Topic 3.2 Notes – Implicit Differentiation

1. What Implicit Differentiation Is

An equation is implicit when $y$ is not isolated. For example:

• Explicit: $y = 3x^2 - 1$
• Implicit: $x^2 + xy + y^2 = 7$

Even if you can’t solve for $y$ easily, it still depends on $x$. That’s the key idea. So when you differentiate with respect to $x$, every $y$ must be treated as a function of $x$.

The foundation is the chain rule.

If $y$ depends on $x$, then:

$$\frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx}$$

That extra $\frac{dy}{dx}$ appears every time you differentiate a $y$-term.

Notation

• Use $\frac{dy}{dx}$ or $y'$
• Use $\frac{dy}{dx}$ notation rather than $f'(x)$, since the relationship is defined implicitly

2. Derivative Rules You Must Apply Inside Implicit Differentiation

Implicit differentiation is just regular derivative rules plus the chain rule layered in.

Power Rule with Chain Rule

• $\frac{d}{dx}(x^4) = 4x^3$
• $\frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx}$

Only $y$-terms get the extra factor.

Product Rule

Very common on quizzes and FRQs.

If you see something like $xy$, use:

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

You’re differentiating two functions of $x$:

• $x$
• $y(x)$

Missing one term here is one of the fastest ways to lose points.

Other Functions

Differentiate normally, then multiply by $\frac{dy}{dx}$ if the input is $y$.

Examples:

• $\frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx}$
• $\frac{d}{dx}(e^y) = e^y \frac{dy}{dx}$

If it’s $\sin x$ or $e^x$, no extra factor.

3. The Implicit Differentiation Process

Here’s what it looks like in action.

Suppose: $$x^2 + y^2 = 10$$

Step-by-step

1. Differentiate both sides: $$2x + 2y \frac{dy}{dx} = 0$$
2. Solve for $\frac{dy}{dx}$:

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

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

That’s it. Most problems follow this pattern:

• Differentiate everything.
• Collect all $\frac{dy}{dx}$ terms.
• Factor.
• Solve.

This equation represents a circle centered at the origin with radius $\sqrt{10}$. The slope at any point on that circle is given by $-\frac{x}{y}$.

Notice how the tangent line’s slope changes depending on the point. The formula $-\frac{x}{y}$ uses both coordinates, which is typical for implicit derivatives.

4. Finding Slopes and Tangent Lines

Once you have $\frac{dy}{dx}$, treat it like any derivative.

Slope at a Point

Using the circle example:

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

At the point $(1, 3)$ (which satisfies $1^2 + 3^2 = 10$):

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

Always check that the point satisfies the original equation. AP graders do.

Equation of the Tangent Line

Use point-slope form:

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

At $(1,3)$ with slope $-\frac{1}{3}$:

$$y - 3 = -\frac{1}{3}(x - 1)$$

Very common structure on FRQs:

• Part (a): find $\frac{dy}{dx}$
• Part (b): find the tangent line

Differentiate first. Plug in after.

5. Common Mistakes and Exam Traps

Forgetting the chain rule

Writing $\frac{d}{dx}(y^3) = 3y^2$
Correct answer: $3y^2 \frac{dy}{dx}$

Every $y$ produces a $\frac{dy}{dx}$.

Missing the product rule

If you see $xy$, $x^2y$, or similar, product rule is almost guaranteed.

Stopping before isolating $\frac{dy}{dx}$

You must solve for it completely. Leaving it mixed in costs points on FRQs.

Plugging in too early

Do not substitute coordinates until after you isolate $\frac{dy}{dx}$. Algebra is much cleaner that way.