#AP Calculus AB/BC: Derivatives - The Ultimate Study Guide 🚀

Hey there, future calculus master! This guide is your go-to resource for acing the derivatives section of the AP Calculus exam. Let's break down everything you need to know, making it super clear and easy to remember. Time is precious, so let's get right to it!

This unit is crucial! It makes up 10-12% of the Calculus AB exam and 4-7% of the Calculus BC exam. Nail this, and you're setting yourself up for success. 💪

#1. What is a Derivative? 🤔

#1.1 The Idea of Instantaneous Change

Imagine Michael Phelps swimming. His average speed is easy to calculate, but what about his speed at a single moment? That's where derivatives come in!

We use a function to represent his position over time. The slope of this function over an interval gives us average speed. To find his speed at a specific instant, we shrink that interval down to almost zero. This process is the essence of a derivative.

#1.2 Formal Definition

Let's get a bit formal. If we have a function s(t) representing position over time, the derivative (instantaneous speed) is:

$$\frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{s(t + \Delta t) - s(t)}{\Delta t}$$

Here, ds/dt is the derivative of s with respect to t. The d notation indicates we're dealing with infinitesimally small changes—derivatives!

In the xy-plane, for a function f(x), the derivative is:

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

This can also be written as f'(x) or y'. While these notations are useful, dy/dx helps you remember that you're finding the rate of change of y with respect to x.

#1.3 Derivative as the Slope of the Tangent Line

Imagine zooming in on a curve until it looks like a straight line. That line is the tangent line, and its slope is the derivative. The tangent line touches the curve at just one point (though it might cross it elsewhere).

#1.4 When Derivatives Don't Exist

Not all functions have derivatives everywhere. If a function has a sharp corner, a cusp, or a vertical tangent, the derivative doesn't exist at that point. If you can draw multiple tangent lines, that's a red flag 🚩. Smooth, continuous functions are your friends here—they usually have derivatives.

Here's an example of a function with a vertical tangent line:

#2. Computing Derivatives 🧮

Now, let's get into the rules that make finding derivatives much easier.

#2.1 Basic Rules

• Constant Rule: The derivative of a constant is always 0. (e.g., if f(x) = 5, then f'(x) = 0)

• Constant Multiple Rule: If you have a constant multiplied by a function, the derivative is the constant times the derivative of the function. (e.g., d/dx(3x) = 3 * d/dx(x))

• Sum Rule: The derivative of a sum is the sum of the derivatives. (e.g., if f(x) = sin(x) + 3x, then f'(x) = d/dx(sin(x)) + d/dx(3x))

• Product Rule: If h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x). (Remember: first times derivative of the second + second times derivative of the first)

• Quotient Rule: If h(x) = f(x)/g(x), then h'(x) = (f'(x)g(x) - f(x)g'(x))/(g(x))². (Remember: low d high minus high d low, over low squared)

#2.2 Power Rule

This rule is derived from the limit definition of the derivative, but you can use it directly to save time. For instance, if f(x) = x³, then f'(x) = 3x².

#2.3 Special Derivatives

Here are some derivatives you should memorize:

#2.4 Example

Let's find the derivative of f(x) = x³ + 3x ln(x) + 5:

Solution:

$$\begin{aligned} f'(x) &= \frac{d}{dx}(x^3) + \frac{d}{dx}(3x \ln(x)) + \frac{d}{dx}(5) \\ &= 3x^2 + 3(\ln(x) + x \cdot \frac{1}{x}) + 0 \\ &= 3x^2 + 3\ln(x) + 3 \end{aligned}$$

#Final Exam Focus 🎯

Okay, you're almost there! Here's what to prioritize in your final review:

• Key Concepts: Understand the definition of a derivative, the relationship between derivatives and tangent lines, and when derivatives don't exist.

• Rules: Master the power rule, product rule, quotient rule, and derivatives of common functions. Practice, practice, practice!

• Applications: Derivatives are used to find rates of change, slopes, and optimization problems. Be ready to apply what you know.

• Common Pitfalls: Watch out for vertical tangents and points where functions are not smooth or continuous. Double-check your algebra and trig identities.

#Practice Questions

Practice Question

#Multiple Choice Questions

1. If $f(x) = 4x^3 - 5x^2 + 2x - 7$, then $f'(x)$ is: (A) $12x^2 - 10x + 2$ (B) $12x^2 - 10x - 7$ (C) $4x^2 - 5x + 2$ (D) $12x^3 - 10x^2 + 2x$

2. The derivative of $y = \frac{x^2}{x+1}$ is: (A) $\frac{2x}{1}$ (B) $\frac{x^2+2x}{(x+1)^2}$ (C) $\frac{x^2+2x}{(x+1)}$ (D) $\frac{x^2-2x}{(x+1)^2}$

3. If $f(x) = \sin(x) \cos(x)$, then $f'(\frac{\pi}{4})$ is: (A) 0 (B) 1 (C) -1 (D) $\frac{1}{2}$

#Free Response Question

Let $f(x) = x^3 - 6x^2 + 9x + 1$.

(a) Find $f'(x)$. (b) Find all values of $x$ where the tangent line to the graph of $f$ is horizontal. (c) Find $f''(x)$. (d) Determine the intervals on which $f$ is concave up and concave down.

#Scoring Breakdown:

(a) 2 points: 1 point for each correct term in the derivative. $f'(x) = 3x^2 - 12x + 9$

(b) 3 points: 1 point for setting $f'(x) = 0$, 1 point for factoring or using the quadratic formula, 1 point for correct values of $x$. $3x^2 - 12x + 9 = 0 \implies 3(x^2 - 4x + 3) = 0 \implies 3(x-1)(x-3) = 0 \implies x = 1, 3$

(c) 1 point: Correct second derivative. $f''(x) = 6x - 12$

(d) 3 points: 1 point for setting $f''(x) = 0$, 1 point for the correct interval, 1 point for correct concavity. $6x - 12 = 0 \implies x = 2$. Concave down: $(-\infty, 2)$. Concave up: $(2, \infty)$

#You've Got This! 🎉

Remember, you've got the tools and the knowledge to succeed. Stay calm, stay focused, and trust in your preparation. You're going to do great! Good luck on your exam!