Problem 106 - Calculus and its applications

Question

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\frac{x^{5}+x}{x^{2}} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) is \( 3x^2 - \frac{1}{x^2} \).

Step 1: Simplify the Expression

First, simplify the given function by dividing each term in the numerator by the denominator. The function is \( y = \frac{x^5 + x}{x^2} \). Divide each term separately: \( y = \frac{x^5}{x^2} + \frac{x}{x^2} = x^{5-2} + x^{1-2} = x^3 + x^{-1} \).

Step 2: Differentiate each term separately

Differentiate each term using the power rule. For the term \( x^3 \), the derivative is \( 3x^{3-1} = 3x^2 \). For the term \( x^{-1} \), the derivative is \( -1x^{-1-1} = -x^{-2} \).

Step 3: Combine the derivatives

Combine the derivatives found in Step 2 to find \( \frac{dy}{dx} \). Combining these results gives \( \frac{dy}{dx} = 3x^2 - x^{-2} \).

Step 4: Simplify the derivative if necessary

Check if any further simplification of the expression is needed. In this case, the derivative \( 3x^2 - \frac{1}{x^2} \) is already in its simplest form.

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus used for finding the derivative of polynomial functions. It is a straightforward technique that involves decreasing the exponent of a term by one and multiplying the coefficient by the original exponent. This rule is expressed in the simple form: if you have a function of the form \( f(x) = x^n \), its derivative \( f'(x) \) is \( nx^{n-1} \).

In the context of the given problem, once we have simplified the function to \( y = x^3 + x^{-1} \), the power rule becomes very handy. It allows us to differentiate each term separately.

• For \( x^3 \), applying the power rule results in the derivative \( 3x^{2} \).
• For \( x^{-1} \), the power rule gives us \( -1x^{-2} \).

This step-by-step application of the power rule makes it easy to handle more complex polynomial terms in calculus problems.

Algebraic Simplification

Before applying calculus techniques such as differentiation, it's often necessary to simplify algebraic expressions. Simplification can make differentiation much more manageable. In the original exercise, algebraic simplification involves dividing the terms in the numerator by the denominator to express the function in a simpler form.

Here's how it was done: Starting with \( y = \frac{x^5 + x}{x^2} \), you simplify by dividing each term:

• The term \( x^5 \) divided by \( x^2 \) simplifies to \( x^{5-2} = x^3 \).
• The term \( x \) divided by \( x^2 \) becomes \( x^{1-2} = x^{-1} \).

This results in the simpler function \( y = x^3 + x^{-1} \), which is much easier to differentiate. Efficient algebraic simplification is a crucial skill, as it lays the groundwork for correct and simple differentiation.

Calculus Problem Solving

Solving calculus problems often involves a blend of algebraic manipulation and the application of differentiation techniques. The steps you take can significantly affect the ease of finding a solution. In this problem, identifying the need for algebraic simplification upfront was critical for successful problem solving.

The process begins by simplifying a complex expression, making it suitable for rule-based differentiation. Afterward, using the power rule helps to find derivatives in a methodical way.

• First, simplify the expression through basic algebra.
• Second, apply the power rule to each separate term.
• Finally, combine the derived terms to get \( \frac{dy}{dx} \).

It's also important to assess whether the derived expression can be simplified further. In this exercise, the end result was already simplified, demonstrating the effectiveness of our approach. Combining both algebraic skills and a solid grasp of calculus rules allows one to tackle a wide range of differentiation problems with confidence.