Answer
$$\dfrac{\partial u}{\partial \alpha} =4e^{-4} \\ \dfrac{\partial u}{\partial \beta} =-7e^{-4} \\ \dfrac{\partial u}{\partial \gamma} =-24e^{-4} $$
Work Step by Step
$du=e^{-4} dx-2e^{-4}dy+8e^{-4} dt$
Now, $\dfrac{\partial u}{\partial \alpha} =e^{-4} \dfrac{\partial x}{\partial \alpha} -2e^{-4}\dfrac{\partial y}{\partial \alpha} +8e^{-4} \dfrac{\partial t}{\partial \alpha} $ and $\dfrac{\partial u}{\partial \beta} =e^{-4} \dfrac{\partial x}{\partial \alpha} -2e^{-4}\dfrac{\partial y}{\partial \beta} +8e^{-4} \dfrac{\partial t}{\partial \beta} $ and $\dfrac{\partial u}{\partial \gamma} =e^{-4} \dfrac{\partial x}{\partial \gamma} -2e^{-4}\dfrac{\partial y}{\partial \gamma} +8e^{-4} \dfrac{\partial t}{\partial \gamma} $
When $\alpha=-1;\beta=2; \gamma=1$, we have
$$ \dfrac{\partial u}{\partial \alpha} =4e^{-4}
\\ \dfrac{\partial u}{\partial \beta} =-7e^{-4} \\ \dfrac{\partial u}{\partial \gamma} =-24e^{-4} $$
Our answers are: $$\dfrac{\partial u}{\partial \alpha} =4e^{-4} \\ \dfrac{\partial u}{\partial \beta} =-7e^{-4} \\ \dfrac{\partial u}{\partial \gamma} =-24e^{-4} $$
