Implicit differentiation and derivatives of inverse functions

Q: How do you differentiate a term where \( x \) and \( y \) are multiplied, such as \( xy \)?

You must use the Product Rule : \( \frac{d}{dx}[xy] = y + x\frac{dy}{dx} \).

AP Calculus AB · AP (Advanced Placement) · 5 min read · Updated: 7 Sept 2026

Welcome to the World of Hidden Derivatives!

Up until now, you have probably been working with "explicit" functions—equations like \( y = x^2 + 5 \) where \( y \) is sitting all by itself, waiting to be differentiated. But what happens when \( x \) and \( y \) are tangled together like a pair of knotted earbuds? That is where Implicit Differentiation and Inverse Functions come in.

These techniques are like "secret weapons" that allow you to find the slope of curves that aren't even functions (like circles!) or find derivatives for functions that are hard to undo. Don't worry if this seems a bit abstract at first; once you see the patterns, it becomes a very logical step-by-step process!

3.2 & 5.12: Implicit Differentiation

An implicit relation is an equation where \( y \) is not isolated. For example, the equation of a circle: \( x^2 + y^2 = 25 \). To find the derivative \( \frac{dy}{dx} \), we don't need to solve for \( y \). We just differentiate everything exactly where it stands.

The Golden Rule of Implicit Differentiation

Every time you differentiate a term containing \( y \), you must multiply by \( \frac{dy}{dx} \). This is because of the Chain Rule: we are assuming \( y \) is a function of \( x \).

Step-by-Step Process:

1. Differentiate both sides of the equation with respect to \( x \).
2. Apply the Power Rule, Product Rule, or Chain Rule as needed.
3. Collect all terms with \( \frac{dy}{dx} \) on one side of the equation.
4. Factor out the \( \frac{dy}{dx} \).
5. Divide to isolate \( \frac{dy}{dx} \).

Example: Find \( \frac{dy}{dx} \) for \( x^2 + y^3 = 5x \).

1. Differentiate: \( \frac{d}{dx}[x^2] + \frac{d}{dx}[y^3] = \frac{d}{dx}[5x] \)
2. Apply rules: \( 2x + 3y^2 \cdot \frac{dy}{dx} = 5 \)
3. Isolate the \( \frac{dy}{dx} \) term: \( 3y^2 \cdot \frac{dy}{dx} = 5 - 2x \)
4. Solve: \( \frac{dy}{dx} = \frac{5 - 2x}{3y^2} \)

Quick Review: Why did we add \( \frac{dy}{dx} \)? Because the derivative of "something cubed" is \( 3(\text{something})^2 \cdot (\text{derivative of the something}) \). Here, the "something" is \( y \).

Common Mistake: The Product Rule Trap

When you see a term like \( xy \), you must use the Product Rule!
\( \frac{d}{dx}[xy] = (1)(y) + (x)(\frac{dy}{dx}) = y + x\frac{dy}{dx} \).
Many students forget this and just write \( 1 \cdot \frac{dy}{dx} \). Don't let the exam catch you on that!

Horizontal and Vertical Tangents

On the AP Exam, you may be asked to find where a curve has a horizontal or vertical tangent line:
- Horizontal Tangents: Occur when the numerator of \( \frac{dy}{dx} \) equals zero (and the denominator is not zero).
- Vertical Tangents: Occur when the denominator of \( \frac{dy}{dx} \) equals zero (and the numerator is not zero).

Key Takeaway: Implicit differentiation allows us to find the slope at any point \( (x, y) \) on a curve, even if the curve isn't a function.

3.3: Differentiating Inverse Functions

If we know the derivative of a function \( f(x) \), we can find the derivative of its inverse \( f^{-1}(x) \) without actually finding the inverse's formula!

The Inverse Derivative Formula

If \( g(x) \) is the inverse of \( f(x) \), then the derivative of the inverse at a point \( (a, b) \) is the reciprocal of the derivative of the original function at the "flipped" point \( (b, a) \).

Formal Notation: \( (f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))} \)

Memory Trick: The "Flipped Point" Table

If you are given that \( f(2) = 5 \) and asked for the derivative of the inverse at \( x = 5 \):
1. Identify the point on the original: \( (2, 5) \).
2. Identify the point on the inverse: \( (5, 2) \).
3. The derivative at the inverse point is: \( \frac{1}{f'(2)} \).

Did you know? This works because inverse functions are reflections across the line \( y = x \). If a hill has a steepness of \( 3 \), its reflection will have a steepness of \( 1/3 \)!

3.4: Differentiating Inverse Trigonometric Functions

You are required to memorize the derivatives of inverse trig functions. The three most common ones on the AP Calculus AB exam are \( \arcsin(x) \), \( \arccos(x) \), and \( \arctan(x) \).

The Formulas (using the Chain Rule with \( u \)):

1. Inverse Sine: \( \frac{d}{dx} [\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \)
2. Inverse Cosine: \( \frac{d}{dx} [\arccos(u)] = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \) (Just like sine, but negative!)
3. Inverse Tangent: \( \frac{d}{dx} [\arctan(u)] = \frac{1}{1+u^2} \cdot \frac{du}{dx} \)

Pro-Tip: Notice that the derivative of \( \arctan(x) \) has no square root. This makes it easy to distinguish from \( \arcsin(x) \).

Example with Chain Rule:

Find the derivative of \( f(x) = \arctan(e^x) \).
Using the formula: \( f'(x) = \frac{1}{1+(e^x)^2} \cdot \frac{d}{dx}[e^x] \)
Simplify: \( f'(x) = \frac{e^x}{1+e^{2x}} \)

3.6: Higher-Order Derivatives (Implicitly)

Sometimes you need to find the second derivative \( \frac{d^2y}{dx^2} \) for an implicit equation. This involves a "substitution" step.

Step 1: Find the first derivative \( \frac{dy}{dx} \).
Step 2: Differentiate \( \frac{dy}{dx} \) using the Quotient Rule or Product Rule.
Step 3: You will end up with a \( \frac{dy}{dx} \) inside your second derivative expression. Substitute your answer from Step 1 into that spot to get the final answer in terms of \( x \) and \( y \).

Key Takeaway: Don't leave \( \frac{dy}{dx} \) in your final answer for a second derivative. Always plug the first derivative back in!

Summary Checklist for the Exam

- [ ] Did I remember \( \frac{dy}{dx} \) every time I differentiated a \( y \)?
- [ ] Did I use the Product Rule for terms like \( xy \) or \( x^2y^3 \)?
- [ ] For inverse functions, did I use the reciprocal of the derivative at the correct point?
- [ ] Do I have \( \arcsin \) and \( \arctan \) derivatives memorized? (Remember: "Co" functions like \( \arccos \) are negative!)
- [ ] On Free Response Questions, did I show the setup (like setting the denominator to zero for vertical tangents) before calculating?

Calculus can be challenging, but mastering these rules gives you the power to analyze almost any curve imaginable. Keep practicing, and these "implicit" steps will become "explicitly" easy!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is an implicit relation, and how does it differ from an explicit function?

An implicit relation is an equation where \( y \) is not isolated (e.g., \( x^2 + y^2 = 25 \)), whereas an explicit function has \( y \) written directly in terms of \( x \) (e.g., \( y = f(x) \)).

What is the "Golden Rule" when differentiating a term containing \( y \) with respect to \( x \)?

Every time you differentiate a term containing \( y \), you must multiply by \( \frac{dy}{dx} \) because of the Chain Rule.

What are the general steps for performing implicit differentiation?

1. Differentiate both sides; 2. Apply rules (Power, Product, Chain); 3. Collect \( \frac{dy}{dx} \) terms; 4. Factor out \( \frac{dy}{dx} \); 5. Isolate \( \frac{dy}{dx} \).

How do you differentiate a term where \( x \) and \( y \) are multiplied, such as \( xy \)?

You must use the Product Rule: \( \frac{d}{dx}[xy] = y + x\frac{dy}{dx} \).

How do you find the locations of horizontal and vertical tangent lines for an implicit curve?

Horizontal tangents occur where the numerator of \( \frac{dy}{dx} = 0 \); vertical tangents occur where the denominator of \( \frac{dy}{dx} = 0 \).

What is the derivative of an inverse function \( f^{-1} \) at a point \( (a, b) \)?

\( (f^{-1})'(a) = \frac{1}{f'(b)} \), which is the reciprocal of the original function's derivative at the "flipped" point.

What are the derivatives of \( \arcsin(u) \) and \( \arccos(u) \)?

\( \frac{d}{dx} [\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \cdot u' \) and \( \frac{d}{dx} [\arccos(u)] = \frac{-1}{\sqrt{1-u^2}} \cdot u' \).

What is the derivative of \( \arctan(u) \)?

\( \frac{d}{dx} [\arctan(u)] = \frac{1}{1+u^2} \cdot u' \).

How do you find a second derivative (\( \frac{d^2y}{dx^2} \)) using implicit differentiation?

Find \( \frac{dy}{dx} \), differentiate it again, and then substitute the first derivative's expression back into the result.

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