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When we talk about the instantaneous rate of change, we refer to how a function changes at a specific point. Imagine if you were looking at the speedometer of a car and you wanted to know the exact speed at a given instant. Similarly, in calculus, the instantaneous rate of change at a point is found using the derivative.

For the cost function given, this is equivalent to the derivative evaluated at a particular production level. The derivative represents the slope of the tangent line at any point on a curve, which gives us the exact rate of change at that specific point. In the context of this exercise, the instantaneous rate of change tells us how the cost is changing exactly when 100 units are produced.

To find this, you differentiate the cost function \( C(x) = 5000 + 10x + 0.05x^2 \), giving \( C'(x) = 10 + 0.1x \). Then, you plug \( x = 100 \) into this derivative to find \( C'(100) = 20 \). This means the cost increases by $20 for each additional unit produced around 100 units.

Marginal cost is a concept closely related to the instantaneous rate of change. It signifies the cost of producing one extra unit of a product at a given production level. It's an important measure for businesses because it indicates how the total cost is changing with respect to production volume.

In economic terms, marginal cost is the derivative of the cost function with respect to the number of products. It helps businesses decide how many items to produce. If the marginal cost is less than the price at which the product sells, producing more could lead to more profit.

In our example, the marginal cost at \( x = 100 \) units, which we found through differentiation, is \(20. This indicates that the cost of producing the 101st unit is about \)20, underlining how vital marginal cost is in production planning.

Differentiation is a fundamental concept in calculus, used to compute the derivative of a function. Derivatives help us understand how a function behaves by providing us with rates of change. When we differentiate a function, we find a new function that gives the slope of the original function at any point.

This is useful in a myriad of applications, such as finding velocities, predicting future values, and analyzing changing rates when it comes to economics and engineering. Calculating derivatives often involves rules and shortcuts, such as the power rule, which allows us to differentiate polynomials easily. For a function like \( C(x) = 5000 + 10x + 0.05x^2 \), applying differentiation rules provides \( C'(x) = 10 + 0.1x \), which is simple enough to evaluate for any \( x \).

Understanding and applying differentiation is, therefore, crucial for solving real-world problems, like determining instant rates of change or costs, as we did in the exercise above.