Overview

The chain rule allows you to differentiate composite functions — functions within functions. Implicit differentiation is a powerful technique for finding derivatives when a function is not solved explicitly for one variable. Both are essential for solving real-world calculus problems.

Key Concepts and Structures

• Chain Rule: \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)
• Recognizing Composite Functions: For example, \sin(x^2) is f(g(x)) with f(u) = \sin(u), g(x) = x^2
• Step-by-Step Chain Rule:
  • Differentiate the outer function
  • Multiply by the derivative of the inner function
• Implicit Differentiation: When y is mixed with x, treat y as a function of x and use the chain rule: \frac{d}{dx}(y^n) = n y^{n-1} \cdot \frac{dy}{dx}
• Example: If x^2 + y^2 = 25, then \frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}
• Solve for \frac{dy}{dx} after differentiating both sides

Quick Tip

Practice Problems

1. Use the chain rule to find \frac{d}{dx}(\cos(x^2))
   Show Solution

   Step 1: Let f(u) = \cos(u) and u = x^2

   Step 2: f'(u) = -\sin(u) and du/dx = 2x

   Final Answer: \frac{d}{dx}(\cos(x^2)) = -\sin(x^2) \cdot 2x

2. Use implicit differentiation to find \frac{dy}{dx} if x^2 + y^2 = 25
   Show Solution

   Step 1: Differentiate both sides with respect to x:

   • \frac{d}{dx}(x^2) = 2x
   • \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}

   So: 2x + 2y \cdot \frac{dy}{dx} = 0

   Step 2: Solve for \frac{dy}{dx}:

   2y \cdot \frac{dy}{dx} = -2x → \frac{dy}{dx} = -x/y

   Final Answer: \frac{dy}{dx} = -x/y