Problem 19 - Single Variable Calculus

Question

Find the instantaneous rate of change of the given function when \(x=a .\) \(f(x)=\frac{2}{x}+x ; \quad a=1\)

Step-by-Step Solution

Answer

The instantaneous rate of change of the function \(f(x) = \frac{2}{x} + x\) when \(x = a = 1\) is -1.

Step 1: Differentiate the function

Recall that the derivative of a function gives us its rate of change at any point. So, we need to find the derivative of \(f(x) = \frac{2}{x} + x\). To differentiate, we'll use the power rule: \(\frac{d}{dx}[x^n] = nx^{n-1}\) and the chain rule. First, we rewrite the function as \(f(x) = 2x^{-1} + x\) Now, we differentiate each term with respect to x. \(f'(x) = \frac{d}{dx}[2x^{-1}] + \frac{d}{dx}[x]\)

Step 2: Use power rule and derivative rules

Applying the power rule for each term and simplifying the results: \(f'(x) = -2x^{-2} + 1\) Now, rewrite the derivative in a more familiar form: \(f'(x) = -\frac{2}{x^2} + 1\)

Step 3: Evaluate derivative at x = a

Now that we have our derivative, we can find the instantaneous rate of change of the function when \(x = a\), which is \(a = 1\). Substitute \(a = 1\) into the derivative function: \(f'(a) = -\frac{2}{1^2} + 1 = -2 + 1 = -1\) So, the instantaneous rate of change when \(x = a = 1\) is -1.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. Consider a function, such as the one in the exercise: \( f(x) = \frac{2}{x} + x \). To determine how this function changes as \( x \) changes, we need to find its derivative, denoted by \( f'(x) \).

The derivative functions as a window into the behavior of the original function, showing how steeply it rises or falls at any given point. This is especially valuable when we want to find specific characteristics of the function, such as local minima and maxima, and where the function is increasing or decreasing.

When you perform differentiation, you essentially apply a set of rules to systematically discover the derivative. Let's explore some crucial differentiation rules, using the exercise as a guide.

Power Rule

The power rule is a straightforward and commonly used method for differentiating functions. It simplifies the process of finding derivatives when dealing with power terms like \( x^n \). The power rule states that if you have a function in the form of \( x^n \), its derivative is \( nx^{n-1} \).

For instance, in the function \( f(x) = \frac{2}{x} + x \), we rewrite \( \frac{2}{x} \) as \( 2x^{-1} \) to make it easier to apply the power rule. Then, the differentiation process uses the power rule separately on each part:

• For \( 2x^{-1} \), take the derivative: \( \frac{d}{dx}[2x^{-1}] = -2x^{-2} \).
• For \( x \), since it's \( x^1 \), the derivative is simply \( 1\).

The power rule offers a fast track to finding derivatives without complication. It's especially handy when you have polynomial or simple power functions.

Chain Rule

The chain rule is another crucial differentiation tool, especially useful when differentiating composite functions. A composite function appears when one function is nested within another, and handling its derivative requires special attention to each layer.

In our example, the chain rule isn't employed directly, but it's essential to understand for more complex cases than \( f(x) = \frac{2}{x} + x \). When applying the chain rule, you take the derivative of the outer function and multiply it by the derivative of the inner function.

For practical applications:

• If you have a function \( g(h(x)) \), find \( g'(h(x)) \ldots\) then multiply by \( h'(x) \).
• Helps in differentiating expressions like \( (3x + 2)^5 \), where you first deal with the outer exponent and then with the internal linear polynomials.

Understanding the chain rule broadens your differentiation capabilities, making complex derivatives manageable.

Derivative Evaluation

Once you have found the derivative of a function, evaluating it at a specific point gives you the instantaneous rate of change at that point. In our exercise, the function's derivative \( f'(x) = -\frac{2}{x^2} + 1 \) is evaluated at \( x = a = 1 \).

Plugging \( a = 1 \) into the derivative gives \( f'(1) = -\frac{2}{1^2} + 1 = -2 + 1 = -1 \). Hence, the instantaneous rate of change when \( x = 1 \) is \(-1\).

This result shows how quickly the function value changes around \( x = 1 \). Derivative evaluation is a powerful method to analyze behaviors at specific intervals. Through evaluating derivatives at different points, one can gain insights into trends, velocities, or the gradient of curves across varied contexts.