The chain rule is a powerful tool for differentiating composite functions. It allows us to break down complex functions into simpler parts, making differentiation more manageable. This technique is crucial for solving real-world problems involving rates of change.

Mastering the chain rule opens up a world of possibilities in calculus. From trigonometric functions to exponentials and logarithms, this rule helps us tackle a wide range of composite functions. It's a key skill for any calculus student to develop.

Derivatives of Composite Functions

Chain rule for composite functions

• Differentiates composite functions (function of a function)
• If $f(x) = h(g(x))$, then $f'(x) = h'(g(x)) \cdot g'(x)$
  • Identifies outer function $h(x)$ and inner function $g(x)$
  • Differentiates outer function, keeping inner function as a variable
  • Multiplies result by derivative of inner function
• Examples:
  • If $f(x) = (3x^2 + 1)^5$, then $f'(x) = 5(3x^2 + 1)^4 \cdot 6x$
  • If $f(x) = \sqrt{2x - 1}$, then $f'(x) = \frac{1}{2\sqrt{2x - 1}} \cdot 2$

Order in chain rule application

• Applies chain rule to complex composite functions by working from outside in
  • Differentiates outermost function first, keeping inner functions as variables
  • Multiplies result by derivative of next inner function
  • Continues process until all functions differentiated
• Example: If $f(x) = \sin(e^{x^2 + 1})$, then:
  1. $f'(x) = \cos(e^{x^2 + 1}) \cdot \frac{d}{dx}(e^{x^2 + 1})$
  2. $f'(x) = \cos(e^{x^2 + 1}) \cdot e^{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1)$
  3. $f'(x) = \cos(e^{x^2 + 1}) \cdot e^{x^2 + 1} \cdot 2x$

Derivatives of complex composite functions

• Trigonometric functions:
  • If $f(x) = \sin(g(x))$, then $f'(x) = \cos(g(x)) \cdot g'(x)$
  • If $f(x) = \cos(g(x))$, then $f'(x) = -\sin(g(x)) \cdot g'(x)$
  • If $f(x) = \tan(g(x))$, then $f'(x) = \sec^2(g(x)) \cdot g'(x)$
• Exponential functions:
  • If $f(x) = e^{g(x)}$, then $f'(x) = e^{g(x)} \cdot g'(x)$
  • If $f(x) = a^{g(x)}$ ($a > 0$, $a \neq 1$), then $f'(x) = a^{g(x)} \cdot \ln(a) \cdot g'(x)$
• Logarithmic functions:
  • If $f(x) = \ln(g(x))$, then $f'(x) = \frac{1}{g(x)} \cdot g'(x)$
  • If $f(x) = \log_a(g(x))$ ($a > 0$, $a \neq 1$), then $f'(x) = \frac{1}{g(x) \cdot \ln(a)} \cdot g'(x)$

Chain rule in real-world applications

• Finds rates of change in real-world problems using composite functions
• Steps to solve:
  1. Identifies composite function relating quantities in problem
  2. Uses chain rule to differentiate composite function
  3. Substitutes given values into derivative to find rate of change
• Example: Volume of sphere increasing at 10 cm³/min, find rate radius increasing when radius is 5 cm
  • Volume of sphere: $V = \frac{4}{3}\pi r^3$
  • $\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$
  • Substitute values: $10 = 4\pi (5)^2 \cdot \frac{dr}{dt}$
  • Solve for $\frac{dr}{dt}$ to find rate of change of radius