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Understanding the concept of a derivative is foundational in calculus. It essentially measures how a function changes as its input changes, or the "slope" of the function at any given point. Knowing how to calculate the derivative allows us to determine the rate of change, which is vital in fields like physics, engineering, and economics.
The derivative of a function at a specific point is defined as the limit:

• \( \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \)

This formula helps find the instantaneous rate of change, contrasting with average rate of change over intervals. With the example of inverse trigonometric functions, knowing how to differentiate them establishes understanding that's crucial for solving many types of real-world problems.

The Chain Rule is a fundamental technique for differentiating compositions of functions. When you have a function nested inside another function, the chain rule is the key to finding its derivative.
In mathematical terms, if you have two functions, say \( f(g(x)) \), the chain rule states:

• \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \)

This rule ensures that you first differentiate the outer function while keeping the inner function intact, and then multiply it by the derivative of the inner function. In our problem, applying the chain rule helps us differentiate the inverse tangent function involving the expression \( \frac{6x\sqrt{x}}{1-9x^3} \).
It's crucial for complex functions, as it breaks them down into manageable pieces.

The Quotient Rule is an essential method when differentiating a function that's the quotient of two functions - that is, one function divided by another.
The general formula for the quotient rule is:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

Where \( u \) and \( v \) are functions of \( x \). This rule comes in handy for the expression \( \frac{6x\sqrt{x}}{1-9x^3} \) in our problem, where we needed to find the derivative of a numerator and denominator separately.
This technique is a step forward in developing fluency in calculus, especially when manipulating fractions of functions.

Inverse trigonometric functions allow for the reversal of regular trigonometric functions, letting us find angles from their sine, cosine, or tangent values. These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).

• The derivative of an inverse tangent (\( \tan^{-1}(x) \)) is particularly useful and is given by \( \frac{d}{dx} [\tan^{-1}(x)] = \frac{1}{1+x^2} \).

These derivatives help solve calculus problems where you need to integrate or differentiate inverse trigonometric functions. In this exercise, the familiarity with \( \tan^{-1}(x) \)'s derivative aids in finding \( g(x) \) by applying it within the chain and quotient rules.