2.3 – Classwork (Basic Rules & Context)

Learning Objectives

By the end of this section, you should be able to:

  • Apply derivative rules to solve mixed practice problems
  • Connect derivative concepts to real-world applications
  • Solve problems involving multiple derivative rules
  • Interpret derivatives in various contexts
  • Build confidence through extensive practice
Section Progress

Derivative Calculator

Enter a function to find its derivative:

The derivative will appear here.

Motion Problem Solver

Enter position function to find velocity and acceleration:

Velocity: -

Acceleration: -

Times at rest: -

Related Rates Calculator

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Practice Problem Sets

Problem 1: Basic Differentiation

Find the derivative of \( f(x) = 4x^3 - 2x^2 + 7x - 5 \)

\[ f'(x) = \frac{d}{dx}[4x^3] - \frac{d}{dx}[2x^2] + \frac{d}{dx}[7x] - \frac{d}{dx}[5] \]

\[ f'(x) = 12x^2 - 4x + 7 \]

Problem 2: Motion Application

A ball is thrown upward with position function \( s(t) = -16t^2 + 64t + 80 \) feet. Find:

(a) The velocity function

(b) When the ball reaches its maximum height

(c) The velocity when the ball hits the ground

(a) Velocity function:

\[ v(t) = s'(t) = -32t + 64 \]

(b) Maximum height occurs when velocity is zero:

\[ -32t + 64 = 0 \]

\[ t = 2 \text{ seconds} \]

(c) Ball hits ground when \( s(t) = 0 \):

\[ -16t^2 + 64t + 80 = 0 \]

\[ t^2 - 4t - 5 = 0 \]

\[ (t-5)(t+1) = 0 \]

\[ t = 5 \text{ seconds} \]

\[ v(5) = -32(5) + 64 = -96 \text{ ft/s} \]

Problem 3: Related Rates

A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?

Step 1: Set up the relationship

Let \( x \) = distance from wall to bottom of ladder

Let \( y \) = height of top of ladder

\[ x^2 + y^2 = 10^2 \]

Step 2: Differentiate with respect to time

\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]

Step 3: Find values when \( x = 6 \)

\[ 6^2 + y^2 = 100 \]

\[ y^2 = 64 \]

\[ y = 8 \text{ ft} \]

Step 4: Solve for \( \frac{dy}{dt} \)

\[ 2(6)(1) + 2(8)\frac{dy}{dt} = 0 \]

\[ 12 + 16\frac{dy}{dt} = 0 \]

\[ \frac{dy}{dt} = -\frac{12}{16} = -0.75 \text{ ft/s} \]

The top is sliding down at 0.75 ft/s.

Quick Concept Check

Question 1: What is the derivative of \( f(x) = 5x^4 - 3x^2 + 2 \)?

\( 20x^3 - 6x \)
\( 5x^3 - 3x \)
\( 20x^3 - 6x + 2 \)
\( 4x^3 - 2x \)
✓ Correct! You applied the power rule correctly to each term.
✗ Remember: The derivative of a constant is 0, and use the power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \)

Question 2: If \( s(t) \) represents position, what does \( s'(t) \) represent?

Acceleration
Velocity
Jerk
Distance
✓ Correct! Velocity is the derivative of position.
✗ Remember: Position → Velocity → Acceleration → Jerk (each is the derivative of the previous)