Problem 18 - Calculus Early Transcendentals

Question

Find \(y^{\prime}(1)\) $$ y=\frac{x^{3 / 2}+2}{x} $$

Step-by-Step Solution

Answer

The value of \( y'(1) \) is \(-\frac{3}{2}\).

Step 1: Understand the given function

The function given is \( y = \frac{x^{3/2} + 2}{x} \). It is a rational function and can be simplified for differentiation.

Step 2: Rewrite the function

Rewrite the function \( y = \frac{x^{3/2} + 2}{x} \) as \( y = x^{3/2 - 1} + \frac{2}{x} \). This becomes \( y = x^{1/2} + 2x^{-1} \).

Step 3: Differentiate the function

Differentiate each term separately. For \( y = x^{1/2} + 2x^{-1} \), use the power rule. The derivative of \( x^{1/2} \) is \( \frac{1}{2}x^{-1/2} \) and the derivative of \( 2x^{-1} \) is \( -2x^{-2} \). So, \( y' = \frac{1}{2}x^{-1/2} - 2x^{-2} \).

Step 4: Evaluate the derivative at x = 1

Substitute \( x = 1 \) into the derivative to find \( y'(1) \). We have \( y'(1) = \frac{1}{2}(1)^{-1/2} - 2(1)^{-2} \). Simplify this to \( y'(1) = \frac{1}{2} - 2 = -\frac{3}{2} \).

Key Concepts

Rational Function

A rational function is a function that can be expressed as the ratio of two polynomials. In simpler terms, it's a fraction where both the numerator and the denominator are polynomials. Rational functions can sometimes involve variables in their exponents or other complex forms.
For example, in the original function \( y = \frac{x^{3/2} + 2}{x} \):

• "\( x^{3/2} + 2 \)" is the numerator.
• "\( x \)" is the denominator.

Dealing with rational functions often involves simplifying them to make differentiation easier. You can rewrite the original function in a more straightforward form by separating the terms, which transforms it into simpler expressions that are easier to differentiate.

Power Rule

The power rule is a useful tool in calculus for differentiating functions where the variable is raised to a power. It's an essential rule to grasp as it allows you to find the derivative of power functions quickly. The power rule states:\[ \frac{d}{dx} \left( x^n \right) = nx^{n-1}\]Here's how it works:

• Take the exponent \( n \) and multiply it by the coefficient.
• Then decrease the exponent by one to get the new power.

For instance, when differentiating \( x^{1/2} \), apply the power rule:

• The exponent here is \( \frac{1}{2} \).
• Applying the rule gives: \( \frac{1}{2}x^{-1/2} \).

It's straightforward once you get the hang of it and it makes taking derivatives of such functions much more approachable.

Derivative Evaluation

Evaluating the derivative at a specific point involves substituting a given value into the derivative function. It's a practical way of finding the rate of change of a function at a specific point. Here, after differentiating the expression, you substitute \( x = 1 \) into the derivative.Consider the function given: the derived function is \( y' = \frac{1}{2}x^{-1/2} - 2x^{-2} \).

• Substitute \( x = 1 \): \( y'(1) = \frac{1}{2}(1)^{-1/2} - 2(1)^{-2} \).
• This simplifies to \( y'(1) = \frac{1}{2} - 2 \).
• Finally, calculate the result to find \( y'(1) = -\frac{3}{2} \).

This value represents the derivative, and it shows the rate at which the original function changes at \( x = 1 \). By evaluating it at specific values, you make the derivative applicable to real problems and scenarios.

Simplifying Expressions

Simplifying expressions is a fundamental skill necessary for working with complex functions. Simplification aids in making the differentiation process more manageable. In the given exercise, the transformation from a hard-to-differentiate form to an easier one is crucial.Taking the original function \( y = \frac{x^{3/2} + 2}{x} \), you begin by rewriting it:

• The term \( x^{3/2} \) over \( x \) simplifies to \( x^{3/2 - 1} \) which equals \( x^{1/2} \).
• Similarly, \( \frac{2}{x} \) becomes \( 2x^{-1} \).

This simplifies the function to \( y = x^{1/2} + 2x^{-1} \), making it ready for differentiation. Simplification often precedes calculus operations and can also serve in solving algebraic issues, reducing the number of steps in calculations, and avoiding potential mistakes in complex expressions.