I am dealing with a mean curvature type equation as following: $\displaystyle{\sum_{i,j=1}^{2}}(\delta_{ij}\frac{u_{i}u_{j}}{1+Du^{2}})u_{ij}=(1+Du^{2})^{\frac{1}{2}\frac{1}{2\alpha}}$, where $\alpha>1$ fixed. $u$ is convex and defined on the entire $R^{2}$ suppose when $x$ is large, $C_{1}x^{\alpha}\leqDu(x)\leq C_{2}x^{\alpha}$, where $C_{1}$ and $C_{2}$ are fixed positive constants. Then is there such estimate that: when $x$ is large $D^{2}u\leq C_{3}x^{\alpha1}$ for some fixed constant $C_{3}$.
You can find out the right power in the radial case, $u=f(\sqrt{x^2+y^2})$. Your equation becomes
$$
\frac{\displaystyle 1+ \left(\frac{1}{r^2}1\right)(f^\prime)^2}{\displaystyle 1+\frac{1}{r^2}(f^\prime)^2} f^{\prime\prime} + \frac{1}{r} f^{\prime}
=
\left(1+\frac{1}{r^2}(f^\prime)^2\right)^{\frac{1}{2}\frac{1}{2\alpha}}
$$
Now for $r$ large, the assumption $C_1<r^{\alpha} f^\prime <C_2$ means, if you write $g(r)= r^{\alpha} f^\prime $,
$$
(1 + o(1)) {r^2} f^{\prime\prime} + r^{\alpha1} g = r^{\displaystyle \frac{(\alpha1)^2}{\alpha}} (g^2(r)+o(1)),
$$
Since $\alpha>1$, $\alpha1 > \frac{(\alpha1)^2}{\alpha}$, and therefore 

