Recall the definition of slope semistability, taken from section 1.2 of Huybrechts and Lehn's "Geometry of Moduli Spaces of Sheaves" book. Let $X$ be a projective $\mathbb{C}$-scheme and $E \in Coh(X)$ of dimension $d = \dim(X)$. The rank of $E$ is defined to be \begin{equation} rk(E) = \frac{\alpha_d(E)}{\alpha_d(\mathcal{O}_X)}, \end{equation} and the degree of $E$ is defined to be \begin{equation} \deg E = \alpha_{d-1}(E) - rk(E) \alpha_{d-1}(\mathcal{O}_X). \end{equation} The slope of $E$ is defined to be $\mu(E) := \frac{\deg(E)}{rk(E)}$. Then we say $E$ is slope semistable iff for all subsheaves $F \subset E$ with $0 < rk(F) < rk(E)$ we have $rk(E) \deg(F) \leq rk(F) \deg (E)$.

My question is the following. Let $E$ be a slope semistable vector bundle on, say, nonsingular projective variety $X$. Then is the dual $E^*$ also slope semistable? The natural thing to do here is to take a $F \subset E^*$, then dualize, and use additivity of Hilbert polynomials (note the above inequality for slope semistable is equivalent to the inequality $\alpha_d(E) \alpha_{d-1}(F) \leq \alpha_d(F) \alpha_{d-1}(E)$). But when $F$ is not locally free, you run into issues of the dual not being a surjection of sheaves...