91Ó°ÊÓ

Derivatives are a fundamental concept in calculus, representing how a function changes as its input changes. You can think of them as the "rate of change" or the "slope" at a particular point on the graph of a function. When you differentiate a function, you are finding its derivative. In mathematics, differentiating is like peeling back the layers of what the function does.
In this exercise, we start with the function \( f(x) = (\sqrt{x} + 2)(\sqrt{x} - 2) \) and are asked to find its derivative using the Product Rule. This requires understanding not only what derivatives are but also how to compute them when functions are multiplied together, which is where the Product Rule comes in.
To find a derivative effectively:

• Identify the functions involved, as we did with \( u(x) = \sqrt{x} + 2 \) and \( v(x) = \sqrt{x} - 2 \).
• Find their respective derivatives.
• Apply the Product Rule formula to find the derivative of the product.

Function differentiation involves finding the derivative of expressions, whether they are single functions or combinations like products or quotients. Differentiation can be direct for simple functions or more complex when functions are combined, like in the product or chain rule.
The exercise focuses on differentiating products of functions, which is where the Product Rule is indispensable. This rule says if you have two functions, \( u(x) \) and \( v(x) \), their derivative when multiplied together, \( f(x) = u(x) \cdot v(x) \), is computed as:
\[ f'(x) = u'(x) v(x) + u(x) v'(x) \]By applying this rule, we methodically increase our toolkit for dealing with more complex expressions. Each piece of the function, \( \sqrt{x} + 2 \) and \( \sqrt{x} - 2 \), is differentiated separately, paving the way for their product's differentiation.

Simplification in calculus involves condensing the expression you get after differentiating a function. This creates a more manageable form of the derivative, often simplifying the task of recognizing how the function behaves.
In our exercise, after initially finding the derivative using the Product Rule, the result was:
\[ f'(x) = \left( \frac{1}{2} - \frac{1}{\sqrt{x}} \right) + \left( \frac{1}{2} + \frac{1}{\sqrt{x}} \right) \]The simplification process involves:

• Combining like terms: In this case, \( \frac{1}{2} + \frac{1}{2} \) and recognizing that negative and positive \( \frac{1}{\sqrt{x}} \) terms cancel each other.
• A simplified expression, \( f'(x) = 1 \), emerges from organizing these components, showing how simplification can clarify the derivative's form.

Simplification not only makes derivatives easier to interpret but also highlights the deeper structure of the function's behavior.