Answer
$\beta^2 \cos (\gamma) d\alpha+2 \alpha \cos (\gamma) \beta d \beta-\alpha \beta^2 \sin (\gamma) d \gamma$
Work Step by Step
We are given that $R=\alpha \beta^2 \cos \gamma$
The differential form can be written as follows:
$dR=(\dfrac{\partial R}{\partial \alpha}) d\alpha +(\dfrac{\partial R}{\partial \beta}) d\beta+(\dfrac{\partial R}{\partial \gamma}) d\gamma$
Calculate the partial derivatives with respect to $u$, $v$ and $w$.
$dR=\dfrac{\partial R}{\partial \alpha}) d\alpha +(\dfrac{\partial R}{\partial \beta}) d\beta+(\dfrac{\partial R}{\partial \gamma}) d\gamma=\beta^2 \cos (\gamma) d\alpha+2\alpha \cos \gamma \beta d \beta+(-\alpha) \beta^2 \sin \gamma d \gamma$
or, $dR=\beta^2 \cos (\gamma) d\alpha+2 \alpha \cos (\gamma) \beta d \beta-\alpha \beta^2 \sin (\gamma) d \gamma$
