91Ó°ÊÓ

The Product Rule is a fundamental principle of differentiation. It allows us to take the derivative of a product of two functions. When you have a function that can be written as the product of two simpler functions, say \(u(x)\) and \(v(x)\), the product rule states that:

• \(D_x(u \, v) = u' \, v + u \, v'\)

This means to differentiate the product, you take the derivative of the first function times the second function and add it to the first function times the derivative of the second function.
In our exercise, the function \(f(x) \cdot \frac{1}{g(x)}\) was differentiated using this rule. Here, \(f(x)\) is \(u\) and \(\frac{1}{g(x)}\) serves as \(v\). So, we compute \(f'(x) \frac{1}{g(x)} + f(x) D_x\left(\frac{1}{g(x)}\right)\).
This step is crucial because it transforms our problem of differentiating a quotient into one that focuses on products.

The Chain Rule is another cornerstone of calculus, used primarily when dealing with composite functions. If you have a composite function \(y = f(g(x))\), the chain rule helps you find the derivative by taking:

• \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\)

In simpler terms, you first differentiate the outer function with respect to the inner function and then multiply it by the derivative of the inner function.
Within our context, the Chain Rule is applied to find the derivative of \(\frac{1}{g(x)}\). Consider \(h(x) = \left(g(x)\right)^{-1}\). Here, \(h'(x)\) involves differentiating the power of \(-1\). This gives us \(-g(x)^{-2} \cdot g'(x)\), or \(\frac{-g'(x)}{g(x)^2}\).
This application simplifies the derivative of the reciprocal, a critical step in verifying the Quotient Rule.

Differentiation is the process of finding the derivative of a function, which is central in calculus. The derivative represents the rate at which a function is changing at any point, often interpreted as the function's slope.

• It converts functions into derivatives, capturing how values change.
• Critical in solving real-world problems involving change.

For the given exercise, differentiation is employed using both the product and chain rules. We start with the function \(\frac{f(x)}{g(x)}\) and reformulate it as a product: \(f(x) \cdot \frac{1}{g(x)}\). Then, by using these rules, we arrive at:
\[D_{x}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\]This formulation reflects the classic Quotient Rule, which provides a formula for differentiating a quotient of two functions. Understanding differentiation deeply is pivotal, as it enables the correct and effective use of rules like the Product and Chain Rule in various scenarios.