Welcome to Unit 3: Mastering the "Hidden" Derivatives!

In the last unit, you learned how to differentiate basic functions like \(x^2\) or \(\sin(x)\). But what happens when functions are tucked inside one another, or when \(x\) and \(y\) are all mixed up together? That is exactly what we are going to tackle here!

Unit 3 is all about expanding your toolkit. By the end of these notes, you will be able to peel back the layers of complex functions just like an onion. Don’t worry if this seems tricky at first—calculus is a skill, and with these steps, you’ll be a pro in no time.

3.1 The Chain Rule: Dealing with "Functions within Functions"

The Chain Rule is arguably the most important rule in calculus. We use it when we have a composite function—basically, one function sitting inside another, like \(f(g(x))\).

The Onion Analogy

Think of a composite function like an onion. To get to the center, you have to peel the outer layer first, then the next layer, and so on. In calculus terms: Differentiate the outside first (leaving the inside alone), then multiply by the derivative of the inside.

The Formula

If \(y = f(g(x))\), then the derivative is:
\( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

Step-by-Step Process:

1. Identify the "Outside" and "Inside": For \(\sin(x^2)\), the outside is \(\sin(\square)\) and the inside is \(x^2\).
2. Differentiate the Outside: The derivative of \(\sin\) is \(\cos\). Keep the inside the same: \(\cos(x^2)\).
3. Differentiate the Inside: The derivative of \(x^2\) is \(2x\).
4. Multiply them together: \( \cos(x^2) \cdot 2x \).

Quick Review: Always ask yourself, "Is there something other than just a plain \(x\) inside the parentheses?" If yes, you must use the Chain Rule!

Common Mistake: Students often forget to "leave the inside alone" during the first step. For \((5x + 1)^3\), the derivative is not \(3(5)^2\). It is \(3(5x+1)^2 \cdot 5\).

3.2 Implicit Differentiation: When \(x\) and \(y\) are Tangled

Most functions you've seen look like \(y = \dots\). These are explicit. But what if you have something like \(x^2 + y^2 = 25\)? This is implicit because \(y\) isn't isolated.

The "Tag-Along" Rule

When you differentiate terms with \(x\), do it normally. When you differentiate terms with \(y\), do it normally but always multiply by \( \frac{dy}{dx} \) (or \(y'\)).

Why?

Because of the Chain Rule! We treat \(y\) as a function of \(x\). So, the derivative of \(y^2\) is \(2y\) (the outside) times \( \frac{dy}{dx} \) (the inside).

How to Solve:

1. Differentiate both sides of the equation with respect to \(x\).
2. Remember to add \( \frac{dy}{dx} \) every time you differentiate a \(y\) term.
3. Use algebra to get all the terms with \( \frac{dy}{dx} \) on one side and everything else on the other.
4. Factor out \( \frac{dy}{dx} \) and solve.

Key Takeaway: Implicit differentiation is just the Chain Rule in disguise! Every \(y\) is an "inside function."

3.3 Differentiating Inverse Functions

If you know the derivative of a function \(f\), you can find the derivative of its inverse \(f^{-1}\) without even knowing the formula for the inverse!

The Reciprocal Relationship

If the point \((a, b)\) is on the graph of \(f\), then the point \((b, a)\) is on the graph of \(f^{-1}\). The slopes at these points are reciprocals.

The Formula

Let \(g(x)\) be the inverse of \(f(x)\). Then:
\( g'(a) = \frac{1}{f'(g(a))} \)

Memory Aid: "The slope of the inverse is one over the slope of the original." Just make sure you are plugging the correct \(x\) and \(y\) values into the right functions!

Step-by-Step:
1. Find the point on the original function. If you need the derivative of the inverse at \(x = 5\), find where the original function equals 5.
2. Find the derivative of the original function \(f'(x)\).
3. Plug your \(x\) value into \(f'(x)\).
4. Flip that number (take the reciprocal)!

3.4 Differentiating Inverse Trigonometric Functions

These formulas are "must-memorize" for the AP Exam. While they look scary, they follow a pattern.

The Big Three:

1. Arcsine: \( \frac{d}{dx}(\arcsin u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \)
2. Arccosine: \( \frac{d}{dx}(\arccos u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \) (It's just the negative of sine!)
3. Arctangent: \( \frac{d}{dx}(\arctan u) = \frac{1}{1+u^2} \cdot \frac{du}{dx} \)

Did you know? You don't need to memorize the formulas for arccosecant, arcsecant, or arccotangent as often, but they follow the same "co-" rule: the "co-" version is always the negative of the regular version.

Quick Tip: Always remember the Chain Rule here! If you are differentiating \(\arctan(5x)\), your \(u\) is \(5x\), so you must multiply by \(5\) at the end.

3.5 Selecting Procedures for Derivatives

By now, you have a lot of rules: Power, Product, Quotient, Chain, and Implicit. The hardest part is often knowing which one to use first.

Strategy Guide:

- Is \(y\) alone? If no, use Implicit Differentiation.
- Is one function inside another? Use the Chain Rule.
- Are two things multiplying? Use the Product Rule.
- Is there a fraction? Use the Quotient Rule.

Pro Tip: Sometimes you can simplify an expression using algebra or log properties before you differentiate to make the work much easier!

3.6 Higher-Order Derivatives

Sometimes the AP exam will ask for the second derivative (\( \frac{d^2y}{dx^2} \)) of an implicit equation. This is a common "trick" spot for students.

How to do it:

1. Find the first derivative (\( \frac{dy}{dx} \)) using implicit differentiation.
2. Differentiate that expression again.
3. The Substitution Step: Your second derivative will likely contain the term \( \frac{dy}{dx} \). You must substitute your answer from Step 1 back into the equation to get the final answer in terms of \(x\) and \(y\).

Summary: Don't stop at the second derivative! Always look to plug the first derivative back into the second derivative to simplify.

Unit 3 Final Review Box

Chain Rule: Derivative of outside \(\times\) derivative of inside.
Implicit: Add a \( \frac{dy}{dx} \) every time you touch a \(y\).
Inverse: Slopes are reciprocals at reflected points.
Inverse Trig: Memorize \(\arcsin\) and \(\arctan\) formulas; they appear frequently!
Higher Order: Substitute the first derivative into the second.

You've got this! Unit 3 is the bridge to the rest of Calculus. Master these "layers" and the rest of the course will feel much smoother!