Answer
$dR=\beta^2 \cos \gamma d\alpha+2\alpha \cos \gamma \beta d \beta-\alpha \beta^2 \sin \gamma d \gamma$
Work Step by Step
Write the differential form can be evaluated as follows:
$dR=\dfrac{\partial R}{\partial \alpha} d\alpha +\dfrac{\partial R}{\partial \beta} d\beta+\dfrac{\partial R}{\partial \gamma} d\gamma$
We are given that $R=\alpha \beta^2 \cos \gamma$
$dR=\dfrac{\partial R}{\partial \alpha} d\alpha +\dfrac{\partial R}{\partial \beta} d\beta+\dfrac{\partial R}{\partial \gamma} d\gamma\\\\=\beta^2 \cos \gamma d\alpha+2\alpha \cos \gamma \beta d \beta+(-\alpha) \beta^2 \sin \gamma d \gamma\\\\=\beta^2 \cos \gamma d\alpha+2\alpha \cos \gamma \beta d \beta-\alpha \beta^2 \sin \gamma d \gamma$
Hence, $dR=\beta^2 \cos \gamma d\alpha+2\alpha \cos \gamma \beta d \beta-\alpha \beta^2 \sin \gamma d \gamma$
