Differentials and Increments in Business Calculus

Section 6: Differentials

Introduction to Differentials

Differentials provide a method for approximating small changes in a function's value based on its derivative. This concept is essential in business calculus for estimating changes in cost, revenue, and profit functions when the input variable changes by a small amount.

Increments and the Difference Quotient

Definition of Increments

• Increment in x (Δx): The change in the independent variable x, denoted as Δx = x_2 - x_1.

• Increment in y (Δy): The corresponding change in the dependent variable y, given by Δy = f(x_2) - f(x_1) = f(x_1 + Δx) - f(x_1).

• The difference quotient is written as $\frac{f(x + \Delta x) - f(x)}{\Delta x}$, which approaches the derivative as Δx approaches 0.

Example: For $y = f(x) = x^3$, if x changes from 2 to 2.1, then $\Delta x = 0.1$ and $\Delta y = f(2.1) - f(2) = 9.261 - 8 = 1.261$.

Graphical Interpretation of Increments

Increments can be visualized on the graph of a function. The secant line connects two points on the curve, and its slope is given by the difference quotient. The tangent line at a point approximates the curve near that point, and its slope is the derivative.

Differentials: Definition and Calculation

Definition of the Differential

• If $y = f(x)$ is differentiable, the differential $dy$ is defined as $dy = f'(x) dx$, where $dx$ is a small change in x.

• The differential $dy$ approximates the actual change $\Delta y$ for small $dx$.

Symbolically:

$dy = f'(x) dx$

or

$df = f'(x) dx$

Interpretation of Differentials

• Δx and dx both represent a change in x.

• Δy is the actual change in y, while dy is the approximate change in y, estimated using the derivative.

• For small changes, $\Delta y \approx dy = f'(x) dx$.

Example: Calculating Differentials

Find $dy$ for $f(x) = x^2 + 3x$ at $x = 2$ and $dx = 0.1$.

• $f'(x) = 2x + 3$

• $dy = f'(x) dx = (2x + 3) dx$

• For $x = 2$, $dy = (2 \times 2 + 3) \times 0.1 = 0.7$

Comparing Increments and Differentials

Example: Quadratic Function

Let $y = f(x) = x^2 - 6x$. Find $\Delta y$ and $dy$ for $x = 2$ and various values of $\Delta x$.

• $f'(x) = 2x - 6$

• At $x = 2$, $f'(2) = 2 \times 2 - 6 = -2$

• $dy = f'(2) dx = -2 dx$

The following table compares $\Delta y$ and $dy$ for different values of $\Delta x$:

Δx	Δy	dy
0.1	0.19	0.2
0.2	0.36	0.4
0.3	0.51	0.6

Applications in Business: Cost, Revenue, and Profit

Using Differentials for Approximation

Differentials are useful for estimating changes in business functions such as cost, revenue, and profit when production levels change slightly.

• Cost function: $C(x) = 5,000 + 2x$

• Revenue function: $R(x) = 8,000x - x^2$

• Profit function: $P(x) = R(x) - C(x) = 8,000x - x^2 - (5,000 + 2x)$

Example: Estimate the change in revenue and profit if production increases from 2,000 to 2,010 units per week ($dx = 10$).

• $R'(x) = 8,000 - 2x$

• At $x = 2,000$, $R'(2,000) = 8,000 - 4,000 = 4,000$

• $dR = R'(2,000) dx = 4,000 \times 10 = 40,000$

• $P'(x) = R'(x) - C'(x) = (8,000 - 2x) - 2$

• At $x = 2,000$, $P'(2,000) = 8,000 - 4,000 - 2 = 3,998$

• $dP = P'(2,000) dx = 3,998 \times 10 = 39,980$

Thus, the differential provides a quick estimate of the change in revenue and profit for a small increase in production.

Additional info: The use of differentials is foundational in business calculus for making rapid, approximate calculations that inform managerial decisions, especially when exact computation is complex or unnecessary for small changes.