Derivative of $\ln\left(x - 1\right)$

Find $\frac{d}{dx} \left(\ln\left(x - 1\right)\right)$.

The function $\ln\left(x - 1\right)$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = \ln\left(u\right)$ and $g{\left(x \right)} = x - 1$.

Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

The derivative of the natural logarithm is $\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$:

Return to the old variable:

The derivative of a sum/difference is the sum/difference of derivatives:

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

The derivative of a constant is $0$:

Thus, $\frac{d}{dx} \left(\ln\left(x - 1\right)\right) = \frac{1}{x - 1}$.

$\frac{d}{dx} \left(\ln\left(x - 1\right)\right) = \frac{1}{x - 1}$A