Business Calculus Quiz Study Notes: Differentiation and Applications

Differentiation and Applications

Finding Horizontal Tangents Using the Chain Rule

This topic explores how to use the chain rule to differentiate composite functions and determine where a curve has horizontal tangent lines. Horizontal tangents occur where the derivative of the function equals zero.

• Chain Rule: The chain rule is used to differentiate composite functions. If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

• Horizontal Tangents: These occur at points where $\frac{dy}{dx} = 0$.

• Example: For $y = x^4(x^2 - 4x)^3$, apply the product and chain rules to find $\frac{dy}{dx}$, set it equal to zero, and solve for $x$ to find the locations of horizontal tangents.

Steps:

1. Let $y = x^4(x^2 - 4x)^3$.

2. Let $u = x^2 - 4x$, so $y = x^4 u^3$.

3. Use the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x^4) \cdot u^3 + x^4 \cdot \frac{d}{dx}(u^3)$.

4. $\frac{d}{dx}(x^4) = 4x^3$.

5. $\frac{d}{dx}(u^3) = 3u^2 \cdot \frac{du}{dx} = 3(x^2 - 4x)^2 \cdot (2x - 4)$.

6. Combine: $\frac{dy}{dx} = 4x^3(x^2 - 4x)^3 + x^4 \cdot 3(x^2 - 4x)^2(2x - 4)$.

7. Set $\frac{dy}{dx} = 0$ and solve for $x$ to find horizontal tangents.

Application: The points where $\frac{dy}{dx} = 0$ are the $x$-values of horizontal tangents. Substitute these $x$-values back into the original function to find the corresponding $y$-values, giving the coordinates of the horizontal tangents.

Derivative of Composite Functions (Chain Rule)

This topic covers how to differentiate functions defined in terms of another variable, using the chain rule.

• Chain Rule for Composite Functions: If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

• Example: For $y = 5u^2 \cdot 3u$, where $u = x^2 + 1$, first find $\frac{dy}{du}$ and $\frac{du}{dx}$, then multiply.

Steps:

1. Let $y = 5u^2 \cdot 3u = 15u^3$.

2. $\frac{dy}{du} = 45u^2$.

3. $u = x^2 + 1$, so $\frac{du}{dx} = 2x$.

4. $\frac{dy}{dx} = 45u^2 \cdot 2x = 90x(u^2)$.

5. Substitute $u = x^2 + 1$ to get $\frac{dy}{dx} = 90x(x^2 + 1)^2$.

Application: This process is essential for differentiating functions in business calculus, especially when dealing with cost, revenue, or profit functions that are composed of other functions.

Evaluating Composite Functions

This topic involves evaluating composite functions, which is a common operation in calculus and its applications.

• Composite Function: If $f(x)$ and $g(x)$ are functions, then $f(g(x))$ is the composite function.

• Example: Given $f(x) = \frac{x^4 + 1}{x^2 + 1}$ and $g(x) = \sqrt{x}$, find $f(g(4))$.

Steps:

1. Find $g(4) = \sqrt{4} = 2$.

2. Substitute into $f(x)$: $f(2) = \frac{2^4 + 1}{2^2 + 1} = \frac{16 + 1}{4 + 1} = \frac{17}{5}$.

Application: Composite functions are used in business calculus to model scenarios where one quantity depends on another, such as demand depending on price, and price depending on time.