Question

Find \(\partial w / \partial s\) and \(\partial w / \partial t\) by using the appropriate Chain Rule. \(w=z e^{x / y}, \quad x=s-t, \quad y=s+t, \quad z=s t\)

Short answer

The partial derivatives \(\partial w / \partial s\) and \(\partial w / \partial t\) are found by applying the chain rule to the given function \(w=z e^{x / y},\) with \(x=s-t\), \(y=s+t\), and \(z=s t\). This gives us: \(\partial w / \partial s = e^{x / y}[s - (s t)/y] + z e^{x / y} / y\) and \(\partial w / \partial t = e^{x / y}[-1 - xs/y] + s e^{x / y}\).

Step-by-step solution

Write Down the Given Functions and Relationship

The function to be differentiated is \(w=z e^{x / y}\), with \(x=s-t\) and \(y=s+t\), \(z=s t\).

Apply Chain Rule for Partial Differentiation of w with respect to s

To compute \(\partial w / \partial s\), apply the chain rule: \(\partial w / \partial s = (\partial w / \partial x)(\partial x / \partial s) + (\partial w / \partial y)(\partial y / \partial s) + (\partial w / \partial z)(\partial z / \partial s)\). You can find these individual derivatives as \(\partial w / \partial x = z e^{x / y} / y\), \(\partial w / \partial y = -z x e^{x/y} / y^2\) and \(\partial w / \partial z = e^{x / y}\). The derivatives of \(x, y,\) and \(z\) are \(\partial x / \partial s = 1\), \(\partial y / \partial s = 1\), and \(\partial z / \partial s = t\). Substituting these into the chain rule gives the derivative of \(w\) with respect to \(s\).

Apply Chain Rule for Partial Differentiation of w with respect to t

Similarly for \(\partial w / \partial t\), we have \(\partial w / \partial t = (\partial w / \partial x)(\partial x / \partial t) + (\partial w / \partial y)(\partial y / \partial t) + (\partial w / \partial z)(\partial z / \partial t)\). The derivatives of \(x, y,\) and \(z\) with respect to \(t\) are \(\partial x / \partial t = -1\), \(\partial y / \partial t = 1\), and \(\partial z / \partial t = s\). Substituting these into the chain rule gives the derivative of \(w\) with respect to \(t\).

Simplify the Expressions

After substituting the intermediate derivatives into the chain rule expressions, simplifying gives us the final expressions for \(\partial w / \partial s\) and \(\partial w / \partial t\)