## Thomas' Calculus 13th Edition

$W_P=V$ and $W_V=P+\dfrac{\delta v^2}{ 2g}$ $W_{\delta}=\dfrac{V v^2}{ 2g}$; $W_{v}=\dfrac{V \delta v}{g}$; $W_{g}=-\dfrac{ V \delta v^2}{ 2g^2}$;
We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to one of its variables, by keeping all as a constant. $W_P=V$ and $W_V=P+\dfrac{\delta v^2}{ 2g}$ $W_{\delta}=\dfrac{V v^2}{ 2g}$; $W_{v}=\dfrac{2 V \delta v}{ 2g}=\dfrac{V \delta v}{g}$; $W_{g}=-\dfrac{ V \delta v^2}{ 2g^2}$