Question

find \(f^{\prime}(x)\). $$ f(x)=\frac{4 x^{3}-3 x^{2}+2 x+5}{x^{2}} $$

Short answer

The derivative of the given function is \(f'(x) = 4 - \frac{2+10x}{x^3}\).

Step-by-step solution

Rewrite the Function in a Simplified Form

Rewrite the function in a way so that each term has its own denominator. This leads to: \[f(x) = 4x^3x^{-2} -3x^2x^{-2} + 2xx^{-2} + 5x^{-2}\] which further simplifies to: \[f(x) = 4x - 3 + \frac{2}{x} + \frac{5}{x^2}\] As a results, the function now appears as a sum of simple functions.

Calculate the Derivative Using the Power Rule

Now, take the derivative of each term separately using the power rule. The power rule for derivative states that \(d/dx[x^n] = n*x^{n-1}\). Thus, for \(f'(x)\), we have: \[f'(x) = 4 - 0 - \frac{2}{x^2} - \frac{10}{x^3}\] After rearranging the terms: \[f'(x) = 4 - \frac{2}{x^2} - \frac{10}{x^3}\]

Represent the Derivative in the Standard Form

Combine the terms with same degree of \(x\) for a cleaner appearance. It yields: \[f'(x) = 4 - \frac{2+10x}{x^3}\] This is the derivative of the initial function \(f(x)\).

Key concepts

Power Rule

The power rule is a fundamental concept in calculus that allows us to easily find the derivative of functions that are powers of variables. It is a versatile and efficient rule to remember when dealing with polynomial expressions.

The power rule states that if you have a function in the form of \(x^n\), its derivative will be \(n \cdot x^{n-1}\). For example:

• If your function is \(x^3\), the derivative is \(3x^2\).
• Similarly, if it is \(x^{5}\), the derivative would be \(5x^4\).

In the original exercise, we used the power rule to find the derivative of each term in the function. This included terms like \(4x\) and \(\frac{5}{x^2}\), which needed rewriting for the rule. The function \(\frac{1}{x^n}\) can be rewritten as \(x^{-n}\), making it possible to apply the power rule directly. For example, \(\frac{5}{x^2}\) becomes \(5x^{-2}\), and its derivative is \(-10x^{-3}\).

Understanding the power rule not only streamlines the process of calculating derivatives but also builds a foundation for handling more complicated rational functions and their derivatives.

Rational Function

A rational function is a fraction of two polynomials. It is a vital concept in calculus as it composes a significant part of differentiation and integration tasks. The general form of a rational function is \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials.

In our exercise, the function \(f(x) = \frac{4x^3 - 3x^2 + 2x + 5}{x^2}\) is a rational function because it involves a polynomial numerator and a polynomial denominator. The function required rearranging so that the derivative could be handled term by term effectively.

• We transformed the function into individual terms like \(4x - 3 + \frac{2}{x} + \frac{5}{x^2}\) using polynomial long division.
• This step is crucial as it simplifies the process of applying the power rule in differentiation.

Rational functions can sometimes lead to complicated derivatives. Hence, rewriting them into simpler terms helps in taking derivatives more systematically using rules such as the power rule.

Simplification of Functions

Simplifying functions is a crucial step in solving calculus problems. It makes functions easier and more manageable to work with, especially for finding derivatives. Simplification fundamentally transforms complex-looking terms into simpler or more standard forms.

In the original problem, the function \(f(x) = \frac{4x^3 - 3x^2 + 2x + 5}{x^2}\) was rewritten to separate out terms for easier differentiation, yielding a simplified expression:

• Instead of computing the derivative of the rational expression as a whole, we broke it down to simpler terms like \(4x\) and \(\frac{2}{x}\), and more.
• This allowed the use of the power rule straightforwardly on each term.

The simplification also aids in obtaining the derivative in a standard form, making further mathematical manipulations easier. Always aim to simplify functions where possible to facilitate every step in calculus analyses, from differentiation to integration.