Problem 27 - Single Variable Calculus

Question

Find the derivative of the function. \(A=0.001 x^{2}-0.4 x+5+\frac{200}{x}\)

Step-by-Step Solution

Answer

The derivative of the function \(A(x) = 0.001x^2 - 0.4x + 5 + \frac{200}{x}\) is \(\frac{dA}{dx} = 0.002x - 0.4 - 200x^{-2}\).

Step 1: Differentiating polynomial terms

To differentiate the polynomial terms, we apply the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \). The polynomial terms are \(0.001x^2\) and \(-0.4x\), so their derivatives are: \(\frac{d}{dx}(0.001x^2) = 2(0.001)x^{2-1} = 0.002x\) \(\frac{d}{dx}(-0.4x) = (-0.4)x^{1-1} = -0.4\)

Step 2: Differentiating the fraction term

To differentiate the fraction term \(\frac{200}{x}\), we rewrite it as a power of x: \(200x^{-1}\). Applying the power rule as before, we get: \(\frac{d}{dx}(200x^{-1}) = (-1)(200)x^{-1-1} = -200x^{-2}\)

Step 3: Combining the results

Now, we combine the derivatives of each term to get the derivative of the entire function: \(\frac{dA}{dx} = 0.002x - 0.4 - 200x^{-2}\) This is the derivative of the given function A(x).

Key Concepts

Power Rule

The power rule is a fundamental tool in calculus for differentiating functions that contain powers of a variable, typically denoted as \( x \). If you have a function of the form \( x^n \), the power rule states that its derivative is \( nx^{n-1} \). Essentially, you multiply the variable's exponent by the coefficient, then reduce the exponent by one. This makes calculating derivatives straightforward for terms in polynomial functions.
Applying the power rule to the terms \( 0.001x^2 \) and \( -0.4x \) involves:

• For \( 0.001x^2 \), the derivative is \( 2 \times 0.001 \times x^{2-1} = 0.002x \).
• For \( -0.4x \), the derivative is \( 1 \times -0.4 \times x^{1-1} = -0.4 \).

Notice how simple it becomes to differentiate when you apply the power rule correctly.

Polynomial Functions

Polynomial functions are expressions involving sums of powers of a variable and constant coefficients. These are typically in the form \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), where each \( a_i \) is a constant, and \( n \) is a non-negative integer. Tackling derivatives becomes manageable since each term can be handled separately using the power rule.

In the original exercise, the polynomial function terms include \( 0.001x^2 \) and \( -0.4x \). Each of these can be differentiated independently, making it easy to sum their derivatives and solve part of the problem without hassle. Polynomials often appear in calculus problems because their predictable structure facilitates straightforward manipulation and solution.

Fractional Functions

Fractional functions, or rational functions, include expressions where a variable is in the denominator. They often take the form \( \frac{c}{x^m} \), a fraction involving constants and powers of a variable. Rewriting these as negative exponents, like \( cx^{-m} \), allows for easy application of the power rule.

In the exercise, the fractional term is \( \frac{200}{x} \), which becomes \( 200x^{-1} \). To find its derivative, apply the power rule:

• The derivative is \( -1 \times 200 \times x^{-1-1} = -200x^{-2} \).

Understanding fractional functions in this rewritten form streamlines differentiation, making complex problems easier to digest and solve.

Derivative Calculation

Calculating derivatives involves determining the rate at which a function changes. It's a cornerstone of calculus that requires techniques like the power rule to simplify. Each function component's derivative is calculated separately, and the results are then summed.

For the function \( A=0.001 x^{2}-0.4 x+5+\frac{200}{x} \), the steps to calculate the derivative are:

• Identify each component: polynomial terms \( 0.001x^2 \) and \( -0.4x \), and the fractional term \( \frac{200}{x} \).
• Differentiate each part: apply the power rule to polynomial terms, and rewrite for fractional terms.
• Combine all derivatives: \( 0.002x - 0.4 - 200x^{-2} \).

Bringing these steps together delivers a whole derivative, facilitating deeper understanding and more informed application of calculus principles.