What Am I Missing in the Chain Rule Calculation?

[yungman]

$$\hbox { Let }\; u(x,y)=v(x^2-y^2,2xy) \;\hbox { and let }\; t=x^2-y^2,\;s=2xy$$

$$u_x = 2xv_t \;+\; 2yv_s$$

$$u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}$$

The $u_{yy}$ can be done the same way and is not shown here.



According to Chain Rule:

$$u_x = \frac{\partial v}{\partial x} \;=\; \frac{\partial v}{\partial t}\frac{\partial t}{\partial x} \;+\; \frac{\partial v}{\partial s}\frac{\partial s}{\partial x} \;=\; 2x\frac{\partial v}{\partial t} \;+\; 2y\frac{\partial v}{\partial s}$$

$$u_{xx} = \frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x}[2x\frac{\partial v}{\partial t} \;+\; 2y\frac{\partial v}{\partial s}] = 4x^2\frac{\partial^2 v}{\partial t^2} \;+\; 2\frac{\partial v}{\partial t} \;+\; 4y^2 \frac{\partial^2 v}{\partial s^2}$$

Where

$$\frac{\partial y}{\partial x} = 0$$

Please tell me what I am doing wrong? How do I miss $8xyv_{ts}$ term?

 

[Wizlem]

$$\frac{\partial}{\partial x} \frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial t^2}\frac{\partial t}{\partial x} + \frac{\partial^2 v}{\partial s \partial t}\frac{\partial s}{\partial x}$$

 

[yungman]

I see, thanks.