So I'm trying to follow the derivation in a book on Analytic combinatorics. We have the function $$P(z,u)=(1-z)^{-u}$$ and I need to take the partial derivative with respect to $u$, then evaluate at $u=1$. So, my attempt: $$\ln (P(z,u))=-u\ln(1-z)$$ $$\frac {P_u(z,u)}{P(z,u)}=-\ln (1-z)$$ $$P_u(z,u)=-(1-z)^{-u}\ln (1-z)$$ then evaluating at $u=1$, ending up with $$-(1-z)^{-1}\ln (1-z)$$
The only problem is, the author has no negative sign in his expression....did I make a mistake in the chain rule or something?
No, you didn't make a mistake. Your answer is correct. That's clearly a typo in the book.