Questions: Consider the following. [ w=x y z, quad x=s+4 t, quad y=s-4 t, quad z=s t^2 ] (a) Find (fracpartial wpartial s) and (fracpartial wpartial t) by using the appropriate Chain Rule. (Enter your answers in terms of (s) and (t).) [ fracpartial wpartial s=square ] [ fracpartial wpartial t=square ]

Solution Steps

To find the partial derivatives \(\frac{\partial w}{\partial s}\) and \(\frac{\partial w}{\partial t}\), we will use the chain rule for multivariable functions. First, express \(w\) in terms of \(s\) and \(t\) by substituting the expressions for \(x\), \(y\), and \(z\) into \(w\). Then, differentiate \(w\) with respect to \(s\) and \(t\) separately, applying the product rule as necessary.

We start by defining the variables \(x\), \(y\), and \(z\) in terms of \(s\) and \(t\): \[ x = s + 4t, \quad y = s - 4t, \quad z = st^2 \]

Next, we express \(w\) as: \[ w = xyz = (s + 4t)(s - 4t)(st^2) \]

Using the product rule, we find the partial derivative of \(w\) with respect to \(s\): \[ \frac{\partial w}{\partial s} = st^2(s - 4t) + st^2(s + 4t) + t^2(s - 4t)(s + 4t) \] This simplifies to: \[ \frac{\partial w}{\partial s} = st^2(s - 4t + s + 4t) + t^2(s^2 - 16t^2) = 2s t^2 + t^2(s^2 - 16t^2) \]

Now, we calculate the partial derivative of \(w\) with respect to \(t\): \[ \frac{\partial w}{\partial t} = 4st^2(s - 4t) - 4st^2(s + 4t) + 2st(s - 4t)(s + 4t) \] This simplifies to: \[ \frac{\partial w}{\partial t} = 4st^2(s - 4t - s - 4t) + 2st(s^2 - 16t^2) = -8st^2 + 2st(s^2 - 16t^2) \]

Final Answer