Problem 65 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. $$ f(x)=2 x^{2}-\frac{3 x-1}{x^{3}} $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = 4x - \frac{6}{x^3} + \frac{3}{x^4}\).

Step 1: Differentiate the First Term

The first term in the function is \(2x^2\). Use the power rule to differentiate this term. The power rule states that if \(f(x) = ax^n\), then \(f'(x) = anx^{n-1}\).Therefore, the derivative of \(2x^2\) is \(2 \cdot 2x^{2-1} = 4x\).

Step 2: Simplify the Second Term

Rewrite the second term \(\frac{3x-1}{x^3}\) as \(\frac{3x}{x^3} - \frac{1}{x^3}\). This can be simplified to \(3x^{-2} - x^{-3}\) using the property of exponents \( \frac{a}{b} = ab^{-1} \).

Step 3: Differentiate Each Part of the Second Term

Differentiate \(3x^{-2}\) using the power rule. This becomes \(3 \cdot (-2)x^{-2-1} = -6x^{-3}\).Now, differentiate \(-x^{-3}\). Using the power rule, this becomes \(-3x^{-3-1} = -3(-1)x^{-4} = 3x^{-4}\).

Step 4: Combine the Derivatives

Add the derivatives obtained from each of the steps: the derivative of the first term \(4x\), and the derivatives of parts of the second term \(-6x^{-3}\) and \(+3x^{-4}\).Thus, the derivative is:\(f'(x) = 4x - 6x^{-3} + 3x^{-4}\).

Step 5: Finalize

Express the final answer in a more readable format if desired. This gives:\[f'(x) = 4x - \frac{6}{x^3} + \frac{3}{x^4}\]

Key Concepts

Power Rule

In calculus, the power rule is one of the most fundamental tools used to find derivatives. It's a simple, yet powerful tool that can make differentiating polynomials much easier. The power rule states that if you have a function of the form \( f(x) = ax^n \), its derivative is \( f'(x) = anx^{n-1} \). This means you take the exponent "\(n\)", multiply it by the coefficient "\(a\)", and subtract 1 from the exponent.

• For \( f(x) = 2x^2 \), applying the power rule gives us: \( f'(x) = 2 \cdot 2x^{2-1} = 4x \).
• This rule is especially useful because it applies in a straightforward way to any polynomial term.

Using the power rule, differentiating terms becomes a matter of simple multiplication and subtraction. It's essential for dealing with more complex functions as well.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. In simpler terms, it's the slope of the function at any given point. Calculating derivatives is a central concept in calculus and is widely used in various fields, including physics, engineering, and economics to model and predict changes.When finding derivatives:

• The function can be broken down into individual terms, each to be differentiated separately.
• Like slicing a loaf of bread, each slice can be tackled on its own, then combined for the full loaf.

For the function \( f(x)=2x^2-\frac{3x-1}{x^3} \), derivatives for components were calculated separately and then compiled into a single expression. This modular approach helps in managing complexity and ensures accuracy in differentiation.

Calculus Steps

Finding the derivative of a function involves specific steps and rules from calculus. These steps break down complex problems into manageable parts:

• Simplify: First, rewrite complex terms to make them easier to differentiate, like transforming \( \frac{3x-1}{x^3} \) into \( 3x^{-2} - x^{-3} \).
• Apply Rules: Use rules like the power rule for each term separately. This is done for each term such as \( 3x^{-2} \) and \( -x^{-3} \).
• Combine: After finding the derivative of each term, combine them to find the overall derivative. Here, we combine to get \( f'(x) = 4x - 6x^{-3} + 3x^{-4} \).

These steps allow you to methodically process each part of a function, ensuring that nothing is overlooked. Taking it step by step ensures clarity and correctness when working through calculus problems.