Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=x^{12}+5 x^{-2}-\pi x^{-10} $$

Step-by-Step Solution

Answer

The derivative \( D_{x} y = 12x^{11} - 10x^{-3} + 10\pi x^{-11} \).

Step 1: Identify Each Term of the Function

The given function is \( y = x^{12} + 5x^{-2} - \pi x^{-10} \). It consists of three terms: \( x^{12} \), \( 5x^{-2} \), and \( -\pi x^{-10} \).

Step 2: Apply the Power Rule to Derive Each Term

The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). For each term, we will apply this rule:- The derivative of \( x^{12} \) is \( 12x^{11} \).- The derivative of \( 5x^{-2} \) is \( 5 \times (-2)x^{-3} = -10x^{-3} \).- The derivative of \( -\pi x^{-10} \) is \( -\pi \times (-10)x^{-11} = 10\pi x^{-11} \).

Step 3: Combine the Derivatives to Find \( D_{x} y \)

Now, we combine the derivatives of each term to form the derivative of the whole function \( y \). Thus, \( D_{x} y = 12x^{11} - 10x^{-3} + 10\pi x^{-11} \).

Key Concepts

Power Rule in Calculus

The power rule is one of the fundamental rules of differentiation in calculus. It makes finding derivatives of polynomial terms a simple process. This rule states that if you have a term in the form of \( x^n \), where \( n \) is any real number, the derivative is found by multiplying \( n \) by the term and decreasing the exponent by one.
For example:

• For \( x^{12} \), using the power rule, the derivative is \( 12x^{11} \).
• For \( x^{-2} \), the derivative is \( -2x^{-3} \).
• For \( x^{-10} \), the derivative is \( -10x^{-11} \).

To apply this rule effectively:

• Identify the exponent of every term in the polynomial.
• Multiply the exponent by the coefficient and then subtract one from the exponent for the new term.

The rule works seamlessly across all real numbers and is a handy tool for quickly finding derivatives of polynomial expressions.

Derivative of Polynomial Functions

Polynomial functions are expressions composed of variables raised to positive or negative whole number powers and can include coefficients. Differentiating polynomial functions involves applying the power rule to each term individually. Remember that when applying differentiation to a constant coefficient, it remains unaffected by the differentiation process itself, as it is simply multiplied by the derived term.
For instance, consider a simple polynomial:

• \( y = 3x^4 - 2x^3 + x^2 - x + 6 \)

To differentiate this function, you apply the power rule to each term:

• The derivative of \( 3x^4 \) becomes \( 12x^3 \).
• \( -2x^3 \) becomes \( -6x^2 \).
• \( x^2 \) becomes \( 2x \).
• \( -x \) becomes \( -1 \).
• The constant \( 6 \) becomes 0, since the rate of change of a constant is zero.

Combining these individual derivatives results in the derivative of the entire polynomial, \( 12x^3 - 6x^2 + 2x - 1 \).

Differentiation Techniques

Differentiation is a core concept in calculus used to find the rate at which a function is changing at any given point. There are various techniques utilized in the differentiation process beyond the basic power rule, including:

• The Product Rule
• The Quotient Rule
• The Chain Rule

These rules handle more complex derivatives such as:

• Multiplication of multiple functions (Product Rule).
• Division of two functions (Quotient Rule).
• Compositions of functions (Chain Rule).

Each technique has specific rules to follow, adjusting the basics of differentiation for their particular structure:

• The Product Rule states: if \( u(x) \) and \( v(x) \) are differentiable functions, then \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).
• The Quotient Rule states: \( \frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).
• The Chain Rule is used when differentiating composite functions: \( \frac{d}{dx}[f(g(x))] = f '(g(x))g'(x) \).

Understanding when and how to use these differentiation techniques is crucial in successfully solving calculus problems involving more complex functions.