Question

Verify the natural \(\log\) rule \(\int \frac{1}{x} d x=\ln |x|+C\) by showing that the derivative of \(\ln |x|+C\) is \(1 / x\).

Short answer

The derivative of \(\ln |x| + C\) is indeed \(\frac{1}{x}\), which confirms the natural logarithm rule.

Step-by-step solution

Identify the function to derive

Recognize that the function to be derived is \(\ln |x| + C\). Here C is an arbitrary constant. Because the derivative of any constant is zero, we need to only take the derivative of \(\ln |x|\).

Use the derivative rule for natural logarithm

We will apply the derivative rule for the natural logarithm, which states the derivative of \(\ln u\) where \(u\) is any function of \(x\) is \(\frac{1}{u} \cdot u'\), where \(u'\) is the derivative of \(u\). In our case \(u = |x|\), and the derivative of \(|x|\) with respect to \(x\) is its sign function, which yields 1 for \(x > 0\) and -1 for \(x < 0\).

Simplify

The derivative of \(\ln |x| + C\) based on the chain rule becomes \(\frac{1}{|x|} \cdot sign(x)\). However, dividing by the absolute value instead of the original value results in the same output due to the multiplication with the sign of \(x\). Therefore, the derivative simplifies to \(\frac{1}{x}\).