Let $u$ be quasiconcave and differentiable at $x$. If $u(x) \geq u(y)$, then how to show that $\frac{\partial u((1-t)x+ty)}{\partial t} \mid_{t =0} \geq 0$?
$u$ is quasiconcave means that for all $\lambda \in [0,1]$,$u(\lambda a + (1 - \lambda)b) \geq \min \{u(a), u(b)\}$.
I don´t know why this is true. Suppose $u(x) = x$, which is a quasiconcave function. But if $a > b$, then $\frac{(1-t)a+tb}{\partial t} \mid_{t =0} =b-a<0$.