If we choose a unit square in which the four points where the function $f$ is known are as shown in the diagram the interpolation formula for $f$ is:\begin{equation}\begin{split}f(x+u, y+v)\approx (1-u)(1-v)f(x,y)+u(1-v)f(x+\Delta x,y)\\&\hspace{-7cm}+(1-u)vf(x,y+\Delta y)+uvf(x+\Delta x,y+\Delta y).\end{split}\end{equation}The above formula is equivalent to the following Taylor expansion series\begin{equation}f(x+u, y+v) \approx f(x,y) + u \frac{\partial{f}}{\partial x}(x,y) + v\frac{\partial{f}}{\partial y}(x,y)+uv \frac{\partial^2{f}}{\partial x \partial y}(x,y).\end{equation}The derivative of $f$ w.r.t. $x$ is given by:\begin{equation}\frac{\partial f}{\partial x}(x+u,y+v)= \frac{\partial{f}}{\partial x}(x,y)+u\frac{\partial^2{f}}{\partial x^2}(x,y) + v\frac{\partial{f}}{\partial x\partial y}(x,y)+uv \frac{\partial^3{f}}{\partial^2 x \partial y}(x,y)\end{equation}In a previous post I stated the approximate derivatives as follows:\begin{equation}\frac{\partial{f}}{\partial x}(x,y)= \frac{f(x+\Delta x, y)-f(x,y)}{\Delta x}\end{equation}\begin{equation}\frac{\partial^2{f}}{\partial^2 x}(x,y)= \frac{f(x+2\Delta x, y)-2f(x+\Delta x,y)+f(x,y)}{\Delta^2 x}\end{equation}\begin{equation}\frac{\partial^2{f}}{\partial x \partial y}(x,y)= \frac{f(x+\Delta x, y+\Delta y)-f(x, y+\Delta y)-f(x+\Delta x,y)+f(x,y)}{\Delta x\Delta y}\end{equation}\begin{equation}\frac{\partial^3{f}}{\partial^2 x \partial y}(x,y) = \frac{f(x+2\Delta x, y+\Delta y)-2f(x+\Delta x, y+\Delta y)+f(x, y +\Delta y)}{\Delta^2 x\Delta y}\\\frac{-f(x+2\Delta x,y)+2f(x+\Delta x,y)-f(x,y)}{\Delta^2 x\Delta y}\end{equation}Is it true to say that the $\frac{\partial^2{f}}{\partial^2 x}$ and $\frac{\partial^3{f}}{\partial^2 x \partial y}$ are equal to zero?Because we have to move $2\Delta x$ and bilinear interpolation involves 4 points. 
Derivatives involved in Bilinear interpolation
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