Implicit differentiation and derivatives of inverse functions

Q: What is the difference between an explicit function and an implicit function?

An explicit function has the dependent variable isolated (e.g., \(y = f(x)\)), while an implicit function has variables mixed together in an equation (e.g., \(x^2 + y^3 = 5xy\)).

Q: How do you handle a term like \(xy\) during implicit differentiation?

You must use the Product Rule : \(\frac{d}{dx}[xy] = (1)(y) + (x)(\frac{dy}{dx})\).

Q: How do you identify which inverse trig derivatives are negative?

Any inverse trig function starting with "co" (arccos, arccot, arccosec) has a negative derivative.

Q: What is the crucial final step when finding a second derivative (\(\frac{d^2y}{dx^2}\)) implicitly?

You must substitute the expression for \(\frac{dy}{dx}\) back into the second derivative so the final answer is in terms of \(x\) and \(y\).

AP Calculus BC · AP (Advanced Placement) · 5 min read · Updated: Sep 7, 2026

Welcome to Implicit and Inverse Differentiation!

Up until now, you have mostly dealt with functions where \(y\) is all by itself, like \(y = x^2 + 3x\). This is called an explicit function. But what happens when \(x\) and \(y\) are tangled together in a messy equation like \(x^2 + y^3 = 5xy\)? Or what if you need the slope of an inverse function without actually solving for the inverse?

In this chapter, we will learn the tools to "untangle" these derivatives. These techniques are essential for the AP exam, appearing in both multiple-choice and free-response sections. Don't worry if it feels a bit "alphabet-heavy" at first—once you see the pattern, it becomes a very predictable process!

3.2 Implicit Differentiation

Implicit differentiation is used when we have a relationship between \(x\) and \(y\) that isn't easily solved for \(y\). Instead of rearranging the equation, we differentiate everything just as it is.

The Golden Rule

The most important thing to remember is that \(y\) is a function of \(x\). This means every time you take the derivative of a term containing \(y\), you must apply the Chain Rule and multiply by \(\frac{dy}{dx}\) (often written as \(y'\)).

Analogy: Think of \(\frac{dy}{dx}\) as a "tax" you have to pay every time you differentiate a \(y\) variable. If you derive \(x^2\), it's just \(2x\). If you derive \(y^2\), you pay the tax: \(2y \cdot \frac{dy}{dx}\).

Step-by-Step Process

1. Differentiate both sides of the equation with respect to \(x\).
2. Apply the Power Rule, Product Rule, or Quotient Rule as needed.
3. Every time you differentiate a term with \(y\), attach a \(\frac{dy}{dx}\).
4. Collect all terms with \(\frac{dy}{dx}\) on one side and all other terms on the other.
5. Factor out \(\frac{dy}{dx}\) and solve.

Example: Find \(\frac{dy}{dx}\) for \(y^2 + x^2 = 25\).
1. Differentiate: \(\frac{d}{dx}[y^2] + \frac{d}{dx}[x^2] = \frac{d}{dx}[25]\).
2. Use the "tax": \(2y \frac{dy}{dx} + 2x = 0\).
3. Move \(2x\): \(2y \frac{dy}{dx} = -2x\).
4. Solve: \(\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}\).

Quick Review: Common Mistakes

Forget the Constant: Remember that the derivative of a constant (like \(25\) or \(7\)) is \(0\). Students often accidentally leave the constant as it is.
Product Rule Trap: In a term like \(5xy\), you must use the Product Rule. Treat it as \((5x) \cdot (y)\). Its derivative is \(5y + 5x\frac{dy}{dx}\).

Key Takeaway: Treat \(y\) as an inside function. \(\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}\).

3.3 Differentiating Inverse Functions

Sometimes we know everything about a function \(f(x)\), but we need to find the derivative of its inverse, \(f^{-1}(x)\), at a specific point. We don't actually need to find the inverse formula to do this!

The Relationship

If a function \(f\) contains the point \((a, b)\), then its inverse \(f^{-1}\) contains the point \((b, a)\). The slopes at these "reflected" points are reciprocals of each other.

The Formula:
\((f^{-1})'(b) = \frac{1}{f'(a)}\) where \(f(a) = b\).

The "Table Method" for Success

To avoid getting \(x\) and \(y\) values mixed up, create a small table:

1. Original Function \(f\): Point \((a, b)\), Slope \(f'(a)\).
2. Inverse Function \(f^{-1}\): Point \((b, a)\), Slope \(\frac{1}{f'(a)}\).

Example: If \(f(3) = 5\) and \(f'(3) = 2\), find the derivative of the inverse at \(x = 5\).
Since \(f\) has point \((3, 5)\), \(f^{-1}\) has point \((5, 3)\). The derivative \((f^{-1})'(5) = \frac{1}{f'(3)} = \frac{1}{2}\).

Did you know? The graph of a function and its inverse are reflections across the line \(y = x\). This geometric flip is why the slopes are reciprocals!

Key Takeaway: The derivative of the inverse at \(x = b\) is \(1\) divided by the derivative of the original at \(x = a\). Just remember: "Flip the point, then flip the slope!"

3.4 Differentiating Inverse Trigonometric Functions

You are required to know the derivatives of the inverse trig functions. While there are six, the AP exam focuses heavily on arcsin, arccos, and arctan.

The Formulas

Assuming \(u\) is a function of \(x\) (so we use the Chain Rule):

1. \(\frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\)
2. \(\frac{d}{dx}[\arccos(u)] = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\)
3. \(\frac{d}{dx}[\arctan(u)] = \frac{1}{1+u^2} \cdot \frac{du}{dx}\)

Memory Aids & Tricks

The "CO" Rule: Just like with regular trig derivatives, any function starting with "co" (arccos, arccot, arccosec) has a negative derivative.
Arcsin vs. Arctan: Notice that arcsin has a square root and a minus sign, while arctan has no square root and a plus sign. Think: "Tan is a plus!"

Example: Find the derivative of \(y = \arctan(3x)\).
Here, \(u = 3x\) and \(u' = 3\).
\(\frac{dy}{dx} = \frac{1}{1+(3x)^2} \cdot 3 = \frac{3}{1+9x^2}\).

Key Takeaway: Memorize these three structures. They appear frequently in both Unit 3 (Differentiation) and Unit 6 (Integration).

3.6 Higher-Order Derivatives

On the AP exam, you might be asked to find the second derivative (\(\frac{d^2y}{dx^2}\)) using implicit differentiation. This requires two steps of differentiation.

The Process

1. Find \(\frac{dy}{dx}\) using implicit differentiation as usual.
2. Differentiate \(\frac{dy}{dx}\) again with respect to \(x\).
3. Crucial Step: Your second derivative will likely contain a \(\frac{dy}{dx}\) term. You must substitute your result from Step 1 back into the equation to get the final answer in terms of \(x\) and \(y\) only.

Example: If \(\frac{dy}{dx} = \frac{x}{y}\), then \(\frac{d^2y}{dx^2} = \frac{(y)(1) - (x)(\frac{dy}{dx})}{y^2}\). Replace \(\frac{dy}{dx}\) with \(\frac{x}{y}\) to finish the problem.

Key Takeaway: Never leave a \(\frac{dy}{dx}\) inside your final expression for a second derivative. Always substitute!

Final Chapter Summary

Implicit Differentiation: Used when \(x\) and \(y\) are mixed. Remember to attach \(\frac{dy}{dx}\) to every \(y\) derivative.
Inverse Derivatives: Use the reciprocal slope at the reflected point: \((f^{-1})'(b) = \frac{1}{f'(a)}\).
Inverse Trig: Memorize the formulas for arcsin, arccos, and arctan. Use the Chain Rule (\(u'\)) every time.

Quick check

Can you answer these now?

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

What is the difference between an explicit function and an implicit function?

An explicit function has the dependent variable isolated (e.g., \(y = f(x)\)), while an implicit function has variables mixed together in an equation (e.g., \(x^2 + y^3 = 5xy\)).

What is the "Golden Rule" of implicit differentiation?

Whenever you differentiate a term containing \(y\), you must apply the Chain Rule and multiply by \(\frac{dy}{dx}\).

What are the basic steps for solving an implicit differentiation problem?

1. Differentiate both sides with respect to \(x\); 2. Attach \(\frac{dy}{dx}\) to \(y\) terms; 3. Collect \(\frac{dy}{dx}\) terms on one side; 4. Factor and solve for \(\frac{dy}{dx}\).

How do you handle a term like \(xy\) during implicit differentiation?

You must use the Product Rule: \(\frac{d}{dx}[xy] = (1)(y) + (x)(\frac{dy}{dx})\).

What is the relationship between the slope of a function and the slope of its inverse?

The slopes at reflected points \((a, b)\) and \((b, a)\) are reciprocals: \((f^{-1})'(b) = \frac{1}{f'(a)}\).

What is the "Table Method" for finding inverse derivatives?

Create a table mapping the original point \((a, b)\) and slope \(f'(a)\) to the inverse point \((b, a)\) and slope \(\frac{1}{f'(a)}\).

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

\(\frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\)

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

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

How do you identify which inverse trig derivatives are negative?

Any inverse trig function starting with "co" (arccos, arccot, arccosec) has a negative derivative.

What is the crucial final step when finding a second derivative (\(\frac{d^2y}{dx^2}\)) implicitly?

You must substitute the expression for \(\frac{dy}{dx}\) back into the second derivative so the final answer is in terms of \(x\) and \(y\).

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