Problem 43 - Calculus and its applications

Question

Find \(f^{\prime}(x)\). $$ f(x)=-0.01 x^{2}+0.4 x+50 $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -0.02x + 0.4 \).

Step 1: Understand the problem

We are asked to find the derivative of the function \( f(x) = -0.01x^2 + 0.4x + 50 \). This means we need to find \( f'(x) \), which is the rate at which the function changes with respect to \( x \).

Step 2: Differentiate each term

Differentiation involves finding the derivative of each term separately. The derivative of \( ax^n \) is \( n \cdot ax^{n-1} \). Here, we apply this to each term of the function.

Step 3: Differentiate the quadratic term

Differentiate \( -0.01x^2 \). Using the power rule, the derivative is \( 2 \cdot (-0.01)x^{2-1} = -0.02x \).

Step 4: Differentiate the linear term

Differentiate \( 0.4x \). The derivative of \( x \) is 1, so \( 0.4x \) becomes \( 0.4 \).

Step 5: Differentiate the constant term

The constant term \( 50 \) has a derivative of 0 because constants do not change with \( x \).

Step 6: Combine the derivatives

Add the results of the differentiated terms together to form the derivative of the function: \( f'(x) = -0.02x + 0.4 + 0 \). Simplifying this gives \( f'(x) = -0.02x + 0.4 \).

Key Concepts

Power Rule

The power rule is one of the most basic and essential tools in calculus, used to find derivatives of polynomial functions.
It provides a straightforward method to determine how each term in a polynomial behaves as its variable changes.
To apply the power rule:

• Identify the term in the format of \( ax^n \), where \( a \) is a constant, and \( n \) is a positive integer exponent.
• Use the formula \( \frac{d}{dx}[ax^n] = n \cdot ax^{n-1} \) to differentiate.

In the exercise, we have the term \(-0.01x^2\). Applying the power rule, the derivative is calculated as follows:

• Multiplying the exponent 2 by the coefficient \(-0.01\) results in \(-0.02\).
• The power of \( x \) is reduced by one, from \( 2 \) to \( 1 \), giving us \(-0.02x\).

Thus, the derivative of \(-0.01x^2\) is \(-0.02x\). This shows how powerful the power rule is in simplifying the differentiation process.

Differentiation

Differentiation is the process of finding the derivative, which represents the rate of change of a function with respect to a variable.
It is a crucial concept in calculus because it gives us insight into how functions behave at any point.
The basic principles of differentiation include:

• Function's behavior: The derivative tells us whether a function is increasing or decreasing at a specific point.
• Instantaneous rate of change: Unlike average rate of change, which looks at two points, the derivative provides the rate at a single point.
• Slope of the tangent line: The derivative at a point is the slope of the tangent line to the curve at that point.

In the original exercise, using differentiation allows us to transform the function \( f(x) = -0.01x^2 + 0.4x + 50 \) into its derivative \( f'(x) = -0.02x + 0.4 \). By differentiating each term separately, we effectively determine how the entire function changes as \( x \) changes.
This practice of breaking down a function and finding its derivative piece by piece captures the essence of differentiation.

Calculus Concepts

Calculus is an expansive field of mathematics focused on continuous change.
Its two main branches are differential calculus, which concerns derivatives, and integral calculus, which deals with integrals.
Key concepts include:

• Limits: A fundamental idea in calculus that helps in defining derivatives and integrals, describing behavior as values approach certain points.
• Derivatives: Represent the rate of change, helping analyze functions for optimization, motion, and other dynamic systems.
• Integrals: Opposite of differentiation, used to find areas under curves and accumulate quantities.

In the context of this exercise, our focus is on derivatives. Derivatives allow us to understand the behavior of a function locally and to make predictions about its trend.
The derivative \( f'(x) = -0.02x + 0.4 \) tells us how \( f(x) \) behaves near any point \( x \):

• If \( f'(x) > 0 \), the function is increasing at that point.
• If \( f'(x) < 0 \), the function is decreasing.

Understanding the broader calculus concepts gives context to the specific calculation, demonstrating how powerful a tool calculus is in mathematics and its applications across different fields.