Answer
$\dfrac{\partial u}{\partial \alpha} =4e^{-4} $
and $\dfrac{\partial u}{\partial \beta} =-7e^{-4} $
and
$\dfrac{\partial u}{\partial \gamma} =-24e^{-4} $
Work Step by Step
Here, we have $du=e^{-4} dx-2e^{-4}dy+8e^{-4} dt$
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} $
When $\alpha=-1;\beta=2; \gamma=1$, we have
$\dfrac{\partial u}{\partial \alpha} =4e^{-4} $
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} $
When $\alpha=-1;\beta=2; \gamma=1$, we have
$\dfrac{\partial u}{\partial \beta} =-7e^{-4} $
$\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 \gamma} =-24e^{-4} $
Hence, we have $\dfrac{\partial u}{\partial \alpha} =4e^{-4} $
and $\dfrac{\partial u}{\partial \beta} =-7e^{-4} $
and
$\dfrac{\partial u}{\partial \gamma} =-24e^{-4} $
