Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{3}{x^{3}}+x^{-4} $$

Step-by-Step Solution

Answer

The derivative of \( y \) is \( D_x y = -9x^{-4} - 4x^{-5} \).

Step 1: Simplify the expression using exponent rules

First, rewrite the expression \( y = \frac{3}{x^3} + x^{-4} \) in terms of powers of \( x \). The expression can be rewritten as \( y = 3x^{-3} + x^{-4} \).

Step 2: Differentiate each term separately

Apply the power rule for differentiation to each term. For a term of the form \( ax^n \), the derivative is \( ax^{n-1} \).

Step 3: Differentiate the first term

For the term \( 3x^{-3} \), apply the power rule: \( D_x (3x^{-3}) = 3(-3)x^{-4} = -9x^{-4} \).

Step 4: Differentiate the second term

Next, differentiate the term \( x^{-4} \). Applying the power rule gives: \( D_x (x^{-4}) = (-4)x^{-5} = -4x^{-5} \).

Step 5: Combine the derivatives

Combine the derivatives of both terms: \( D_x y = -9x^{-4} - 4x^{-5} \).

Key Concepts

Power Rule for Differentiation

Differentiation is a crucial concept in calculus used to calculate the rate at which a function is changing. The power rule is one of the most straightforward and widely used rules for finding derivatives, especially with polynomials. It states that for any function in the form of \( ax^n \), where \( a \) is a constant and \( n \) is any real number, the derivative is \( anx^{n-1} \).

Let's break this down further:

• If you have \( x^3 \), applying the power rule would give you \( 3x^{3-1} = 3x^2 \).
• For \( 5x^{-2} \), the rule transforms it into \( 5(-2)x^{-3} = -10x^{-3} \).

The power rule simplifies the process of differentiation, especially when dealing with polynomials and terms with negative or fractional exponents.

When you use the power rule, it's important to carefully manage the coefficients and exponents. If a term has a coefficient, such as \( 3 \) in \( 3x^{-3} \), multiply the coefficient by the new exponent after decreasing it by one.

Exponents and Powers

Understanding exponents and powers is fundamental to algebra, and they frequently appear in calculus problems too. When we talk about powers or exponents, we refer to expressions like \( x^n \), where \( x \) is the base and \( n \) is the exponent.

Some key rules about exponents include:

• \( x^a \times x^b = x^{a+b} \): Multiplying like bases adds the exponents.
• \( (x^a)^b = x^{a\cdot b} \): An exponent raised to another exponent multiplies the exponents.
• \( x^{-a} = \frac{1}{x^a} \): A negative exponent indicates a reciprocal.

In calculus, managing these types of terms is crucial; especially when simplifying expressions. Rewriting fractions into a form using negative exponents, like transforming \( \frac{3}{x^3} \) into \( 3x^{-3} \), is beneficial for subsequently applying differentiation rules.

Simplification of Expressions

Simplification is the process of transforming mathematical expressions into a more concise or recognizable form. Before differentiation, simplifying the expression can make the process more straightforward and reduce the likelihood of errors.

In our example, simplifying involved rewriting \( \frac{3}{x^3} + x^{-4} \) as \( 3x^{-3} + x^{-4} \). This step is essential because:

• It aligns all terms to be in a consistent format, like expressing them in terms of powers of \( x \).
• This consistency allows the power rule to be applied seamlessly.
• It reduces complex fractions or product expressions into simpler terms.

Effective simplification ensures better clarity and results in easier application of calculus rules. By expressing functions clearly, you create less room for mistakes during differentiation and further calculus operations.