See THIS ANSWER for a detailed development of the evaluation of $\int_\gamma \frac1z'\,dz'$, where $\gamma$ is any rectifiable curve in $\mathbb{C} \setminus\{0\}$ from $1$ to $z$.

If the logarithm function is defined on the sheet for which $-\pi+2k\pi <\arg(z)\le \pi+2k\pi$, then

$$\log(z)=\log(|z|)+i(\arg(z))$$

and

$$\frac{d \log(z)}{dz}=\frac1z$$

for $-\pi+2k\pi <\arg(z)\le \pi+2k\pi$.

Inasmuch as $\log(z)$ is defined, but discontinuous on the branch cut, it is clearly not differentiable there.

No, it means that $1/z$ isn't the derivative of $\log z$ on the nonpositive axis. The derivative of $\log z$ isn't defined when $\log z$ isn't differentiable.

This is usually why we have

$$\int\frac1xdx=\ln|x|+c$$

And by practical means, since it is impossible to integrate over $x=0$, this definition works fine.