal Lines, and Curve Analysis

Intermediate

Recall: a normal line at a point is perpendicular to the tangent line at that point.

  1. Find the equation of the tangent line to $x^2 + xy + y^2 = 7$ at the point $(2, 1)$.
  2. Find the equation of the normal line to $y^2 = x^3 - x$ at $(2, \sqrt{6})$.
  3. At what points does $\dfrac{x^2}{4} + y^2 = 1$ have vertical tangent lines?
  4. Find all points on $x^2 - 2xy + 4y^2 = 4$ where the tangent line is horizontal.
  5. Show that the tangent line to $x^{2/3} + y^{2/3} = a^{2/3}$ (astroid) at point $(x_0, y_0)$ has $x$-intercept $a^{2/3}x_0^{1/3}$ and $y$-intercept $a^{2/3}y_0^{1/3}$.

F. Exam-Level Questions

AP / University Exam Style

These problems combine multiple skills or involve reasoning steps that are commonly tested under time pressure.

  1. Let $F(x,y) = x^3 y - y^3 x = 0$ define $y$ implicitly as a function of $x$. (a) Find $dy/dx$. (b) Find $d^2y/dx^2$. (c) Determine whether the curve is concave up or down at the point $(1, 0)$ — or explain why the test fails.
  2. The curve $C$ is defined by $e^y - y = x^2 - 1$. (a) Show that the point $(1, 0)$ lies on $C$. (b) Find the slope of $C$ at $(1, 0)$. (c) Find the second derivative at $(1, 0)$ and classify the concavity.
  3. Suppose $x$ and $y$ satisfy $x^2 + 4xy - y^2 = 8$. Using implicit differentiation, find all values of $x$ at which the curve has a horizontal tangent, assuming real solutions exist.
  4. (Free Response style) The equation $x \sin y + y \cos x = \pi/2$ implicitly defines $y$ as a function of $x$ near the point $\left(\pi/2, \pi/2\right)$. (a) Verify the point is on the curve. (b) Find the slope of the tangent line. (c) Write the equation of the tangent line.
  5. (Multiple concept) A particle moves along the curve $x^2 y^2 = 16$. At the point $(2, 2)$, $x$ is increasing at $3$ units/sec. Find the rate of change of $y$ at that instant. (Differentiate implicitly with respect to time $t$.)

G. Challenge Set

Challenge

For students who want to push beyond standard course expectations. These problems require deeper structural insight or more involved algebra.

  1. For $x^4 + y^4 = 1$, find the coordinates of all points where the curvature is maximized. (Curvature $\kappa = \dfrac{|y''|}{(1+(y')^2)^{3/2}}$; first find $y'$ and $y''$ implicitly.)
  2. The equation $y^5 + y = x$ cannot be solved for $y$ explicitly. Find a formula for $d^ny/dx^n$ in terms of $y$ and lower derivatives, for $n = 1, 2, 3$.
  3. Show that if $F(x,y) = 0$ defines $y$ as a differentiable function of $x$, then $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$, where $F_x$ and $F_y$ are partial derivatives. Apply this to $F = x^2 + 3xy - y^3 + 7$ and verify your result against the standard implicit method.

Error Diagnosis Problems

Each problem below shows a student's incorrect work. Identify the error, explain why it is wrong, and produce the correct solution.

  1. Student's work: For $x^2 + y^2 = 1$, they wrote $2x + 2y = 0$, then $y = -x$.
    Identify the error and correct it.
  2. Student's work: For $xy = 5$, they wrote $\dfrac{d}{dx}[xy] = 1 \cdot 1 = 1$, so $1 = 0$, a contradiction.
    Identify the error and correct it.
  3. Student's work: For $\cos(y) = x^2$, they wrote $-\sin(y) = 2x$, giving $y = -\arcsin(2x)$.
    Is this correct? If not, identify the error.

Conceptual Reasoning Questions

These questions test understanding rather than computation. Full sentences are expected.

  1. Explain in words (without formulas) why differentiating $y^3$ with respect to $x$ produces $3y^2 \dfrac{dy}{dx}$ rather than just $3y^2$. What principle is at work?
  2. A student claims: "If a curve has a horizontal tangent at a point, then $dy/dx = 0$ at that point, so the numerator of my $dy/dx$ expression must be zero — but I don't need to check whether the denominator is also zero." Is this reasoning sound? Construct a counterexample or confirm it.
  3. Why does the result of implicit differentiation typically contain both $x$ and $y$, rather than just $x$ alone? What does this mean geometrically when evaluating slope at a specific point?
  4. Suppose you correctly find $\dfrac{dy}{dx} = \dfrac{2x - y}{x - 2y}$. At what points on the curve does the formula fail to be defined, and what geometric feature do those points represent?
  5. The equation $x^2 + y^2 = -4$ has no real solutions. What happens if you try to apply implicit differentiation to it? What does the resulting expression for $dy/dx$ tell you, if anything?

Real-World Modeling Problems

Physics: Planetary Orbit

The orbit of a planet around the Sun is modeled by the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes in AU (astronomical units).

  1. Show that the slope of the orbit at any point $(x,y)$ is $\dfrac{dy}{dx} = -\dfrac{b^2 x}{a^2 y}$.
  2. For Earth's orbit, $a = 1.000$ AU and $b = 0.9998$ AU. At the point where $x = 0.5$ AU and $y > 0$, find $y$ (to 4 decimal places) and then compute the slope of Earth's orbital path. Interpret the sign of the slope in the context of Earth's motion.

Engineering: Implicit Constraint Surface

In a pressurized gas system, the relationship between pressure $P$ (in atm) and volume $V$ (in liters) for a Van der Waals gas satisfies:

$$\left(P + \frac{a}{V^2}\right)(V - b) = nRT$$

where $a$, $b$, $n$, $R$, $T$ are constants.

  1. Differentiate both sides with respect to $V$, treating $P$ as an implicitly defined function of $V$. Solve for $\dfrac{dP}{dV}$.
  2. Interpret the sign of $\dfrac{dP}{dV}$ physically. Under what algebraic condition (on $a$, $b$, $P$, $V$) would $\dfrac{dP}{dV} = 0$? What physical state does this represent?

Economics: Production Isoquant

A firm's production of $Q$ units is modeled by the Cobb-Douglas isoquant $K^{0.4} L^{0.6} = Q_0$, where $K$ is capital and $L$ is labor, and $Q_0$ is a fixed output level.

  1. Find $\dfrac{dK}{dL}$ (the marginal rate of technical substitution, MRTS) using implicit differentiation. Express in terms of $K$ and $L$.
  2. Interpret the MRTS economically: what does $\dfrac{dK}{dL} = -2$ mean in this context, and why is it negative?
  3. At the production point $K = 100$, $L = 200$, compute the MRTS numerically.

Biology: Population Constraint

Two competing species have population levels $x$ and $y$ (in thousands) that satisfy the constraint $x^2 + xy + y^2 = 3Q$ for some ecological constant $Q > 0$.

  1. Find $\dfrac{dy}{dx}$ and interpret it as the rate at which species $y$'s population changes with respect to species $x$'s population along the ecological boundary.
  2. At what population ratios $x:y$ is $\dfrac{dy}{dx} = -1$? What does this mean for the two species?

Optics: Refraction Curve

The boundary of an optical lens is described by the curve $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$. A light ray strikes the lens at the point $(3, 3.2)$.

  1. Verify that $(3, 3.2)$ is on the lens curve (approximately). Find the slope of the tangent to the lens at that point. This tangent gives the local orientation of the lens surface.
  2. Find the slope of the normal to the lens at that point. The normal determines the angle of incidence for the incoming light ray.

Exam Traps and Strategic Advice

Trap 1 — The Missing $dy/dx$
The most frequent error at every level: differentiating $y^n$ and writing $ny^{n-1}$ without the trailing $dy/dx$ factor. On AP free-response, this costs the "correct application of the chain rule" scoring point — often 1 of 4 available points on the problem. Train yourself to write $dy/dx$ as a reflex the moment you touch a $y$-term.
Trap 2 — Forgetting the Product Rule on Mixed Terms
A term like $x^2 y^3$ requires both the product rule and the chain rule. Students in a hurry write $2xy^3$ and omit the $3x^2y^2\,dy/dx$ part. Before differentiating any term, ask: "Are $x$ and $y$ multiplied here?" If yes, product rule first.
Trap 3 — Not Substituting the Point
Many problems ask for the slope at a specific point. Students who correctly find $dy/dx = f(x,y)$ and then stop have only completed half the work. Substitute the given $(x_0, y_0)$ into the derivative expression. This step is required for full credit and is surprisingly often forgotten under pressure.
Trap 4 — Horizontal vs. Vertical Tangent Confusion
Horizontal tangent: slope = 0, so set the numerator of $dy/dx$ to zero. Vertical tangent: slope is undefined, so set the denominator to zero. These are swapped by a distressing number of students every year. Commit the logic to memory, not the rule-of-thumb.
Trap 5 — The Second Derivative Shortcut Illusion
Finding $d^2y/dx^2$ requires differentiating $dy/dx$ again, substituting the expression for $dy/dx$ into the result. Many students stop after the first differentiation and present it as the second derivative. Check your work: $d^2y/dx^2$ should be expressible in $x$ and $y$ only, not in $dy/dx$.
AP Strategy: Show All Steps
AP graders award points for process, not just the final answer. Even if you make an algebraic error isolating $dy/dx$, you can earn the point for correctly differentiating both sides if those steps are visible. Write out every intermediate line. Never skip from "differentiated equation" to "final answer" in one jump.
Time-Saver: Partial Derivative Formula
If $F(x,y) = 0$, then $dy/dx = -F_x/F_y$. This formula — provable from the implicit function theorem — lets you bypass algebra in many problems. If $F = x^3 + y^3 - 6xy$, then $F_x = 3x^2 - 6y$, $F_y = 3y^2 - 6x$, and $dy/dx = -(3x^2-6y)/(3y^2-6x)$ immediately. Useful under time pressure once mastered.
University-Level: Domain Restrictions
At the university level, you may be asked about the existence of $dy/dx$, not just its value. By the implicit function theorem, $dy/dx$ exists near $(x_0, y_0)$ provided $F_y(x_0, y_0) \neq 0$. If $F_y = 0$ at a point, the implicit function theorem does not guarantee a local function $y(x)$ — a fact occasionally tested in honours calculus.

Mixed Rule Detection

For each problem below, before computing: (i) name every differentiation rule you will need, and (ii) identify which terms require each rule. Then solve.

  1. $x^2 \sin y + y^2 \cos x = 1$
  2. $e^{xy} \ln(x + y) = 3$
  3. $\sqrt{x^2 + y^2} = \arctan(y/x)$
  4. $\left(\dfrac{x}{y}\right)^2 + \ln y = x$
  5. $y^x = x^y$ (take logarithms of both sides first; treat both $y$ and $x$ as variable)
  6. $\cos(e^y) = x^3 + y^2$
  7. $\arcsin(xy) = \dfrac{\pi}{4}$
  8. $x^2\arctan y - y^2\arctan x = 0$

Full Answer Key

Section A — Core Skill Development

A1

Differentiate: $2x + 2y\dfrac{dy}{dx} = 0$

Isolate: $\dfrac{dy}{dx} = -\dfrac{x}{y}$

$\dfrac{dy}{dx} = -\dfrac{x}{y}$
A2

$3 + 5\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = -\dfrac{3}{5}$
A3

$3x^2 + 3y^2\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = -\dfrac{x^2}{y^2}$
A4

$2x - 2y\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = \dfrac{x}{y}$
A5

$\dfrac{1}{2\sqrt{x}} + \dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{\sqrt{y}}{\sqrt{x}}$

$\dfrac{dy}{dx} = -\sqrt{\dfrac{y}{x}}$
A6

$4y^3\dfrac{dy}{dx} = 3x^2$

$\dfrac{dy}{dx} = \dfrac{3x^2}{4y^3}$
A7

$2xy + x^2\dfrac{dy}{dx} = 0$ (product rule on $x^2y$)

$\dfrac{dy}{dx} = -\dfrac{2y}{x}$
A8

$\dfrac{x}{8} + \dfrac{2y}{9}\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = -\dfrac{9x}{16y}$
A9

$\dfrac{1}{3}x^{-2/3} + \dfrac{1}{3}y^{-2/3}\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = -\dfrac{y^{2/3}}{x^{2/3}} = -\left(\dfrac{y}{x}\right)^{2/3}$
A10

$2y\dfrac{dy}{dx} + 2\dfrac{dy}{dx} = 1$, so $\dfrac{dy}{dx}(2y+2) = 1$

$\dfrac{dy}{dx} = \dfrac{1}{2(y+1)}$

Section B — Structural Recognition (Selected)

B11 — $xy^2 + x^2y = 6$

Product rule on both terms:

$\left(y^2 + x\cdot 2y\dfrac{dy}{dx}\right) + \left(2xy + x^2\dfrac{dy}{dx}\right) = 0$

$\dfrac{dy}{dx}(2xy + x^2) = -y^2 - 2xy$

$\dfrac{dy}{dx} = \dfrac{-y^2 - 2xy}{2xy + x^2} = \dfrac{-y(y+2x)}{x(2y+x)}$
B12 — $x^2 e^y = y+1$

Product + chain rule: $2xe^y + x^2 e^y\dfrac{dy}{dx} = \dfrac{dy}{dx}$

$2xe^y = \dfrac{dy}{dx}(1 - x^2 e^y)$

$\dfrac{dy}{dx} = \dfrac{2xe^y}{1 - x^2 e^y}$
B13 — $\ln(xy) = x - y$

Use $\ln(xy) = \ln x + \ln y$. Differentiate: $\dfrac{1}{x} + \dfrac{1}{y}\dfrac{dy}{dx} = 1 - \dfrac{dy}{dx}$

$\dfrac{dy}{dx}\!\left(\dfrac{1}{y}+1\right) = 1 - \dfrac{1}{x}$

$\dfrac{dy}{dx} = \dfrac{y(x-1)}{x(y+1)}$

Section C — Multi-Step (Selected)

C21 — $\sin(xy) + \cos(x+y) = 1$

$\cos(xy)\!\left(y + x\dfrac{dy}{dx}\right) - \sin(x+y)\!\left(1+\dfrac{dy}{dx}\right) = 0$

$\dfrac{dy}{dx}\left[x\cos(xy) - \sin(x+y)\right] = \sin(x+y) - y\cos(xy)$

$\dfrac{dy}{dx} = \dfrac{\sin(x+y) - y\cos(xy)}{x\cos(xy) - \sin(x+y)}$
C22 — $x^2 y^3 - 3xy = 4$

$2xy^3 + 3x^2 y^2\dfrac{dy}{dx} - 3y - 3x\dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx}(3x^2y^2 - 3x) = 3y - 2xy^3$

$\dfrac{dy}{dx} = \dfrac{3y - 2xy^3}{3x^2y^2 - 3x}$
C23 — $e^{x+y} = x^2 + y^2$

$e^{x+y}\!\left(1+\dfrac{dy}{dx}\right) = 2x + 2y\dfrac{dy}{dx}$

$\dfrac{dy}{dx}(e^{x+y} - 2y) = 2x - e^{x+y}$

$\dfrac{dy}{dx} = \dfrac{2x - e^{x+y}}{e^{x+y} - 2y}$

Section D — Second Derivatives (Selected)

D31 — Ellipse $4x^2 + 9y^2 = 36$

First derivative: $8x + 18y\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{4x}{9y}$

Second derivative (quotient rule): $\dfrac{d^2y}{dx^2} = -\dfrac{4\cdot 9y - 4x\cdot 9\dfrac{dy}{dx}}{81y^2} = -\dfrac{36y - 36x(-4x/9y)}{81y^2}$

$= -\dfrac{36y + 16x^2/y}{81y^2} = -\dfrac{36y^2 + 16x^2}{81y^3}$

Since $4x^2 + 9y^2 = 36$, we have $16x^2 = 4(4x^2) = 4(36-9y^2) = 144-36y^2$. Thus numerator $= 36y^2 + 144 - 36y^2 = 144$.

$\dfrac{d^2y}{dx^2} = -\dfrac{144}{81y^3} = -\dfrac{16}{9y^3}$. Since $y > 0$ on the upper half, $d^2y/dx^2 < 0$: concave down. ✓
D32 — $xy = 4$

$y + x\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{y}{x}$

$\dfrac{d^2y}{dx^2} = -\dfrac{\dfrac{dy}{dx}\cdot x - y}{x^2} = -\dfrac{-y/x \cdot x - y}{x^2} = -\dfrac{-2y}{x^2} = \dfrac{2y}{x^2}$

$\dfrac{d^2y}{dx^2} = \dfrac{2y}{x^2}$

Section E — Tangent and Normal Lines (Selected)

E36 — $x^2 + xy + y^2 = 7$ at $(2,1)$

Verify: $4 + 2 + 1 = 7$ ✓. Differentiate: $2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$

At $(2,1)$: $4 + 1 + 2\dfrac{dy}{dx} + 2\dfrac{dy}{dx} = 0 \Rightarrow 5 + 4\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{5}{4}$

Tangent: $y - 1 = -\dfrac{5}{4}(x-2) \Rightarrow y = -\dfrac{5}{4}x + \dfrac{7}{2}$

Tangent line: $y = -\dfrac{5}{4}x + \dfrac{7}{2}$

Section F — Exam-Level (Selected)

F41 — $x^3y - y^3x = 0$

(a) $3x^2y + x^3\dfrac{dy}{dx} - 3y^2\dfrac{dy}{dx}\cdot x - y^3 = 0$. Collect:

$\dfrac{dy}{dx}(x^3 - 3xy^2) = y^3 - 3x^2y$

$\dfrac{dy}{dx} = \dfrac{y^3 - 3x^2y}{x^3 - 3xy^2} = \dfrac{y(y^2-3x^2)}{x(x^2-3y^2)}$

(b) Second derivative requires differentiating again — algebraically intensive; result contains $x$, $y$, and the expression for $y'$, then substituting $y'$.

(c) At $(1,0)$: $\dfrac{dy}{dx} = \dfrac{0\cdot(-3)}{1\cdot 1} = 0$. For $d^2y/dx^2$, substitute $(1,0)$ into the differentiated first-derivative equation. The expression is finite and negative: $d^2y/dx^2 < 0$ at $(1,0)$ — concave down.

$\dfrac{dy}{dx} = \dfrac{y(y^2-3x^2)}{x(x^2-3y^2)}$; concave down at $(1,0)$.
F43 — $x^2 + 4xy - y^2 = 8$, horizontal tangents

Differentiate: $2x + 4y + 4x\dfrac{dy}{dx} - 2y\dfrac{dy}{dx} = 0$. Set $\dfrac{dy}{dx} = 0$:

$2x + 4y = 0 \Rightarrow y = -\dfrac{x}{2}$

Substitute into original: $x^2 + 4x(-x/2) - (x/2)^2 = 8 \Rightarrow x^2 - 2x^2 - x^2/4 = 8 \Rightarrow -\dfrac{5x^2}{4} = 8 \Rightarrow x^2 = -\dfrac{32}{5}$

No real solutions — this curve has no horizontal tangents.

No horizontal tangents exist on this curve (the equation yields $x^2 < 0$).

Error Diagnosis

ED49 — Circle error

Error: The student forgot the $dy/dx$ factor on $y^2$. The correct differentiation gives $2y\,dy/dx$, not $2y$. The equation should be $2x + 2y\,dy/dx = 0$, giving $dy/dx = -x/y$, not $y = -x$.

ED50 — $xy = 5$ error

Error: The student did not apply the product rule. $\dfrac{d}{dx}[xy] = y + x\dfrac{dy}{dx}$, not $1\cdot1$. The correct equation is $y + x\dfrac{dy}{dx} = 0$, giving $\dfrac{dy}{dx} = -\dfrac{y}{x}$.

ED51 — $\cos y = x^2$ error

Error: The student forgot $dy/dx$ on the left side. The chain rule gives $-\sin y\cdot\dfrac{dy}{dx} = 2x$, so $\dfrac{dy}{dx} = -\dfrac{2x}{\sin y}$. The student's step of solving for $y$ directly is also problematic — the implicit result depends on both $x$ and $y$.

Real-World Problems (Selected)

RW57 — Ellipse orbit slope

$\dfrac{2x}{a^2} + \dfrac{2y}{b^2}\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{b^2 x}{a^2 y}$ ✓

RW59 — Van der Waals $dP/dV$

Expand: $P(V-b) + \dfrac{a(V-b)}{V^2} = nRT$. Differentiate w.r.t. $V$:

$\dfrac{dP}{dV}(V-b) + P + a\left[\dfrac{V^2 - 2V(V-b)}{V^4}\right] = 0$

Simplify the $a$-term: $\dfrac{V^2-2V^2+2bV}{V^4} = \dfrac{-V+2b}{V^3}$

$\dfrac{dP}{dV} = -\dfrac{P + a\dfrac{2b-V}{V^3}}{V-b}$. This is negative in normal gas conditions (pressure drops as volume increases), consistent with physical intuition.
RW61 — Cobb-Douglas MRTS

$0.4K^{-0.6}L^{0.6}\,dK + 0.6K^{0.4}L^{-0.4}\,dL = 0$ (treat $Q_0$ as constant)

$\dfrac{dK}{dL} = -\dfrac{0.6K^{0.4}L^{-0.4}}{0.4K^{-0.6}L^{0.6}} = -\dfrac{0.6K}{0.4L} = -\dfrac{3K}{2L}$

MRTS $= -\dfrac{3K}{2L}$. At $K=100, L=200$: MRTS $= -\dfrac{300}{400} = -0.75$. One unit increase in labor replaces $0.75$ units of capital while maintaining output.

Quick Reference Answers — Remaining Problems

B14 $\dfrac{dy}{dx} = \dfrac{x^2 - y\sin x}{y\cos x + 2y\cos x}$ (simplify as needed)
B15 $\sec^2 y\,\dfrac{dy}{dx} = 2x$, so $\dfrac{dy}{dx} = 2x\cos^2 y$
B16 $e^y\dfrac{dy}{dx} + ye^x + e^x\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{ye^x}{e^y+e^x}$
B17 $\sin(x-y)\!\left(1-\dfrac{dy}{dx}\right) = 2y\dfrac{dy}{dx}$; isolate $\dfrac{dy}{dx}$
B18 $\dfrac{y+x\dfrac{dy}{dx}}{2\sqrt{xy}} = 1-\dfrac{dy}{dx}$; solve for $\dfrac{dy}{dx}$
B19 $\ln y + \dfrac{x}{y}\dfrac{dy}{dx} = 2y\dfrac{dy}{dx}$; isolate
B20 Quotient rule: $\dfrac{\dfrac{dy}{dx}(x+y)-y(1+\dfrac{dy}{dx})}{(x+y)^2} = 2x$; solve
C26 $(2x+2y\dfrac{dy}{dx})^2\cdot\dfrac{1}{2\sqrt{\ldots}} = \ldots$ — differentiate LHS as chain + power; set equal to RHS
C27 $y = x^y$: take $\ln$: $\ln y = y\ln x$. Differentiate: $\dfrac{y'}{y} = y'\ln x + \dfrac{y}{x}$
C28 $2\sin x\cos x - 2\cos y\sin y\,\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{\sin x\cos x}{\cos y\sin y}$
D33 $3x^2 - 3y^2\dfrac{dy}{dx}=0$; first der: $x^2/y^2$. Second: $\dfrac{2x\cdot y^2 - x^2\cdot 2y\dfrac{dy}{dx}}{y^4}$ — substitute $y' = x^2/y^2$
D34 $\cos y\,\dfrac{dy}{dx}=1$; $\dfrac{dy}{dx}=\sec y$; $\dfrac{d^2y}{dx^2}=\sec y\tan y\,\dfrac{dy}{dx}=\sec^2 y\tan y$
D35 $y'=\dfrac{3x^2}{2y}$; $y''=\dfrac{3}{2}\cdot\dfrac{2y\cdot 2x - x^2\cdot 2y'}{4y^2}=\dfrac{3(4xy-2x^2 y')}{8y^2}$; substitute $y'$; concave up where $y>0$
E37 Normal to $y^2=x^3-x$ at $(2,\sqrt{6})$: first find slope of tangent, then normal slope = negative reciprocal
E38 Vertical tangents of ellipse $\dfrac{x^2}{4}+y^2=1$: denominator $\rightarrow$ need $dy/dx$ undefined; $y=0$, so $x=\pm 2$
MRD68 $y^x = x^y$: $\ln y - y\ln x/x$... $\dfrac{dy}{dx} = \dfrac{y(y - x\ln y)}{x(x - y\ln x)}$

Frequently Asked Questions

What is implicit differentiation, in plain terms?
It is a method for finding the slope of a curve that is described by an equation in $x$ and $y$, without first solving for $y$. You differentiate both sides of the equation with respect to $x$, apply the chain rule to every term that contains $y$, and then solve algebraically for $dy/dx$.
When should I use implicit differentiation instead of solving for $y$ first?
Use implicit differentiation when $y$ cannot be isolated — or when solving for $y$ would produce a messy expression harder to differentiate than the implicit approach. For relations like $x^3 + y^3 = 6xy$ or $\sin(xy) = y + 1$, explicit isolation of $y$ is impossible, so implicit differentiation is the only viable path.
Why do I need to write $dy/dx$ when differentiating a $y$-term?
Because $y$ depends on $x$ (we assume $y$ is a function of $x$), and the chain rule requires that the derivative of a composition $g(y(x))$ includes the factor $dy/dx$. Omitting it is mathematically equivalent to pretending $y$ is a constant, which it is not.
How do I handle a term like $x^2 y^3$?
Apply the product rule, treating $x^2$ and $y^3$ as the two factors: $\dfrac{d}{dx}[x^2 y^3] = 2x\cdot y^3 + x^2\cdot 3y^2\dfrac{dy}{dx}$. The second part comes from the chain rule on $y^3$. Missing either summand is the most common error with mixed-variable terms.
Is implicit differentiation tested on the AP Calculus exam?
Yes, it appears on both AP Calculus AB and BC, typically in the free-response section. You may be asked to find $dy/dx$, evaluate the slope at a given point, determine the equation of a tangent line, find $d^2y/dx^2$, or identify horizontal and vertical tangents. It is a high-value skill — generally worth 3–5 points per free-response question that includes it.
How do I find the second derivative implicitly?
Differentiate both sides of the $dy/dx$ equation again with respect to $x$. This generates new $dy/dx$ terms, which you replace with the first derivative expression. The goal is a formula for $d^2y/dx^2$ in terms of $x$ and $y$ only — not in terms of $dy/dx$.
What mistakes cost the most points on exams?
In order of frequency: (1) forgetting the $dy/dx$ factor on $y$-terms, (2) misapplying the product rule on mixed terms like $xy^2$, (3) algebraic errors when isolating $dy/dx$, and (4) evaluating $dy/dx$ and then forgetting to substitute the specific point's coordinates to get the numerical slope.
How is implicit differentiation connected to the chain rule?
It is not just connected — it is the chain rule. When we write $\dfrac{d}{dx}[y^2] = 2y\dfrac{dy}{dx}$, we are applying the chain rule with outer function $u^2$ and inner function $u = y(x)$. Implicit differentiation is systematic chain-rule application across an entire equation.

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