What is the derivative of ln(x) with respect to x?

What is the derivative of ln(x) with respect to x?

• A. 1/x
• B. x
• C. ln(x)/x
• D. x^2

Answer: (A)

The derivative of the natural logarithm function, ln(x), with respect to x is 1/x. This result is derived from the rules of calculus, specifically utilizing the concept of limits and the definition of a derivative. To understand why this is the case, we can start from the definition of the derivative, which is the limit of the difference quotient: \[ \frac{d}{dx} \ln(x) = \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h} \] Using the properties of logarithms, we can rewrite the difference of logarithms: \[ \ln(x + h) - \ln(x) = \ln\left(\frac{x + h}{x}\right) = \ln\left(1 + \frac{h}{x}\right) \] As \( h \) approaches 0, we can apply the Taylor series expansion or the property that \( \ln(1 + u) \approx u \) when \( u \) is small. Therefore, we can simplify it further: \[ \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x

The derivative of the natural logarithm function, ln(x), with respect to x is 1/x. This result is derived from the rules of calculus, specifically utilizing the concept of limits and the definition of a derivative.

To understand why this is the case, we can start from the definition of the derivative, which is the limit of the difference quotient:

[

\frac{d}{dx} \ln(x) = \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h}

]

Using the properties of logarithms, we can rewrite the difference of logarithms:

[

\ln(x + h) - \ln(x) = \ln\left(\frac{x + h}{x}\right) = \ln\left(1 + \frac{h}{x}\right)

]

As ( h ) approaches 0, we can apply the Taylor series expansion or the property that ( \ln(1 + u) \approx u ) when ( u ) is small. Therefore, we can simplify it further:

[

\lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x