Differentiation of H = U + pV I was asked to use direct differentiation of "H = U + pV" to find a relationship between:

$$ \left(\frac{\partial H}{\partial U}\right)_{\!P}$$   and   $$\left(\frac{\partial U}{\partial V}\right)_{\!P} $$

So my first step was to take a differential of the equation twice, once in terms of U and once in terms of V:

$$ \left(\frac{\partial}{\partial U}H\right)_{\!P}=\left(\frac{\partial}{\partial U}(U +PV)\right)_{\!P}\rightarrow\left(\frac{\partial H}{\partial U}\right)_{\!P}= 1 + P\left(\frac{\partial V}{\partial U}\right)_{\!P}$$
and

$$ \left(\frac{\partial}{\partial V}H\right)_{\!P}=\left(\frac{\partial}{\partial V}(U +PV)\right)_{\!P}\rightarrow\left(\frac{\partial H}{\partial V}\right)_{\!P}= \left(\frac{\partial U}{\partial V}\right)_{\!P} + P$$

When I tried to substitute one equation into the other, they would simply end up all canceling out. Is there another way?
A way to do it would be:

\begin{align}
\left( \frac{\partial H}{\partial U} \right)_P \left( \frac{\partial U}{\partial V} \right)_P &amp;= \left( \frac{\partial H}{\partial V} \right)_P = \left( \frac{\partial U}{\partial V} \right)_P + P\\
\left( \frac{\partial H}{\partial U} \right)_P &amp;= \frac{\left( \frac{\partial U}{\partial V} \right)_P + P}{\left(\frac{\partial U}{\partial V} \right)_P}\\
\left(\frac{\partial U}{\partial V} \right)_P &amp;= \frac{P}{\left( \frac{\partial H}{\partial U} \right)_P - 1}\\
\end{align}