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 11
 Problem Statement

Given these two statements:
y =v(x,t)
v(x(s,t),t) = y(s,t)
find:
second partial derivative of y in respects to t
 Relevant Equations
 chain rule
I had already calculated the first partial derivative to equal the following:
$$\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial t}$$
Now the second partial derivative I can use the chain rule to do and get to:
$$\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t} \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial x} \frac{\partial}{\partial t} \frac{\partial x}{\partial t} + \frac{\partial^2 v}{\partial t^2}$$
How would I continue? I can easily perform the last too terms:
$$\frac{\partial v}{\partial x} \frac{\partial}{\partial t} \frac{\partial x}{\partial t} = \frac{\partial v}{\partial x} \frac{\partial^2 x}{\partial t^2} $$
and
$$\frac{\partial}{\partial t } \frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial t^2} $$
How would I perform the first term? Again:
$$y =v(x,t)$$
$$v(x(s,t),t) = y(s,t)$$
$$\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial t}$$
Now the second partial derivative I can use the chain rule to do and get to:
$$\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t} \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial x} \frac{\partial}{\partial t} \frac{\partial x}{\partial t} + \frac{\partial^2 v}{\partial t^2}$$
How would I continue? I can easily perform the last too terms:
$$\frac{\partial v}{\partial x} \frac{\partial}{\partial t} \frac{\partial x}{\partial t} = \frac{\partial v}{\partial x} \frac{\partial^2 x}{\partial t^2} $$
and
$$\frac{\partial}{\partial t } \frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial t^2} $$
How would I perform the first term? Again:
$$y =v(x,t)$$
$$v(x(s,t),t) = y(s,t)$$