Using first principle, find the derivative of loge x where x ∈ (0, ∞).

To find the derivative of \( \log_e x \) (also known as the natural logarithm, denoted as \( \ln x \)) using the first principle, we start with the definition of the derivative.

Definition of the Derivative

The derivative of a function \( f(x) \) at a point \( x \) is defined as:

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

Applying the Definition

For \( f(x) = \ln x \), we can substitute into the definition:

\( f'(x) = \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h} \)

Using Logarithmic Properties

We can simplify the expression using the property of logarithms:

• \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \)

This gives us:

\( f'(x) = \lim_{h \to 0} \frac{\ln\left(\frac{x+h}{x}\right)}{h} \)

which simplifies to:

\( f'(x) = \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x}\right)}{h} \)

Changing Variables

Let \( u = \frac{h}{x} \), then as \( h \to 0 \), \( u \to 0 \) as well. Thus, \( h = ux \) and we can rewrite the limit:

\( f'(x) = \lim_{u \to 0} \frac{\ln(1 + u)}{ux} \)

This can be further simplified to:

\( f'(x) = \frac{1}{x} \lim_{u \to 0} \frac{\ln(1 + u)}{u} \)

Evaluating the Limit

It is known that:

\( \lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1 \)

Thus, we have:

\( f'(x) = \frac{1}{x} \cdot 1 = \frac{1}{x} \)

Final Result

The derivative of \( \log_e x \) is:

\( \frac{d}{dx} \ln x = \frac{1}{x} \)

This result holds for \( x \) in the interval \( (0, \infty) \).