Consider the following. \[ w=x y, \quad x=e^{4 t}, \quad y=e^{-6 t} \] (a) Find \( d w / d t \) by using the appropriate Chain Rule. \( \frac{d w}{d t}=\square \) (b) Find \( d w / d t \) by converting \( w \) to a function of \( t \) before differentiating. \( \frac{d w}{d t}=\square \)

To solve the problem, we will first differentiate \( w \) using the Chain Rule and then by expressing \( w \) as a function of \( t \). (a) Using the Chain Rule: We have \( w = x y \), \( x=e^{4t} \), and \( y=e^{-6t} \). We can find \( \frac{dw}{dt} \) using the Chain Rule: \[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \] Calculating the required partial derivatives and derivatives: - \( \frac{\partial w}{\partial x} = y \) - \( \frac{\partial w}{\partial y} = x \) - \( \frac{dx}{dt} = 4e^{4t} \) - \( \frac{dy}{dt} = -6e^{-6t} \) Thus, \[ \frac{dw}{dt} = y \cdot \frac{dx}{dt} + x \cdot \frac{dy}{dt} = e^{-6t} \cdot (4e^{4t}) + e^{4t} \cdot (-6e^{-6t}) \] Simplifying this: \[ \frac{dw}{dt} = 4e^{-6t} e^{4t} - 6e^{4t} e^{-6t} = 4e^{-2t} - 6e^{-2t} = -2e^{-2t} \] So, \[ \frac{dw}{dt} = -2 e^{-2t} \] (b) Expressing \( w \) as a function of \( t \) before differentiating: We substitute \( x \) and \( y \) back into \( w \): \[ w = e^{4t} \cdot e^{-6t} = e^{4t - 6t} = e^{-2t} \] Now, we differentiate \( w \) with respect to \( t \): \[ \frac{dw}{dt} = \frac{d}{dt}(e^{-2t}) = -2 e^{-2t} \] Thus, \[ \frac{dw}{dt} = -2 e^{-2t} \] So in both methods, we reach the same conclusion: \[ \frac{dw}{dt} = -2 e^{-2t} \]