But many equations have $x$ and $y$ mixed together and can't be easily solved for $y$:

• $x^2 + y^2 = 25$ (circle)
• $xy + y^3 = x$
• $\sin(xy) = x + y$

Implicit differentiation lets us find $\frac{dy}{dx}$ WITHOUT solving for $y$ first!

The Basic Idea

When differentiating an equation implicitly:

1. Differentiate both sides with respect to $x$
2. Remember that $y$ is a function of $x$ (even though we don't know the formula)
3. Use the Chain Rule whenever you differentiate a term with $y$
4. Solve for $\frac{dy}{dx}$

💡 Key Insight: When you differentiate a function of $y$, multiply by $\frac{dy}{dx}$ (the Chain Rule!)

Important Derivative Patterns

When differentiating implicitly, remember these patterns:

Pattern 1: Powers of y

$\frac{d}{dx}[y^n] = ny^{n-1} \cdot \frac{dy}{dx}$

Examples:

• $\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$
• $\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx}$
• $\frac{d}{dx}[\sqrt{y}] = \frac{1}{2\sqrt{y}}\frac{dy}{dx}$

Pattern 2: Functions of y

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

$\frac{d}{dx}[e^y] = e^y \cdot \frac{dy}{dx}$

$\frac{d}{dx}[\ln y] = \frac{1}{y} \cdot \frac{dy}{dx}$

Pattern 3: Products of x and y

Use the Product Rule!

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

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

Step-by-Step Process

Example: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$

Step 1: Differentiate both sides with respect to $x$

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

Step 2: Apply derivative rules to each term

Left side:

• $\frac{d}{dx}[x^2] = 2x$
• $\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$ (Chain Rule!)

Right side:

• $\frac{d}{dx}[25] = 0$

So we get: $2x + 2y\frac{dy}{dx} = 0$

Step 3: Solve for $\frac{dy}{dx}$

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

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

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

This tells us the slope at any point $(x, y)$ on the circle!

More Complex Examples

Example with Product Rule

Find $\frac{dy}{dx}$ if $xy + y^3 = 6$

Step 1: Differentiate both sides

$\frac{d}{dx}[xy + y^3] = \frac{d}{dx}[6]$

Step 2: Apply rules

For $xy$, use Product Rule: $y + x\frac{dy}{dx}$

For $y^3$: $3y^2\frac{dy}{dx}$

For $6$: $0$

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

Step 3: Collect $\frac{dy}{dx}$ terms

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

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

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

Finding Slopes of Tangent Lines

We can use implicit differentiation to find slopes at specific points!

Example: Circle Equation

For $x^2 + y^2 = 25$, we found $\frac{dy}{dx} = -\frac{x}{y}$

Find the slope at point $(3, 4)$:

$\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}$

Find the slope at point $(0, 5)$:

$\frac{dy}{dx}\bigg|_{(0,5)} = -\frac{0}{5} = 0$ (horizontal tangent!)

Find the slope at point $(5, 0)$:

$\frac{dy}{dx}\bigg|_{(5,0)} = -\frac{5}{0}$ → undefined (vertical tangent!)

⚠️ Common Mistakes

Mistake 1: Forgetting dy/dx

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

Mistake 2: Not Using Product Rule

❌ $\frac{d}{dx}[xy] = y$ ✅ $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$

Mistake 3: Forgetting to Solve for dy/dx

Don't stop after differentiating - you must isolate $\frac{dy}{dx}$!

Mistake 4: Arithmetic Errors

When solving for $\frac{dy}{dx}$, collect ALL terms with $\frac{dy}{dx}$ on one side.

Special Cases

Horizontal Tangents

Occur when $\frac{dy}{dx} = 0$

Set the numerator equal to zero and solve.

Vertical Tangents

Occur when $\frac{dy}{dx}$ is undefined

Set the denominator equal to zero and solve.

Second Derivatives

You can differentiate implicitly twice to find $\frac{d^2y}{dx^2}$!

After finding $\frac{dy}{dx}$, differentiate the entire equation again (treating $\frac{dy}{dx}$ as a function).

When to Use Implicit Differentiation

Use implicit differentiation when:

1. Equation can't be solved for $y$ (or it's very difficult)
2. Circles, ellipses, hyperbolas - standard conic sections
3. Equations with $y$ on both sides or mixed with $x$
4. Finding tangent lines to implicit curves
5. Related rates problems (coming up next!)

💡 Pro Tip: Even if you CAN solve for $y$, implicit differentiation is often faster!

Why It Works

Implicit differentiation works because of the Chain Rule.

When we write $y$, we really mean $y(x)$ - a function of $x$.

So when we differentiate $y^2$: $\frac{d}{dx}[y^2] = \frac{d}{dx}[(y(x))^2] = 2y(x) \cdot y'(x) = 2y\frac{dy}{dx}$

It's just the Chain Rule in disguise!

📝 Practice Strategy

1. Differentiate both sides with respect to $x$
2. Remember the Chain Rule for every $y$ term
3. Use Product Rule when $x$ and $y$ are multiplied
4. Collect all $\frac{dy}{dx}$ terms on one side
5. Factor out $\frac{dy}{dx}$
6. Solve by dividing
7. Simplify your final answer

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

(Remember: $\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$ by chain rule)

Solve for $\frac{dy}{dx}$:

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

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

Step 2: Differentiate each term

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

$\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$ (Product Rule)

$\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$ (Chain Rule)

$\frac{d}{dx}[7] = 0$

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

Step 3: Solve for $\frac{dy}{dx}$

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

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

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

Step 4: Evaluate at point $(1, 2)$

$\frac{dy}{dx}\bigg|_{(1,2)} = \frac{-2(1) - 2}{1 + 2(2)} = \frac{-4}{5}$

So the slope is $m = -\frac{4}{5}$

Step 5: Write equation using point-slope form

$y - y_1 = m(x - x_1)$

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

$y - 2 = -\frac{4}{5}x + \frac{4}{5}$

$y = -\frac{4}{5}x + \frac{4}{5} + 2$

$y = -\frac{4}{5}x + \frac{14}{5}$

Answer: $y = -\frac{4}{5}x + \frac{14}{5}$ or $y - 2 = -\frac{4}{5}(x - 1)$

Left side:

• $\frac{d}{dx}[x^3] = 3x^2$
• $\frac{d}{dx}[xy^2]$ requires product rule: $(1)(y^2)+(x)(2y\frac{dy}{}$

Right side:

• $\frac{d}{dx}[y^3] = 3y^2\frac{dy}{dx}$ (chain rule)
• $\frac{d}{dx}[1] = 0$

Equation:

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

Collect $\frac{dy}{dx}$ terms:

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

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

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

Can factor if desired:

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

$2x + 2y\frac{dy}{dx} = 4 + 4\frac{dy}{dx}$

Step 2: Solve for $\frac{dy}{dx}$

$2y\frac{dy}{dx} - 4\frac{dy}{dx} = 4 - 2x$

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

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

Step 3: Horizontal tangent when $\frac{dy}{dx} = 0$

This occurs when the numerator equals zero:

$2 - x = 0$

$x = 2$

Step 4: Find corresponding $y$ values

Substitute $x = 2$ into the original equation:

$2^2 + y^2 = 4(2) + 4y$

$4 + y^2 = 8 + 4y$

$y^2 - 4y - 4 = 0$

Using the quadratic formula:

$y = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm \sqrt{32}}{2} = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2}$

Step 5: State the points

The tangent line is horizontal at:

$(2, 2 + 2\sqrt{2})$ and $(2, 2 - 2\sqrt{2})$

Answer: Points are $(2, 2 + 2\sqrt{2})$ and $(2, 2 - 2\sqrt{2})$