Let $F : T \mapsto S e^{\int_0^T{\mu \left(t\right)\mathrm{d}t}}$, $S \in \mathbb{R}_+^*$, and define $y \left(T\right) = \ln \left(\frac{K}{F_T}\right)$, where again $K \in \mathbb{R}_+^*$. Note that this implies $K = F_T e^y = S e^{\int_0^T{\mu \left(t\right)\mathrm{d}t}+y}$.
I want to differentiate $y$ with respect to $T$. Doing so directly yields $$ \partial_T y = \partial_T \left[ \ln \left(K\right) - \ln \left(S\right) - \int_0^T{\mu \left(t\right)\mathrm{d}t} \right] = - \mu \left(T\right) $$
However, in the paper I am reading, the author uses the chain rule and gets $$ \partial_T y = \partial_K y \times \partial_T K = \frac{1}{K} \times \mu \left(T\right) S e^{\int_0^T{\mu \left(t\right)\mathrm{d}t}+y} = \frac{1}{K} \times \mu \left(T\right) \times K = \mu \left(T\right) $$
I have to be missing something since I do not see why I would obtain two different results for the derivative. Any idea?
You indicate that $S$ and $K$ are positive real constants. In this case, it makes no sense to differentiate with respect to $K$. So either you are mistaken about that, or the author is just wrong.
Assuming these are indeed constants, you would have
If $y(x) = \ln(K/F(x)) = \ln K - \ln F(x)$, then $$\frac{dy}{dx} = -\frac{F'(x)}{F(x)}$$ Now $$F(x) = Se^{\int_0^x\mu(t)\;dt}$$ so $$F'(x) = Se^{\int_0^x\mu(t)\;dt}\cdot \frac{d}{dx}\left(\int_0^x\mu(t)\;dt\right)$$ $$=Se^{\int_0^x\mu(t)\;dt}\cdot \mu(x)$$ and therefore $$\frac{dy}{dx} = -\frac{F'(x)}{F(x)} = -\frac{Se^{\int_0^x\mu(t)\;dt}\cdot \mu(x)}{Se^{\int_0^x\mu(t)\;dt}}=\boxed{-\mu(x)}.$$
Question: Should there be a minus sign in the exponent in the definition of $F$?
This would produce the desired result.