Let \(g(x)=\log f(x)\) where \(f(x)\) is a twice differentiable positive function on \((0, \infty)\) such that \(f(x+1)=x f(x)\). Then, for \(N=1,2,3, \ldots\), \(g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=\) (A) \(-4\left\\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N-1)^{2}}\right\\}\) (B) \(4\left\\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N-1)^{2}}\right\\}\) (C) \(-4\left\\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N+1)^{2}}\right\\}\) (D) \(4\left\\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N+1)^{2}}\right\\}\)

Logarithmic differentiation is a powerful technique often employed when differentiating complex functions, particularly those involving products, quotients, or powers where traditional methods would be cumbersome. It's especially useful when differentiating a function of a function (a composite function) and the natural logarithm function comes into play.

When a function is given in the form of \(y = \text{log}(f(x))\), logarithmic differentiation begins by taking the natural logarithm of both sides of the equation, giving us \(\text{ln}(y) = \text{ln}(\text{log}(f(x)))\). Applying properties of logarithms and differentiating implicitly with respect to x leads to \(y' = \frac{f'(x)}{f(x)}\), simplifying the differentiation process.

Applying this technique is beneficial in problems like the one presented, where the function \(g(x) = \text{log}(f(x))\) could be challenging to differentiate directly. By using logarithmic differentiation, we can turn a complex differentiation problem into a more manageable one, applying the chain rule effectively and avoiding potential algebraic mess.

The second derivative of a function, denoted as \(f''(x)\) or \(\frac{d^2}{dx^2}f(x)\), provides information about the curvature of the function's graph. It's the derivative of the derivative and measures how the rate of change of a quantity is itself changing. This is crucial in understanding the concavity of the function - whether the graph of the function is concave up or concave down at a given point.

In the context of our given problem, the second derivative helps us analyze the behavior of the function \(g(x)\). By leveraging the formula \(g''(x) = \frac{f''(x)f(x) - [f'(x)]^2}{[f(x)]^2}\), we can explore how the curvature of \(g(x)\) changes, and this performance can provide insights into physical situations when the math is applied to real-world scenarios.

Furthermore, the exercise tests the student's knowledge on applying techniques to find the second derivative in a complex scenario where a functional equation is also involved. Understanding the second derivative is pivotal for higher-level mathematics, including JEE Advanced Mathematics, as well as for applications in physics and engineering.

A telescoping series is a series where each term cancels out with a subsequent or preceding term. This remarkable feature simplifies the process of finding the sum of the series because many terms eliminate each other sequentially, like the parts of an old-fashioned spyglass sliding together.

In practice, a telescoping series is typically written in the form \(b_1 - b_2 + b_2 - b_3 + b_3 - b_4 + \ldots + b_{n-1} - b_n\), which collapses into \(b_1 - b_n\) when summed. During calculations involved in different topics of mathematics, spotting a telescoping series can save a lot of time and effort, resulting in a simplified expression that allows for easy calculation of otherwise complicated sums.

In the given JEE Advanced Mathematics problem, recognizing that the difference \(g''\big(N + \frac{1}{2}\big) - g''\big(\frac{1}{2}\big)\) results in a telescoping series is key to finding the correct option. Students need to be familiar with this concept to perform the subtraction effectively and reach the required sum involving the squares of odd integers.

Functional equations are equations where the variables are functions rather than simple real numbers. These equations establish relations between values of a function or its derivatives at different points. They are often utilized to describe various mathematical and physical phenomena where the exact form of a function is unknown, and we seek to understand the properties of functions that satisfy certain constraints.

In the provided exercise, the functional equation \(f(x+1) = x f(x)\) plays a critical role. It allows us to interrelate the values of the original function and its derivatives at shifted positions on the domain. By manipulating this equation and employing differentiation, we derive relationships that aid in calculating the various derivatives of the function.

Students need to carefully analyze and skillfully derive the necessary relationships from the given functional equations. Mastery of this concept is not only essential for algebra and calculus but also in solving real-world problems where functions represent processes and changes in systems.