Answer
$W_P=V$
$W_V=P+\dfrac{\delta v^2}{ 2g}$
$W_{\delta}=\dfrac{V v^2}{ 2g}$
$W_{v}=\dfrac{V \delta v}{g}$
$W_{g}=-\dfrac{ V \delta v^2}{ 2g^2}$
Work Step by Step
Our aim is to take the first partial derivative of the given function $W(x,y)$ with respect to one of its variables, by treating all others as a constant:
$W_P=V$ and $W_V=P+\dfrac{\delta v^2}{ 2g}$
$W_{\delta}=\dfrac{V v^2}{ 2g}$; $W_{v}=\dfrac{2 V \delta v}{ 2g}=\dfrac{V \delta v}{g}$;
$W_{g}=-\dfrac{ V \delta v^2}{ 2g^2}$
