Question

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y z, \quad x=t^{2}, \quad y=2 t, \quad z=e^{-t}\)

Short answer

So, the derivative \(dw/dt\) by applying the chain rule and by converting \(w\) to a function of \(t\) before differentiating is \(4t^3e^{-t}\)

Step-by-step solution

Identify the functions

In our problem, \(w\) is a function of \(x\), \(y\) and \(z\), and \(x\), \(y\), and \(z\) are functions of \(t\). Also, let's note that \(dw/dt = (\partial w/\partial x)*(dx/dt) + (\partial w/\partial y)*(dy/dt) + (\partial w/\partial z)*(dz/dt)\), which is the multivariable chain rule.

Differentiate using the Chain Rule

Differentiate \(w\) with respect to \(x\), \(y\), and \(z\) separately. We get \(dz/dt = -e^{-t}\), \(dx/dt = 2t\), and \(dy/dt = 2\). Now, calculate, \(dw/dt = yz(dx/dt) + xz(dy/dt) + xy(dz/dt) = 2t*2t*e^{-t} + t^2*2*e^{-t} + 2t*t^2*(-e^{-t}) = 4t^3e^{-t}\).

Convert \(w\) to a function of \(t\) and Differentiate

We can express \(w\) as a function of \(t\) by substituting \(x\), \(y\), and \(z\) directly as given. Therefore, \(w = t^{2}*(2 t)*e^{-t} = 2t^3e^{-t}\) which we differentiate with respect to \(t\) to give \(dw/dt_{direct} = 4t^3e^{-t}\)