For this problem it looks like we’ll have three 1^st order partial derivatives to compute.

Be careful with product rules with partial derivatives. The first term will only need a product rule for the \(t\) derivative and the second term will only need the product rule for the \(v\) derivative. Do not “overthink” product rules with partial derivatives. Do them when required but make sure to not do them just because you see a product. When you see a product look at the “factors” of the product. Do both “factors” have the variable you are differentiating with respect to or not and use the product rule only if they both do.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial g}}{{\partial s}} & = {g_s} = \frac{{{t^2}}}{{s + 2t}} - 3{s^2}\ln \left( {3v} \right)\\ \frac{{\partial g}}{{\partial t}} & = {g_t} = 2t\ln \left( {s + 2t} \right) + \frac{{2{t^2}}}{{s + 2t}} - 2t\ln \left( {3v} \right)\\ \frac{{\partial g}}{{\partial v}} & = {g_v} = 4\ln \left( {3v} \right) - \frac{{{s^3} + {t^2} - 4v}}{v}\end{align*}}\]

Make sure you can differentiate natural logarithms as they will come up fairly often. Recall that, with the chain rule, we have,

\[\frac{d}{{dx}}\left[ {\ln \left( {f\left( x \right)} \right)} \right] = \frac{{f'\left( x \right)}}{{f\left( x \right)}}\]

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