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Understanding partial derivatives is fundamental in multivariable calculus, especially as they are the building blocks of more complex operations such as the chain rule and implicit differentiation. When dealing with a function that has multiple variables, like U which is a function of x, y, and z, partial derivatives measure how U changes with one variable, while keeping the other variables constant.

Imagine you're at a smoothie shop, where the taste (U) of your smoothie depends on the amount of strawberries (x), bananas (y), and ice (z). The partial derivative with respect to strawberries, for instance, shows how the taste changes if you only add or remove strawberries while keeping the amount of bananas and ice fixed.

In the exercise provided, we calculate the way U changes with respect to each variable individually, resulting in three separate derivatives. A helpful tip for students is to use the power rule and multiplication rule of single-variable calculus, but applied to one variable at a time.

The chain rule in calculus is a method for finding the derivative of a composite function. Picture it as a domino effect: when one variable affects another, which then affects a third, you need the chain rule to understand the complete impact start to finish. This rule is more like a relay race, where each runner passes the baton to the next. In the context of our given problem, each function x(t), y(t), and z(t) is like a team member, each with its own effect on the overall function U.

To apply the chain rule,

Identify the Interconnected Functions

Here, U is composed of functions of t through x, y, and z. You first find out how each of these components changes on their own with respect to t, and then assess how these changes affect U. Simply put, how does changing the time (t) alter the taste of your smoothie (U), by first changing the ingredients individually and then altogether.

Implicit differentiation is a technique used when a function is not stated explicitly. For example, rather than having y explicitly defined as a function of x, such as y = x^2, we could have an equation where y and x are mingled together, like in a circle's equation x^2 + y^2 = r^2. In such scenarios, one has to play detective to tease out the relationship between x and y.

In our exercise, even though U is given explicitly at first glance, it's actually implicit because U is defined in terms of other functions of t. Hence, to find dU/dt, we need to implicitly differentiate with respect to t, while keeping in mind the inter-relations of x, y, and z with t. This clever move allows us to capture the composite effect without laboring through a mess of algebra.