Problem 21 - Calculus Early Transcendentals

Question

Find \(d y /\left.d x\right|_{x=1}\). \(y=\left(\frac{3 x+2}{x}\right)\left(x^{-5}+1\right)\)

Step-by-Step Solution

Answer

The derivative evaluated at \( x = 1 \) is \(-29\).

Step 1: Understand the Problem

We need to find the derivative of the function \( y = \left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right) \) and then evaluate it at \( x = 1 \). This involves using both the product rule and the quotient rule of differentiation.

Step 2: Identify the Product Components

The function is of the form \( f(x) \cdot g(x) \), where \( f(x) = \frac{3x + 2}{x} \) and \( g(x) = x^{-5} + 1 \). We will apply the product rule for derivatives, which is \( (f \cdot g)' = f'g + fg' \).

Step 3: Differentiate the First Component Using Quotient Rule

Apply the quotient rule to \( f(x) = \frac{3x + 2}{x} \). The rule states \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \), where \( u = 3x+2 \) and \( v = x \). Hence, \( f'(x) = \frac{x(3) - (3x+2)(1)}{x^2} = \frac{3x - 3x - 2}{x^2} = \frac{-2}{x^2} \).

Step 4: Differentiate the Second Component

The function \( g(x) = x^{-5} + 1 \) is a simple sum of two terms. Its derivative is \( g'(x) = -5x^{-6} \) because the derivative of a constant is zero and the derivative of \( x^{-5} \) is \( -5x^{-6} \).

Step 5: Apply the Product Rule

Using the product rule \( (f \cdot g)' = f' g + f g' \), substitute the derivatives found in the previous steps:\[\frac{-2}{x^2}(x^{-5} + 1) + \frac{3x+2}{x}(-5x^{-6})\]Simplify it to:\[-\frac{2x^{-5} + 2}{x^2} - \frac{5(3x+2)x^{-6}}{x}\]

Step 6: Simplify the Expression

Combine the terms into a single fraction:\[-\frac{2x^{-5} + 2}{x^2} - \frac{15x^{-6} + 10x^{-6}}{x}\]Further simplification leads to:\[-\frac{2}{x^7} - \frac{2}{x^2} - \frac{25}{x^7}\]

Step 7: Evaluate at x = 1

Substitute \( x = 1 \) into the derivative:\[-\frac{2}{1^7} - \frac{2}{1^2} - \frac{25}{1^7} = -2 - 2 - 25 = -29\]

Step 8: Conclusion

The derivative \( \left. \frac{dy}{dx} \right|_{x=1} \) evaluates to \(-29\).

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used for finding the derivative of a product of two functions. If you have a multiplication of two functions, say \( f(x) \) and \( g(x) \), the product rule states that the derivative of this product is given by \( (f \cdot g)' = f'g + fg' \).
This means you differentiate each function separately, multiply each derivative by the other function (without differentiating it), and finally sum the two resulting expressions.

• First, find \( f'(x) \).
• Then find \( g'(x) \).
• Multiply \( f'(x) \) by \( g(x) \) and \( g'(x) \) by \( f(x) \).
• Add them to get the final derivative.

This rule ensures that you account for changes in both functions, leading to an accurate derivative of the entire product.

Quotient Rule

The quotient rule is essential for differentiating functions that are in the form of a ratio, specifically when you have \( \frac{u(x)}{v(x)} \). The rule states that the derivative of this ratio is:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \]where \( u' \) and \( v' \) are the derivatives of \( u(x) \) and \( v(x) \) respectively.
To apply this rule effectively:

• Determine \( u'(x) \), the derivative of the numerator.
• Find \( v'(x) \), the derivative of the denominator.
• Plug these into the formula: \( v(x)u'(x) - u(x)v'(x) \).
• Divide the result by \( v(x)^2 \), which is the square of the denominator.

This process allows you to find the derivative of a fraction where both the numerator and the denominator are functions of \( x \). It’s crucial for accurately reflecting how changes in \( x \) impact the entire expression.

Function Simplification

Simplifying a function before or after differentiation can result in a much clearer and manageable expression. Simplification can involve combining like terms, reducing fractions, or applying algebraic identities.
When differentiating complex expressions:

• Look for opportunities to combine similar terms or factors.
• Consider multiplying through fractions or breaking up sums and differences to simplify each term.
• Simplify the algebraic expression where possible to facilitate easier evaluation.

In the context of derivatives, especially when using rules like the product or quotient rules, simplifying functions can lead to cleaner and more simplified derivative expressions. It also makes subsequent steps, like evaluation, easier and less prone to error.
A tidy expression ultimately makes it easier to substitute values and interpret the derivative's meaning, minimizing calculation mistakes and enhancing understanding.