Answer:
To find the second partial derivative \(f_{ss}\) of the function \(f(x, y) = 2x^2y\), where \(x = 2s + t\) and \(y = 2s - t\), we need to take two partial derivatives with respect to \(s\).
First, let's find the first partial derivatives of \(f\) with respect to \(s\) and \(t\):
\[
\begin{align*}
f_x &= \frac{\partial f}{\partial x} = 2y \cdot \frac{\partial}{\partial x}(2x^2) = 4xy \\
f_y &= \frac{\partial f}{\partial y} = 2x^2
\end{align*}
\]
Now, let's find the second partial derivative \(f_{ss}\) by taking the partial derivative of \(f_s\) (the derivative of \(f\) with respect to \(s\)) with respect to \(s\) again:
\[
f_{ss} = \frac{\partial}{\partial s}(f_s) = \frac{\partial}{\partial s}(4xy)
\]
To find this second derivative, we treat \(x\) and \(y\) as functions of \(s\) and \(t\) and apply the chain rule. We'll need to compute \(\frac{\partial x}{\partial s}\) and \(\frac{\partial y}{\partial s}\):
\[
\begin{align*}
\frac{\partial x}{\partial s} &= \frac{\partial}{\partial s}(2s + t) = 2 \\
\frac{\partial y}{\partial s} &= \frac{\partial}{\partial s}(2s - t) = 2
\end{align*}
\]
Now, we can apply the chain rule:
\[
\begin{align*}
f_{ss} &= \frac{\partial}{\partial s}(4xy) \\
&= \frac{\partial x}{\partial s} \cdot \frac{\partial}{\partial x}(4xy) + \frac{\partial y}{\partial s} \cdot \frac{\partial}{\partial y}(4xy) \\
&= 2 \cdot (4y \cdot 4x) + 2 \cdot (4x^2) \\
&= 32xy + 8x^2
\end{align*}
\]
So, the general expression for the second partial derivative \(f_{ss}\) is:
\[f_{ss} = 32xy + 8x^2\]
Now, you can substitute the expressions for \(x\) and \(y\) in terms of \(s\) and \(t\) back into this expression to get the final result:
\[f_{ss} = 32(2s + t)(2s - t) + 8(2s + t)^2\]
Explanation: