I'm reading a proof of equation $$\frac{\partial \dot{\overrightarrow{r_i}}}{\partial \dot q_{\alpha}}=\frac{\partial \overrightarrow r_i}{\partial q_{\alpha}}$$ where $\overrightarrow r_i=\overrightarrow r_i(q_1, q_2, \ldots, q_s)$ and all $q_i$ are independent, so $$\frac{\partial \dot q_{\beta}}{\partial \dot q_{\alpha}}=\delta_{\alpha\beta}=\left \{ {{1, \alpha=\beta} \atop {0, \alpha\neq\beta}} \right\} $$ where $\delta_{\alpha\beta}$ is Kronecker delta.
And there are only two steps of proof in which I don't understand the second: $$\frac{d\overrightarrow r_i}{dt}=\dot{\overrightarrow{r_i}}=\frac{\partial \overrightarrow r_i}{\partial q_1}\frac{\partial q_1}{\partial t}+\frac{\partial \overrightarrow r_i}{\partial q_2}\frac{\partial q_2}{\partial t}+\ldots+\frac{\partial \overrightarrow r_i}{\partial q_s}\frac{\partial q_s}{\partial t}=\sum\limits_{\beta=1}^s\frac{\partial \overrightarrow r_i}{\partial q_{\beta}}\dot q_{\beta}$$ $$\frac{\partial \dot{\overrightarrow{r_i}}}{\partial \dot q_{\alpha}}=\sum\limits_{\beta=1}^s\frac{\partial \overrightarrow r_i}{\partial q_{\beta}}\delta_{\alpha\beta}=\frac{\partial \overrightarrow r_i}{\partial q_{\alpha}}$$ In the second step I don't understand how we lost the first term which might be seen in my calculations (if they are correct, of course): $$\frac{\partial \dot{\overrightarrow{r_i}}}{\partial \dot q_{\alpha}}=\frac{\partial \sum\limits_{\beta=1}^s\frac{\partial \overrightarrow r_i}{\partial q_{\beta}}\dot q_{\beta}}{\partial \dot q_{\alpha}}=\sum\limits_{\beta=1}^s(\frac{\partial^2 \overrightarrow r_i}{\partial \dot q_{\alpha} \partial q_{\beta}}\dot q_{\beta}+\frac{\partial \overrightarrow r_i}{\partial q_{\beta}}\frac{\partial \dot q_{\beta}}{\partial \dot q_{\alpha}})=\sum\limits_{\beta=1}^s(\frac{\partial^2 \overrightarrow r_i}{\partial \dot q_{\alpha} \partial q_{\beta}}\dot q_{\beta}+\frac{\partial \overrightarrow r_i}{\partial q_{\beta}}\delta_{\alpha\beta})$$ So were am I wrong?