91Ó°ÊÓ

A cost function like the one given in this exercise is used to understand how production costs change with different levels of output. In our case, the cost function is represented as: \[ C(x) = \sqrt{x^3}(x + 100) + 1000. \] Here, \( x \) represents the quantity of products. The cost function helps businesses determine how much they will spend when production changes. By substituting different values of \( x \), we can calculate the exact cost for that production level.

• The term \( \sqrt{x^3} \) indicates some form of scale economy or capacity effect.
• The part \( (x + 100) \) suggests there might be an added fixed cost component or adjustment for larger production.
• The \(+1000\) is likely a constant, fixed overhead cost, independent of production level.

Finding the derivative of a function involves determining how the function changes as its input changes. For our cost function \( C(x) = \sqrt{x^3}(x + 100) + 1000 \), we need to find \( C'(x) \) in order to use linear approximation. Differentiation is crucial here because it provides the rate at which costs will change for small changes in \( x \). The derivative \( C'(x) \) tells us how sensitive the cost is to changes in production levels. To find \( C'(x) \), we apply the product rule and the chain rule, which are methods used to differentiate complex functions.

• The product rule is used when a function is the product of two simpler functions. It states: if \( u(x) \) and \( v(x) \) are functions, then \( (u \cdot v)' = u'v + uv' \).
• The chain rule is applied when functions are composed; i.e., one function inside another. It helps us find the derivative of the outer function multiplied by the derivative of the inner function.

The slope of the tangent line at a specific point on a curve provides the instantaneous rate of change of the function at that point. In context, at \( x = 900 \), this slope represents how the cost is expected to change for an especially small increase in production. To find this slope, we evaluate the derivative \( C'(x) \) at \( x = 900 \). This involves first finding \( C'(x) \) using the derivative calculation techniques we discussed and then substituting \( x = 900 \) into the resulting expression.

• The tangent line can be thought of as the best linear representation of the function near the chosen point.
• The slope of this line is what gives us the rate of cost change per unit of production around \( x = 900 \).

The product and chain rules are fundamental techniques in calculus used to differentiate more complicated functions such as our cost function \( \sqrt{x^3}(x + 100) \).

Product Rule

In our cost function, \( \sqrt{x^3} \) and \( (x + 100) \) are separate expressions that are multiplied together. Thus, we use the product rule:

• Take the derivative of \( \sqrt{x^3} \) and multiply by \( (x + 100) \).
• Then take the derivative of \( (x + 100) \) and multiply by \( \sqrt{x^3} \).
• Add these two results to get the derivative of the entire product.

Chain Rule

The chain rule is used because \( \sqrt{x^3} \) can be viewed as a composition of functions: an outer square root function and an inner cubic function \( x^3 \).

• Differentiate the outer function \( \sqrt{u} \) where \( u = x^3 \), resulting in \( \frac{1}{2\sqrt{u}} \).
• Then differentiate the inner function, \( x^3 \), which is \( 3x^2 \).
• Multiply these derivatives together to achieve the derivative of the chain.

These rules help us calculate the accurately differentiated form of the given cost function for further analysis and application of linear approximation.