Composition of Functions and the Chain Rule in Business Calculus

Functions and Their Composition

Definition of Function Composition

Function composition involves applying one function to the results of another. If f and g are functions, the composition (f ˆ g)(x) means applying g to x first, then applying f to the result.

• Notation: $(f \circ g)(x) = f(g(x))$

• Order matters: $(f \circ g)(x) \neq (g \circ f)(x)$ in general.

Example 1: Computing Compositions

• Let $f(x) = e^{x^2} + 3$ and $g(x) = -x^2 + 5x - 6$.

• Compute $(f \circ g)(x)$:

  • First, find $g(x)$: $g(x) = -x^2 + 5x - 6$

  • Then, $f(g(x)) = e^{(-x^2 + 5x - 6)^2} + 3$

• Compute $(g \circ f)(x)$:

  • First, find $f(x)$: $f(x) = e^{x^2} + 3$

  • Then, $g(f(x)) = -[e^{x^2} + 3]^2 + 5[e^{x^2} + 3] - 6$

Decomposing Functions into Simpler Functions

Many complex functions can be written as the composition of two or more simpler functions. This is useful for applying the chain rule in differentiation.

• Example 2: Write each function as a composite of two simpler functions:

  • $f(x) = 23(2x-7)^4$

    • Let $u(x) = 2x-7$, $v(u) = 23u^4$

    • Then $f(x) = v(u(x))$

  • $y = \sqrt{4 - 7x} + 9$

    • Let $u(x) = 4 - 7x$, $v(u) = \sqrt{u} + 9$

    • Then $y = v(u(x))$

  • $g(x) = \ln(4x^2 + 9)$

    • Let $u(x) = 4x^2 + 9$, $v(u) = \ln(u)$

    • Then $g(x) = v(u(x))$

  • $f(x) = -2e^{3x^2}$

    • Let $u(x) = 3x^2$, $v(u) = -2e^{u}$

    • Then $f(x) = v(u(x))$

The Chain Rule

Statement of the Chain Rule

The chain rule is a fundamental rule for finding the derivative of a composite function. If $y = f(g(x))$, then the derivative of $y$ with respect to $x$ is:

• Formula:

  $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

• This means: Differentiate the outer function (with respect to the inner function), then multiply by the derivative of the inner function.

Examples: Applying the Chain Rule

• Example 3: Compute the derivative of the functions in Example 2.

  • $f(x) = 23(2x-7)^4$

    • Let $u = 2x-7$, so $f(x) = 23u^4$

    • $\frac{df}{dx} = 23 \cdot 4u^3 \cdot \frac{du}{dx} = 92(2x-7)^3 \cdot 2 = 184(2x-7)^3$

  • $y = \sqrt{4 - 7x} + 9$

    • Let $u = 4 - 7x$, $y = \sqrt{u} + 9$

    • $\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot (-7) = -\frac{7}{2\sqrt{4-7x}}$

  • $g(x) = \ln(4x^2 + 9)$

    • Let $u = 4x^2 + 9$, $g(x) = \ln(u)$

    • $g'(x) = \frac{1}{u} \cdot 8x = \frac{8x}{4x^2 + 9}$

  • $f(x) = -2e^{3x^2}$

    • Let $u = 3x^2$, $f(x) = -2e^{u}$

    • $f'(x) = -2e^{3x^2} \cdot 6x = -12x e^{3x^2}$

Further Examples

• Example 4: Compute the derivative of the following functions:

  • $f(x) = [\ln(e^{x^2} + x)]^5$

    • Let $u = \ln(e^{x^2} + x)$, $f(x) = u^5$

    • $f'(x) = 5u^4 \cdot \frac{1}{e^{x^2} + x} \cdot (2x e^{x^2} + 1)$

  • $f(x) = (x^2 + 1) \log(5x + 8)$

    • Product rule and chain rule needed:

    • $f'(x) = 2x \log(5x + 8) + (x^2 + 1) \cdot \frac{1}{(5x + 8) \ln 10} \cdot 5$

  • $f(x) = e^{3(x + 7)^4}$

    • Let $u = 3(x + 7)^4$, $f(x) = e^{u}$

    • $f'(x) = e^{3(x + 7)^4} \cdot 12(x + 7)^3$

  • $f(x) = \frac{\ln(x^2 + x)}{\sqrt{3x^2 + 1}}$

    • Quotient rule and chain rule needed:

    • $f'(x) = \frac{\frac{1}{x^2 + x}(2x + 1) \sqrt{3x^2 + 1} - \ln(x^2 + x) \cdot \frac{1}{2\sqrt{3x^2 + 1}} \cdot 6x}{3x^2 + 1}$

Applications: Rate of Change in Exponential Models

Example 5: Temperature Change in a Yeast Culture

Suppose a yeast culture at room temperature (68°F) is placed in a refrigerator at 38°F. The temperature $T$ after $t$ hours is given by:

$T = 30e^{-0.241t} + 38, \quad t \geq 0$

• Find the rate of change of temperature at $t = 1$ and $t = 4$ hours.

• Differentiate $T$ with respect to $t$:

  • $\frac{dT}{dt} = 30 \cdot (-0.241) e^{-0.241t} = -7.23 e^{-0.241t}$

• At $t = 1$:

  • $\frac{dT}{dt}\bigg|_{t=1} = -7.23 e^{-0.241 \cdot 1} \approx -7.23 \times 0.786 = -5.69$ (rounded)

• At $t = 4$:

  • $\frac{dT}{dt}\bigg|_{t=4} = -7.23 e^{-0.241 \cdot 4} \approx -7.23 \times 0.380 = -2.75$ (rounded)

• Interpretation: The negative sign indicates the temperature is decreasing. The rate slows as time increases.

Summary Table: Chain Rule Applications

Function	Decomposition	Derivative
$f(x) = 23(2x-7)^4$	$u = 2x-7$, $v(u) = 23u^4$	$184(2x-7)^3$
$g(x) = \ln(4x^2 + 9)$	$u = 4x^2 + 9$, $v(u) = \ln(u)$	$\frac{8x}{4x^2 + 9}$
$f(x) = -2e^{3x^2}$	$u = 3x^2$, $v(u) = -2e^{u}$	$-12x e^{3x^2}$
$T = 30e^{-0.241t} + 38$	$u = -0.241t$, $v(u) = 30e^{u} + 38$	$-7.23 e^{-0.241t}$

Additional info: The above notes expand on the chain rule, function composition, and their applications, including examples relevant to business calculus such as exponential decay in temperature models.