Question

Use the chain rule to find dw/dt. w = xey/z, x = t7, y = 4 − t, z = 2 + 9t

Answer 1

Given


`w = xe^(y/z),\ x = t^7,\ y = 4 - t, \ z = 2 + 9t \\ \\ (dw)/(dt) = (dw)/(dx) \cdot (dx)/(dt) + (dw)/(dy) \cdot (dy)/(dt) + (dw)/(dz) \cdot (dz)/(dt) \\ \\ (dw)/(dx)=e^(y/z) \\ \\ (dw)/(dy)= (x)/(z) e^(y/z) \\ \\ (dw)/(dz)=- (xy)/(z^2) e^(y/z) \\ \\ (dx)/(dt)=7t^6 \\ \\ (dy)/(dt)=-1 \\ \\ (dz)/(dt)=9`

Thus,


`(dw)/(dt)=e^(y/z)\cdot7t^6+(x)/(z) e^(y/z)\cdot(-1)+- (xy)/(z^2) e^(y/z)\cdot(9) \\ \\ =7t^6e^(y/z)-(x)/(z) e^(y/z)-9(xy)/(z^2) e^(y/z) \\ \\ =\left(7t^6-(x)/(z)-9(xy)/(z^2)\right)e^(y/z)`