So, this is clearly a function of \(u\), \(v\), \(p\), and \(t\) and so we’ll have four 1^st order partial derivatives and each of them should be pretty easy to compute.

Just remember that when computing each individual derivative that the other variables are to be treated as constants. So, for instance, when computing the \(u\) partial derivative all \(v\)’s, \(p\)’s and \(t\)’s are treated as constants. This in turn means that, for the \(u\) partial derivative, the second, fourth and fifth terms are considered to be constants (they don’t contain any \(u\)’s) and so differentiate to zero. Dealing with these types of terms properly tends to be one of the biggest mistakes students make initially when taking partial derivatives. Too often students just leave them alone since they don’t contain the variable we are differentiating with respect to.

Here are the four 1^st order partial derivatives for this problem.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial f}}{{\partial u}} & = {f_u} = 16u{t^3}p + 4ut\\ \frac{{\partial f}}{{\partial v}} & = {f_v} = - \frac{1}{2}{v^{ - \,\,\frac{1}{2}}}\,{p^2}{t^{ - 5}} - 1\\ \frac{{\partial f}}{{\partial p}} & = {f_p} = 8{u^2}{t^3} - 2\sqrt v \,p{t^{ - 5}} + 12{p^3}\\ \frac{{\partial f}}{{\partial t}} & = {f_t} = 24{u^2}{t^2}p + 5\sqrt v \,{p^2}{t^{ - 6}} + 2{u^2}\end{align*}}\]

The notation used for the derivative doesn’t matter so we used both here just to make sure we’re familiar with both forms.