Problem 10 - Brief Applied Calculus

Question

For each function, find: a. \(f^{\prime \prime}(x)\) and b. \(f^{\prime \prime}(3)\). $$ f(x)=\frac{x-2}{4 x} $$

Step-by-Step Solution

Answer

\( f''(x) = -\frac{1}{x^3} \); \( f''(3) = -\frac{1}{27} \).

Step 1: Find the First Derivative

The given function is \( f(x) = \frac{x-2}{4x} \). To find the first derivative \( f'(x) \), we'll use the quotient rule: \(\frac{u}{v} \rightarrow f'(x) = \frac{v u' - u v'}{v^2}\). Here, \( u = x - 2 \) and \( v = 4x \). So, \( u' = 1 \) and \( v' = 4 \). Applying the quotient rule:\[ f'(x) = \frac{4x(1) - (x - 2)(4)}{(4x)^2} \]\[ f'(x) = \frac{4x - 4x + 8}{16x^2} \]\[ f'(x) = \frac{8}{16x^2} = \frac{1}{2x^2} \].

Step 2: Find the Second Derivative

Now, find \( f''(x) \), the second derivative by differentiating \( f'(x) = \frac{1}{2x^2} \). This can be rewritten for easier differentiation as \( f'(x) = \frac{1}{2}x^{-2} \). Apply the power rule to differentiate:\[ f''(x) = \frac{1}{2}(-2)x^{-3} \]\[ f''(x) = -x^{-3} \]\[ f''(x) = -\frac{1}{x^3} \].

Step 3: Evaluate the Second Derivative at x=3

Substitute \( x = 3 \) into the expression for \( f''(x) = -\frac{1}{x^3}\):\[ f''(3) = -\frac{1}{3^3} \]\[ f''(3) = -\frac{1}{27} \].

Key Concepts

Understanding the Quotient Rule

When you are dealing with functions involving division, like \( f(x) = \frac{x-2}{4x} \), you will often need to use the quotient rule for differentiation. The quotient rule helps you find the derivative of a function that is expressed as the ratio of two functions. It can be depicted as follows:

• For a function \( \frac{u}{v} \), its derivative \( f'(x) \) is given by the formula: \( \frac{v \, u' - u \, v'}{v^2} \).

Here, \( u \) is the numerator function, and \( v \) is the denominator function. In our example, \( u = x - 2 \) and \( v = 4x \).
To effectively use the quotient rule, you must:

• Find \( u' \), the derivative of the numerator. In this case, \( u' = 1 \) since the derivative of \( x-2 \) is 1.
• Find \( v' \), the derivative of the denominator, so here \( v' = 4 \).

Once these derivatives are obtained, plug them into \( \frac{v \, u' - u \, v'}{v^2} \). With careful calculation, you will determine the first derivative of the function without mistakes.

Essentials of Differentiation

Differentiation is a fundamental concept in calculus that helps us determine the rate of change of a function. It is essential for understanding how a function behaves at any given point. In the context of this exercise, once we have used the quotient rule to find the first derivative, we proceed to find the second derivative.
In our specific case, the first derivative \( f'(x) \) becomes \( \frac{1}{2x^2} \), which simplifies our subsequent differentiation process.

• Rewrite the derivative in a familiar form, \( \frac{1}{2}x^{-2} \), which allows the use of standard differentiation rules.
• Apply the power rule here: Multiply the coefficient \( \frac{1}{2} \) by the exponent \(-2\), yielding \( -1 \), and subtract one from the exponent to get \( x^{-3} \).

Following these steps accurately leads us to \( f''(x) = -\frac{1}{x^3} \). Differentiating correctly is key to solving deeper calculus problems as it prepares your understanding for more complex scenarios.

Mastering Calculus Problem Solving

In calculus, solving problems efficiently requires understanding each concept deeply and knowing when and how to apply the appropriate rules. The exercise provided involves several important steps:
After obtaining the second derivative, the last step is evaluating it at a specific point. We do this to understand the behavior of the function at that exact point, which in our case is \( x = 3 \). This gives insights into the curvature and points of inflection.

• Substitute \( x = 3 \) into your second derivative equation \( -\frac{1}{x^3} \).
• Perform the calculation, i.e., \( f''(3) = -\frac{1}{3^3} = -\frac{1}{27} \).

Evaluating derivatives at specific values helps apply theoretical calculus concepts to practical applications, allowing for the interpretation of how systems change. The gradual buildup from understanding and applying the quotient rule to finding derivatives and evaluating them helps solidify a comprehensive approach to solving calculus problems.