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

Be careful with product rules with partial derivatives. For example, the second term, while definitely a product, will not need the product rule since each “factor” of the product only contains \(u\)’s or \(v\)’s. On the other hand, the first term will need a product rule when doing the \(u\) partial derivative since there are \(u\)’s in both of the “factors” of the product. However, just because we had to product rule with first term for the \(u\) partial derivative doesn’t mean that we’ll need to product rule for the \(v\) partial derivative as only the second “factor” in the product has a \(v\) in it.

Basically, be careful to 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 two 1^st order partial derivatives for this problem.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial f}}{{\partial u}} & = {f_u} = 2u\sin \left( {u + {v^3}} \right) + {u^2}\cos \left( {u + {v^3}} \right) - 4\sec \left( {4u} \right)tan\left( {4u} \right){\tan ^{ - 1}}\left( {2v} \right)\\ \frac{{\partial f}}{{\partial v}} & = {f_v} = 3{v^2}{u^2}cos\left( {u + {v^3}} \right) - \frac{{2\sec \left( {4u} \right)}}{{1 + 4{v^2}}}\end{align*}}\]

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