Implicit Differentiation

∫ Implicit Differentiation

Part 1 of 7 — Differentiating Equations Not Solved for $y$

Topics in This Part

🔑 Key Concept: Not every curve can be written as $y = f(x)$. Implicit differentiation lets us find $\frac{dy}{dx}$ directly from an equation relating $x$ and $y$.

📖 Why Do We Need Implicit Differentiation?

Some curves cannot be expressed as $y = f(x)$:

• Circle: $x^2 + y^2 = 25$ (not a function!)
• Ellipse: $\frac{x^2}{9} + \frac{y^2}{4} = 1$
• Folium of Descartes: $x^3 + y^3 = 6xy$

To find slopes on these curves, we differentiate both sides with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule.

The Technique: Step by Step

$\boxed{\text{Differentiate both sides w.r.t. } x, \text{ using } \frac{d}{dx}[y] = \frac{dy}{dx} \text{ (chain rule)}}$

The Key Rule

Whenever you differentiate a term involving $y$, multiply by $\frac{dy}{dx}$:

• $\frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}$
• $\frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx}$
• $\frac{d}{dx}[\sin y] = \cos y \cdot \frac{dy}{dx}$
• $\frac{d}{dx}[xy] = x\frac{dy}{dx} + y$ (product rule!)

📌 Basic Examples

Example 1: Circle $x^2 + y^2 = 25$

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

Solve: $\frac{dy}{dx} = -\frac{x}{y}$

At $(3, 4)$: slope $= -\frac{3}{4}$. At $(3, -4)$: slope $= \frac{3}{4}$.

Example 2: $x^3 + y^3 = 9xy$

$3x^2 + 3y^2\frac{dy}{dx} = 9y + 9x\frac{dy}{dx}$

$3y^2\frac{dy}{dx} - 9x\frac{dy}{dx} = 9y - 3x^2$

$\frac{dy}{dx}(3y^2 - 9x) = 9y - 3x^2$

$\boxed{\frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x} = \frac{3y - x^2}{y^2 - 3x}}$

Example 3: Finding a Tangent Line

$x^2 + xy + y^2 = 7$ at $(1, 2)$.

$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

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

$\frac{dy}{dx} = \frac{-2x - y}{x + 2y} = \frac{-2(1)-2}{1+4} = \frac{-4}{5}$

Tangent line: $y - 2 = -\frac{4}{5}(x - 1)$

AP Tip: Always verify the point satisfies the original equation: $1 + 2 + 4 = 7$ ✓

Implicit Differentiation Basics 🎯

Tangent Lines from Implicit Equations

Systematic Process

1. Verify the point is on the curve
2. Differentiate implicitly
3. Solve for $\frac{dy}{dx}$
4. Evaluate at the given point
5. Write tangent line: $y - y_0 = m(x - x_0)$

Horizontal & Vertical Tangent Lines

If $\frac{dy}{dx} = \frac{N(x,y)}{D(x,y)}$:

• Horizontal tangent where $N = 0$ and $D \neq 0$
• Vertical tangent where $D = 0$ and $N \neq 0$

Example

$x^2 + xy + y^2 = 3$. We found $\frac{dy}{dx} = \frac{-2x-y}{x+2y}$.

Horizontal tangent: $-2x - y = 0 \implies y = -2x$. Substitute: $x^2 + x(-2x) + (-2x)^2 = 3x^2 = 3 \implies x = \pm 1$.

Points: $(1, -2)$ and $(-1, 2)$.

Vertical tangent: $x + 2y = 0 \implies x = -2y$. Substitute: $4y^2 - 2y^2 + y^2 = 3y^2 = 3 \implies y = \pm 1$.

Points: $(-2, 1)$ and $(2, -1)$.

Implicit Differentiation Drill 🧮

1) $x^2 + y^2 = 169$. Slope at $(5, 12)$: $\frac{dy}{dx} =$ (enter as fraction like -5/12)

2) $2x + 3y = 12$. $\frac{dy}{dx} =$ (enter as fraction)

3) $x^2y = 8$ at $(2, 2)$. $\frac{dy}{dx} =$

Chain Rule Application 🔽

Exit Quiz ✅

∫ Implicit Differentiation — Part 2

Part 2 of 7 — Product & Quotient Rule in Implicit Context

Topics in This Part

📖 Product Rule in Implicit Differentiation

When $x$ and $y$ multiply together, you need both the product rule and the chain rule:

$\frac{d}{dx}[xy] = x\frac{dy}{dx} + y \cdot 1 = x\frac{dy}{dx} + y$

$\frac{d}{dx}[x^2 y^3] = 2xy^3 + x^2 \cdot 3y^2\frac{dy}{dx}$

$\frac{d}{dx}[x\sin y] = \sin y + x\cos y\frac{dy}{dx}$

Strategy: Color-Code or Underline

When differentiating complex expressions:

1. Identify every term containing $y$ → these get $\frac{dy}{dx}$
2. For mixed $xy$ terms → product rule first, chain rule on $y$ part
3. Collect all $\frac{dy}{dx}$ terms on one side
4. Factor out $\frac{dy}{dx}$ and solve

Example: $x^2y + xy^2 = 6$

$\frac{d}{dx}[x^2y] + \frac{d}{dx}[xy^2] = 0$

$(2xy + x^2\frac{dy}{dx}) + (y^2 + 2xy\frac{dy}{dx}) = 0$

$x^2\frac{dy}{dx} + 2xy\frac{dy}{dx} = -2xy - y^2$

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

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

📌 More Complex Examples

Example 1: $e^{xy} = x + y$

Left side (chain rule + product rule): $e^{xy}\cdot(y + x\frac{dy}{dx}) = 1 + \frac{dy}{dx}$

$ye^{xy} + xe^{xy}\frac{dy}{dx} = 1 + \frac{dy}{dx}$

$(xe^{xy} - 1)\frac{dy}{dx} = 1 - ye^{xy}$

$\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} - 1}$

Example 2: $\sin(xy) = \frac{1}{2}$

$\cos(xy)\cdot(y + x\frac{dy}{dx}) = 0$

Since $\cos(xy) \neq 0$ (because $\sin(xy) = 1/2$, not $\pm 1$):

$y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}$

Insight: The $\cos(xy)$ divides out! This happens when the outer derivative cancels.

Example 3: $(x + y)^3 = x^3 + y^3$

$3(x+y)^2(1 + \frac{dy}{dx}) = 3x^2 + 3y^2\frac{dy}{dx}$

Let $S = (x+y)^2$:

$3S + 3S\frac{dy}{dx} = 3x^2 + 3y^2\frac{dy}{dx}$

$\frac{dy}{dx}(3S - 3y^2) = 3x^2 - 3S$

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

Product Rule in Implicit Diff 🎯

Quotient Scenarios

Example: $\frac{x}{y} = x - y$

Multiply both sides by $y$: $x = xy - y^2$

$1 = y + x\frac{dy}{dx} - 2y\frac{dy}{dx}$

$1 - y = (x - 2y)\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1 - y}{x - 2y}$

AP Tip: When quotients appear, it's often easier to cross-multiply first, then differentiate.

When Direct Quotient Rule Is Needed

$\frac{d}{dx}\left[\frac{y}{x}\right] = \frac{x\frac{dy}{dx} - y}{x^2}$

$\frac{d}{dx}\left[\frac{x^2}{y}\right] = \frac{2xy - x^2\frac{dy}{dx}}{y^2}$

Second Derivatives

To find $\frac{d^2y}{dx^2}$, differentiate $\frac{dy}{dx}$ again implicitly:

$y^2 = x \implies \frac{dy}{dx} = \frac{1}{2y}$

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{1}{2y}\right] = -\frac{1}{2y^2}\frac{dy}{dx} = -\frac{1}{2y^2}\cdot\frac{1}{2y} = -\frac{1}{4y^3}$

Product Rule Practice 🧮

1) $xy = 20$ at $(4, 5)$. $\frac{dy}{dx} =$

2) $x^2 + xy = 10$ at $(2, 3)$. $\frac{dy}{dx} =$

3) $x^2y^2 = 16$ at $(2, 2)$. $\frac{dy}{dx} =$

Identify the Correct Differentiation 🔽

Exit Quiz — Part 2 ✅

∫ Implicit Differentiation — Part 3

Part 3 of 7 — Trigonometric & Exponential Implicit Equations

Topics in This Part

📖 Trigonometric Implicit Equations

Key Derivatives with Chain Rule

Example 1: $\sin y + \cos x = 1$

$\cos y \cdot \frac{dy}{dx} - \sin x = 0$

$\frac{dy}{dx} = \frac{\sin x}{\cos y}$

This is undefined when $\cos y = 0$, i.e., $y = \frac{\pi}{2} + n\pi$ — vertical tangent lines!

Example 2: $\tan(x + y) = x$

$\sec^2(x+y)\left(1 + \frac{dy}{dx}\right) = 1$

$1 + \frac{dy}{dx} = \cos^2(x+y)$

$\frac{dy}{dx} = \cos^2(x+y) - 1 = -\sin^2(x+y)$

But since $\tan(x+y) = x$, we can express: $\sin^2(x+y) = \frac{x^2}{1+x^2}$

$\boxed{\frac{dy}{dx} = -\frac{x^2}{1+x^2}}$

Exponential & Logarithmic Implicit Equations

Key Derivatives

Example 1: $e^y = x^2 + y$

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

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

$\frac{dy}{dx} = \frac{2x}{e^y - 1}$

Example 2: $\ln(x^2 + y^2) = \arctan\frac{y}{x}$

$\frac{2x + 2y\frac{dy}{dx}}{x^2 + y^2} = \frac{1}{1+(y/x)^2} \cdot \frac{x\frac{dy}{dx} - y}{x^2}$

Simplifying the right side: $\frac{x\frac{dy}{dx} - y}{x^2 + y^2}$

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

$2x + 2y\frac{dy}{dx} = x\frac{dy}{dx} - y$

$(2y - x)\frac{dy}{dx} = -y - 2x$

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

Example 3: $\ln y + xy = 1$ at the point $(1, 1)$

First confirm the point is on the curve: $\ln 1 + 1 \cdot 1 = 1$. ✓

$\frac{1}{y}\frac{dy}{dx} + y + x\frac{dy}{dx} = 0$

$\frac{dy}{dx}\left(\frac{1}{y} + x\right) = -y \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{-y^2}{1 + xy}$

At $(1, 1)$: $\frac{dy}{dx} = -\frac{1}{2}$.

Pattern: $\ln y$ differentiates to $\frac{1}{y}\frac{dy}{dx}$, and $xy$ needs the product rule. Collect every $\frac{dy}{dx}$ term on one side, then factor.

Trig & Exp Implicit Differentiation 🎯

📌 Combined Examples

Example: $e^y\sin x = 1$

Product rule: $e^y\frac{dy}{dx}\sin x + e^y\cos x = 0$

$e^y(\frac{dy}{dx}\sin x + \cos x) = 0$

Since $e^y > 0$ always: $\frac{dy}{dx}\sin x + \cos x = 0$

$\frac{dy}{dx} = -\frac{\cos x}{\sin x} = -\cot x$

Insight: The $e^y$ cancels completely! The derivative depends only on $x$.

Example: $y = \arctan(x/y)$

$\frac{dy}{dx} = \frac{1}{1 + (x/y)^2} \cdot \frac{y - x\frac{dy}{dx}}{y^2}$

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

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

$(y^2 + x^2 + x)\frac{dy}{dx} = y$

$\frac{dy}{dx} = \frac{y}{x^2 + y^2 + x}$

Trig & Exp Practice 🧮

1) $\cos y = x$ at $(0, \pi/2)$. $\frac{dy}{dx} =$ (enter an integer)

2) $e^y = 2x$ at $(e/2, 1)$. $\frac{dy}{dx} =$ (enter as fraction like a/b)

3) $\ln(xy) = 0$ at $(1, 1)$. $\frac{dy}{dx} =$

Apply the Chain Rule 🔽

Exit Quiz — Part 3 ✅

∫ Implicit Differentiation — Part 4

Part 4 of 7 — Second Derivatives & Higher-Order Implicit Differentiation

Topics in This Part

📖 Finding $\frac{d^2y}{dx^2}$ Implicitly

The Process

1. Find $\frac{dy}{dx}$ using implicit differentiation
2. Differentiate $\frac{dy}{dx}$ again with respect to $x$
3. Substitute the original $\frac{dy}{dx}$ expression back in
4. Simplify (often using the original equation)

Example: $x^2 + y^2 = 25$

Step 1: $\frac{dy}{dx} = -\frac{x}{y}$

Step 2: Differentiate using the quotient rule: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\frac{y(1) - x\frac{dy}{dx}}{y^2}$

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

Step 4: Since $x^2 + y^2 = 25$: $\boxed{\frac{d^2y}{dx^2} = -\frac{25}{y^3}}$

AP Tip: The AP exam loves asking for $\frac{d^2y}{dx^2}$ at a specific point. You can either simplify fully or just plug in the point after Step 3.

Substitution Technique

Why Simplify with the Original Equation?

After differentiating twice, expressions get messy. Use the original equation to simplify:

• $x^2 + y^2 = r^2 \implies$ replace $x^2 + y^2$ with $r^2$
• $xy = k \implies$ replace $xy$ with $k$
• $y = \sin x \implies$ replace trig expressions

Example: $xy = 6$

Step 1: $y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}$

Step 2: Quotient rule: $\frac{d^2y}{dx^2} = -\frac{x\frac{dy}{dx} - y}{x^2} = -\frac{x(-y/x) - y}{x^2} = -\frac{-y - y}{x^2} = \frac{2y}{x^2}$

At $(2, 3)$: $\frac{d^2y}{dx^2} = \frac{6}{4} = \frac{3}{2}$

Alternative: Direct Second Differentiation

From $y + x\frac{dy}{dx} = 0$, differentiate again:

$\frac{dy}{dx} + \frac{dy}{dx} + x\frac{d^2y}{dx^2} = 0$

$x\frac{d^2y}{dx^2} = -2\frac{dy}{dx} = -2(-y/x) = \frac{2y}{x}$

$\frac{d^2y}{dx^2} = \frac{2y}{x^2}$ ✓ Same answer!

Second Derivatives 🎯

📌 AP-Style Problems

Problem 1: Free Response Style

Given $x^3 + y^3 = 9$ and the point $(1, 2)$:

(a) Find $\frac{dy}{dx}$: $3x^2 + 3y^2\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x^2}{y^2}$

At $(1,2)$: $\frac{dy}{dx} = -\frac{1}{4}$

(b) Find the equation of the tangent line: $y - 2 = -\frac{1}{4}(x - 1) \implies y = -\frac{1}{4}x + \frac{9}{4}$

(c) Find $\frac{d^2y}{dx^2}$ at $(1, 2)$: $\frac{d^2y}{dx^2} = -\frac{2xy^2 - x^2 \cdot 2y\frac{dy}{dx}}{y^4} = -\frac{2(1)(4) - 1 \cdot 4(-1/4)}{16} = -\frac{8 + 1}{16} = -\frac{9}{16}$

(d) Is the curve concave up or down at $(1,2)$?

Since $\frac{d^2y}{dx^2} = -9/16 < 0$, the curve is concave down.

Note: On free response, each part (dy/dx, the second derivative, the concavity conclusion) is usually scored separately, so show each.

Second Derivative Practice 🧮

1) $y^2 = 4x$. $\frac{d^2y}{dx^2}$ at $(1, 2) =$ (enter as fraction like -a/b)

2) $xy = 12$ at $(3, 4)$. $\frac{d^2y}{dx^2} =$ (enter as fraction)

Concavity Analysis 🔽

Exit Quiz — Part 4 ✅

∫ Implicit Differentiation — Part 5

Part 5 of 7 — Horizontal & Vertical Tangents

Topics in This Part

📖 Horizontal Tangent Lines

A horizontal tangent occurs where $\frac{dy}{dx} = 0$.

If $\frac{dy}{dx} = \frac{N(x,y)}{D(x,y)}$, set $N(x,y) = 0$ (and verify $D \neq 0$).

Systematic Method

1. Find $\frac{dy}{dx} = \frac{N}{D}$
2. Set $N = 0$ → get a relationship between $x$ and $y$
3. Substitute back into original equation → find actual points
4. Verify $D \neq 0$ at those points

Example: $x^2 + xy + y^2 = 7$

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

Horizontal: $2x + y = 0 \implies y = -2x$

Substitute: $x^2 + x(-2x) + (-2x)^2 = 7$ $x^2 - 2x^2 + 4x^2 = 3x^2 = 7$ $x = \pm\sqrt{7/3}$

Points: $\left(\sqrt{7/3}, -2\sqrt{7/3}\right)$ and $\left(-\sqrt{7/3}, 2\sqrt{7/3}\right)$

Verify: $x + 2y = \sqrt{7/3} + 2(-2\sqrt{7/3}) = -3\sqrt{7/3} \neq 0$ ✓

Finding Vertical Tangent Lines

A vertical tangent occurs where $\frac{dy}{dx}$ is undefined: $D(x,y) = 0$ (and $N \neq 0$).

Example (continued): $x^2 + xy + y^2 = 7$

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

Vertical: $x + 2y = 0 \implies x = -2y$

Substitute: $(-2y)^2 + (-2y)y + y^2 = 7$ $4y^2 - 2y^2 + y^2 = 3y^2 = 7$ $y = \pm\sqrt{7/3}$

Points: $\left(-2\sqrt{7/3}, \sqrt{7/3}\right)$ and $\left(2\sqrt{7/3}, -\sqrt{7/3}\right)$

What If Both $N = 0$ and $D = 0$?

This is a singular point — more analysis needed (could be a cusp, isolated point, or self-intersection).

Example: $y^2 = x^3$ at $(0, 0)$

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

At $(0,0)$: $\frac{0}{0}$ — singular point! This is actually a cusp (pointed corner).

AP Note: The AP exam typically avoids singular points. If you get $0/0$, check if you're at a special geometric feature.

Tangent Line Types 🎯

📌 Implicit Curves: The Ellipse

Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\frac{dy}{dx} = -\frac{b^2x}{a^2y}$

To find any tangent line, compute $\frac{dy}{dx}$ at the point and use point-slope form.

Find Tangent Types 🧮

1) $y^2 = 4x$. How many vertical tangent points? (enter number)

2) $x^2 + 4y^2 = 16$. Slope at $(2, \sqrt{3})$ = (enter as fraction with sqrt, like -1/sqrt3)

Classify Tangent Lines 🔽

Exit Quiz — Part 5 ✅

∫ Implicit Differentiation — Part 6

Part 6 of 7 — Related Connections & Inverse Functions

Topics in This Part

📖 Implicit Differentiation & Inverse Functions

Implicit differentiation provides elegant proofs for inverse function derivatives.

Deriving $\frac{d}{dx}[\arcsin x]$

Let $y = \arcsin x$, so $\sin y = x$.

Differentiate: $\cos y \cdot \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{\cos y}$

Since $\sin y = x$ and $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$:

$\boxed{\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}}$

Deriving $\frac{d}{dx}[\arctan x]$

Let $y = \arctan x$, so $\tan y = x$.

$\sec^2 y \cdot \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$

$\boxed{\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}}$

Why This Matters: These proofs show that implicit differentiation isn't just a technique — it's how fundamental derivative formulas are derived.

Complete Inverse Trig Derivatives

📌 Normal Lines

The normal line is perpendicular to the tangent line at a point.

If $\frac{dy}{dx} = m$ at point $(a, b)$:

• Tangent: $y - b = m(x - a)$
• Normal: $y - b = -\frac{1}{m}(x - a)$

Example: $x^2 + y^2 = 25$ at $(3, 4)$

$\frac{dy}{dx} = -\frac{3}{4}$

• Tangent: $y - 4 = -\frac{3}{4}(x - 3)$
• Normal: $y - 4 = \frac{4}{3}(x - 3)$

Notice the normal line passes through the origin! This is always true for circles centered at the origin: the normal is a radius.

AP Insight: The AP exam occasionally asks for normal lines. The key formula is: normal slope = $-1/m_{tangent}$.

Inverse Functions & Normal Lines 🎯

Implicit Curves & Symmetry

Types of Symmetry

Example: $x^4 + y^4 = 1$ (Superellipse)

$\frac{dy}{dx} = -\frac{x^3}{y^3}$

At $(a, b)$: slope $= -a^3/b^3$ At $(-a, b)$: slope $= a^3/b^3$ (y-axis symmetry: sign flips) At $(a, -b)$: slope $= a^3/b^3$ (x-axis symmetry: sign flips) At $(-a, -b)$: slope $= -a^3/b^3$ (origin symmetry: same slope)

Folium of Descartes: $x^3 + y^3 = 3xy$

This curve has a beautiful self-intersection at the origin and is symmetric about $y = x$.

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

On the line $y = x$: slope $= \frac{x - x^2}{x^2 - x} = -1$ (the tangent is perpendicular to $y = x$).

Normal Line Practice 🧮

1) Normal to $x^2 + y^2 = 100$ at $(6, 8)$. Slope of normal = (enter as fraction)

2) Normal to $xy = 6$ at $(2, 3)$. Slope = (enter as fraction)

Symmetry & Normal Lines 🔽

Exit Quiz — Part 6 ✅

∫ Implicit Differentiation — Part 7

Part 7 of 7 — AP Exam Review & Mixed Practice

Topics in This Part

📖 Implicit Differentiation on the AP Exam

How It Appears

• Multiple Choice: Find $\frac{dy}{dx}$ for a given implicit equation, evaluate at a point
• Free Response: Often embedded in related rates (see the Related Rates topic) or as part of a multi-step problem
• Common pairings: Tangent lines, second derivatives, concavity analysis

Speed Tips

1. Don't solve for $y$ first — differentiate directly
2. Collect $\frac{dy}{dx}$ terms immediately — don't expand unnecessarily
3. Plug in the point early when finding a numerical answer
   • Sometimes you can sub in $(x_0, y_0)$ before solving for $\frac{dy}{dx}$
4. Check your point — verify it satisfies the original equation

The "Plug In Early" Shortcut

For $x^3 + y^3 = 9$ at $(1, 2)$:

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

Plug in immediately: $3(1) + 3(4)\frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{3}{12} = -\frac{1}{4}$

This is faster than solving for $\frac{dy}{dx}$ algebraically first!

Free Response Walkthrough

Sample Problem (AP Style)

Consider the curve defined by $2x^2 + 3y^2 - 4xy = 36$.

(a) Show that $\frac{dy}{dx} = \frac{2y - 2x}{3y - 2x}$.

Differentiate: $4x + 6y\frac{dy}{dx} - 4(y + x\frac{dy}{dx}) = 0$

$4x + 6y\frac{dy}{dx} - 4y - 4x\frac{dy}{dx} = 0$

$(6y - 4x)\frac{dy}{dx} = 4y - 4x$

$\frac{dy}{dx} = \frac{4y - 4x}{6y - 4x} = \frac{2(2y - 2x)}{2(3y - 2x)} = \frac{2y - 2x}{3y - 2x}$ ✓

(b) Find all points where the tangent is horizontal.

$2y - 2x = 0 \implies y = x$

Substitute: $2x^2 + 3x^2 - 4x^2 = x^2 = 36 \implies x = \pm 6$

Points: $(6, 6)$ and $(-6, -6)$

(c) Find $\frac{d^2y}{dx^2}$ at $(6, 6)$.

Since the tangent is horizontal, $\frac{dy}{dx} = 0$ at $(6, 6)$.

$\frac{d^2y}{dx^2} = \frac{(3y-2x)(2\frac{dy}{dx} - 2) - (2y-2x)(3\frac{dy}{dx}-2)}{(3y-2x)^2}$

At $(6,6)$ with $\frac{dy}{dx} = 0$:

$= \frac{(18-12)(0-2) - (12-12)(0-2)}{(18-12)^2} = \frac{6(-2) - 0}{36} = -\frac{12}{36} = -\frac{1}{3}$

Since $\frac{d^2y}{dx^2} < 0$, the curve is concave down at $(6, 6)$.

AP Exam Practice 🎯