Problem 12 - Brief Applied Calculus

Question

For each function, find: a. \(f^{\prime \prime}(x)\) and b. \(f^{\prime \prime}(3)\). $$ f(x)=\frac{1}{12 x^{3}} $$

Step-by-Step Solution

Answer

\( f''(x) = x^{-5}, f''(3) = \frac{1}{243} \).

Step 1: Identify the Function to Differentiate

We are given the function \( f(x) = \frac{1}{12x^3} \). Observe that it is a rational function with a single term in the denominator and a constant numerator. This format hints that it may be more easily differentiated by rewriting it with negative exponents.

Step 2: Rewrite the Function Using Negative Exponents

Rewrite the initial function as \( f(x) = \frac{1}{12} x^{-3} \). This represents the same function but makes the differentiation process simpler due to the power rule.

Step 3: Differentiate to Find the First Derivative

Apply the power rule. The power rule states that \( \frac{d}{dx}[x^n] = n \cdot x^{n-1} \). Thus, the first derivative \( f'(x) \) is given by:\[ f'(x) = \frac{1}{12} \cdot (-3) \cdot x^{-4} = -\frac{1}{4} x^{-4} \].

Step 4: Differentiate Again to Find the Second Derivative

Once again, apply the power rule to the first derivative. For \( f'(x) = -\frac{1}{4} x^{-4} \), differentiate to get:\[ f''(x) = -\frac{1}{4} \cdot (-4) \cdot x^{-5} = x^{-5} \].

Step 5: Evaluate the Second Derivative at x=3

Now, evaluate \( f''(x) = x^{-5} \) at \( x = 3 \). This means calculating \( f''(3) = 3^{-5} \):\[ f''(3) = \frac{1}{3^5} = \frac{1}{243} \].

Key Concepts

Differentiation

Differentiation is a cornerstone concept in calculus. It involves finding the rate at which a function is changing at any point. This process is crucial for understanding patterns and behaviors in various real-world phenomena, from physics to economics. In mathematical terms, differentiation provides you with the derivative. To compute a derivative, you identify how a function’s output value changes as its input value changes infinitesimally.
To begin differentiating, first ensure your function is simplified, making it easier to apply rules like the power rule. Concepts such as rewriting expressions using exponents are useful here, as they allow for straightforward application of differentiation techniques.
Differentiation isn't just about finding the first derivative. It's also about understanding how these derivatives can illustrate the function's behavior, showing whether it is increasing, decreasing, or at a peak or trough.

Power rule

The power rule is a simple yet powerful tool for differentiation, especially when dealing with polynomial expressions. It states that if you have a function in the form of \( x^n \), its derivative is \( n \cdot x^{n-1} \). This rule simplifies the process of finding derivatives significantly and is essential for handling functions with powers.
When you apply the power rule, start by identifying the power of \( x \) in your expression. Multiply the entire term by that power, then reduce the power by one. This adjustment reflects how the function changes.
For instance, if you have \( f(x) = \frac{1}{12} x^{-3} \), applying the power rule gives \( f'(x) = -\frac{1}{4} x^{-4} \). Notice that by bringing down the exponent as a coefficient and reducing the exponent by one, you align with the power rule's mechanism effectively, simplifying your work flow in calculus.

Second derivative

While the first derivative tells you about the slope of the function at any point, meaning whether the function is increasing or decreasing, the second derivative gives deeper insight. It reveals the function's concavity or the rate at which the slope is changing. Essentially, it's a derivative of a derivative, and it is critical for analyzing the shape and nature of the graph.
A positive second derivative indicates that the function is concave up, resembling a U-shape, where the rate of change is increasing. Conversely, a negative second derivative implies the function is concave down, like an upside-down U, showing a decreasing change rate.
Calculating the second derivative follows the same procedure as the first: differentiate the first derivative. Using our example, after differentiating \( f'(x) = -\frac{1}{4} x^{-4} \), we find the second derivative to be \( f''(x) = x^{-5} \). Evaluating at \( x=3 \), this becomes \( f''(3)=\frac{1}{243} \), providing precise measurement of concavity at that point.