91Ó°ÊÓ

Logarithmic functions are incredibly useful in calculus and appear frequently in various types of problems. The logarithmic function with base \(e\) is written as \( \text{ln}(x) \). It is the inverse function of the exponential function \( e^x \). When you differentiate the natural logarithm of a function, the result is \( \frac{1}{x} \), or more generally, \( \frac{1}{u} \frac{du}{dx} \) if you're dealing with a composite function. Always remember these key properties when working with logarithms:

• \text{ln}(ab) = \text{ln}(a) + \text{ln}(b)
• \text{ln}\big( \frac{a}{b} \big) = \text{ln}(a) - \text{ln}(b)
• \text{ln}(a^b) = b \text{ln}(a)

These properties help simplify expressions before differentiating, making it easier and more manageable.

The chain rule is a powerful technique in calculus used to differentiate composite functions. If you have a function within another function, you must employ the chain rule. The generalized formula for the chain rule is \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x) \]. You differentiate the outer function while keeping the inner function untouched, then multiply by the derivative of the inner function.
In our problem, we simplified \( y = \text{ln} \big( \frac{1}{2}x \big) \) to \( y = \frac{1}{2} \text{ln} x \). To differentiate this, apply the chain rule:
\[ \frac{d}{dx}\big(\frac{1}{2} \text{ln} x\big) = \frac{1}{2} \times \frac{d}{dx}(\text{ln} x) \]
We know the derivative of \( \text{ln} x \) is \( \frac{1}{x} \). Thus, after applying the chain rule, we have:
\[ y' = \frac{1}{2} \times \frac{1}{x} = \frac{1}{2x} \] The chain rule simplifies the differentiation of nested functions immensely.

Simplification is the process of making an expression easier to work with. It is often the first step in solving calculus problems. For logarithmic functions, use properties of logarithms to break down complex expressions. In our example, we started with \( y = \text{ln} \big( \frac{1}{2} x \big) \). Simplifying it using the log property \( \text{ln}(a^b) = b \text{ln}(a) \), we converted the expression to:
\[ y = \frac{1}{2} \text{ln} x \]
This simplification allowed us to differentiate easily using the chain rule. Always look out for opportunities to simplify expressions as much as possible before applying differentiation rules. It's a vital skill that keeps the calculation manageable and reduces errors.