Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(f^{\prime}(x) .\) Strategize to minimize your work. For example, \(\frac{x^{2}+3}{3 x}\) does not require the Quotient Rule. \(\frac{x^{2}+3}{3 x}=\frac{x}{3}+\frac{1}{x}=\frac{1}{3} x+x^{-1} .\) This is simpler to differentiate. $$ f(x)=\frac{x-2 x^{2}}{5} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = 0.2x - 0.4x^2\) is \(f'(x) = 0.2 - 0.8x\).

Step 1: Simplify the function

The function \(f(x)\) can be expressed as a sum of two fractions, i.e. \(f(x) = \frac{x}{5} - \frac{2x^2}{5}\). Alternatively, we can write this as \(f(x) = 0.2x - 0.4x^2\). This form is easier to differentiate because we no longer have a quotient.

Step 2: Differentiation

Differentiate the function \(f(x)\) with respect to \(x\). For the first part, \(0.2x\), the derivative is just \(0.2\). For the second part, \(-0.4x^2\), the derivative according to the power rule (which states that the derivative of \(x^n\) is \(nx^{n-1}\)) is \(-0.8x\). Therefore, the derivative of the entire function \(f(x)\) is \(f'(x) = 0.2 - 0.8x\)

Step 3: Final Answer

The derivative of the given function is \(f'(x) = 0.2 - 0.8x\).

Key Concepts

Power Rule

When working with differentiation, one of the most basic and useful rules to become familiar with is the power rule. The power rule is a straightforward technique in calculus used for finding the derivative of functions of the form \(x^n\). The derivative using this rule is \(nx^{n-1}\).

Let's see how this works with an example given in the exercise: For the term \(-0.4x^2\), applying the power rule involves multiplying the exponent (2) by the coefficient (-0.4), resulting in \(-0.8\). This becomes the new coefficient, and you subtract 1 from the original exponent, reducing \(x^2\) to \(x^{1}\), thus giving a simplified derivative of \(-0.8x\).

• For expressions like \(x^n\), the power rule is handy and quick.
• The rule simplifies the process by removing any need for more complex approaches.

Remember, the power rule is only applicable to terms where the base is \(x\) and should not be used for more complex functions requiring other rules like the product or chain rule.

Simplifying Expressions

Simplifying expressions is a vital step before differentiation, as it can transform a complex equation into a simpler form, making differentiation much more manageable. In calculus problems, expressions often appear complicated, but with some algebraic manipulation, you can streamline these expressions.

Consider the original function \(f(x) = \frac{x - 2x^2}{5}\). By splitting this into two separate fractions, we rewrite it as \(f(x) = \frac{x}{5} - \frac{2x^2}{5}\). By further simplifying, this converts to \(f(x) = 0.2x - 0.4x^2\). This simpler version dissolves the quotient, which otherwise might prompt using the Quotient Rule unnecessarily.

• Always aim to express components in the simplest terms before moving on to differentiation.
• Reducing the complexity upfront can effectively minimize computational work.

By simplifying, you ease the workload in differentiation, ensuring precision and efficiency in executing the derivative operation.

Calculus Problem Solving

Approaching calculus problems requires strategic thinking, especially when finding derivatives. Identifying the simplest path to the solution can often mean transforming or simplifying the given function before applying any calculus techniques.

The example provided illustrates minimizing effort by eliminating any unnecessary complex rules like the Quotient Rule when simpler options unite. Calculus problem-solving is not just about doing calculations; it's about making your path as direct and clear as possible. Hence, prepping with simplification and knowing when to employ the right rule, like the power rule, is key.

• Assess whether complex rules can be avoided by simplifying expressions beforehand.
• Understanding the problem fully before diving into math is crucial for effective solutions.

By fostering these approaches, you develop a refined eye for clear paths to solutions, ensuring you not only perform calculations accurately but also efficiently.