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The chain rule is a powerful differentiation technique used when working with composite functions. In simple terms, it helps you find the derivative of a composite function by differentiating the outer function and then the inner function separately.
The chain rule formula can be expressed as follows if you have a composite function: \textbf{Example:}If you have a composite function \textcolor{blu}{(f ∘ g)(x) = f(g(x))}.
The chain rule states: \textbf{Example:}\[ \frac{d}{dx} [f(g(x))] = f'(g(x)) \times g'(x) \]
This rule simplifies finding the derivative by breaking it down into smaller, more manageable steps. For instance, in our problem: \( y = \frac{1}{2 + 3 \ln x} \), we first let \( u = 2 + 3 \ln x \), then differentiate \( \frac{1}{u} \) and multiply by the derivative of \( u \).
The chain rule allows us to compute such derivatives efficiently.

Rational functions are functions represented as the ratio of two polynomials. In our example, \( y = \frac{1}{2 + 3 \ln x} \) is a rational function because the numerator is simply a constant 1, and the denominator is a polynomial involving the natural logarithm of \( x \). When differentiating rational functions, especially those involving composite elements, we can use a combination of basic differentiation rules, the chain rule, and sometimes the quotient rule. In this particular problem, the numerator is a constant, simplifying our use of the chain rule.
Key points to remember about rational functions are:

• The numerator and denominator can be any polynomial. etc.
• They can often be simplified using algebraic manipulation.

In our case, simplifying the denominator using the substitution \( u = 2 + 3 \ln x \) allowed us to differentiate more easily.

A composite function is formed when one function is applied to the result of another function. It is often written as \( (f \, ∘ \, g)(x) = f(g(x)) \). To differentiate composite functions, the chain rule is typically used.
For our given problem: \( y = \frac{1}{2+3 \ln x} \), we identify that the inner function is \( u = 2 + 3 \ln x \) and the outer function is \( y = \frac{1}{u} \).
By differentiating the outer function \( y = \frac{1}{u} \) with respect to \( u \) first, and then the inner function \( u = 2 + 3 \ln x \) with respect to \( x \), we apply the chain rule:

• Differentiate the outer function: \( \frac{d}{du} \left( \frac{1}{u} \right) = -\frac{1}{u^2}\)
• Differentiate the inner function: \( \frac{du}{dx} = \frac{3}{x} \)
• Combine using the chain rule: \( \frac{dy}{dx} = -\frac{1}{u^2} \times \frac{3}{x} \)
• Substitute back \( u = 2 + 3 \ln x \) into the final expression: \( \frac{dy}{dx} = -\frac{3}{x(2 + 3 \ln x)^2}\)

Working with composite functions requires a careful process, but using the chain rule makes it smooth and manageable.