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The chain rule is a fundamental theorem in calculus used to find the derivative of a function based on its composition of other functions. It is especially useful when a variable depends on multiple other variables.
In essence, the chain rule helps us break down complex derivatives into simpler parts that are easier to handle. When you have a composite function, say, where a variable \( u \) depends on \( x \) and \( y \), which in turn depend on \( t \), you can use the chain rule to find \( \frac{du}{dt} \). The formula for the total derivative using the chain rule is:

• \( \frac{du}{dt} = \frac{\text{\textpartial} u}{\text{\textpartial} x} \frac{dx}{dt} + \frac{\text{\textpartial} u}{\text{\textpartial} y} \frac{dy}{dt} \)

In the exercise provided, we have a function \( u = \tan^{-1}\big(\frac{y}{x}\big) \) with \( x = \text{ln} \thinspace t \) and \( y = e^t \).
Using the chain rule helps us systematically compute \( \frac{du}{dt} \) without having to explicitly express \( u \) in terms of \( t \), which can be complex and cumbersome.

Partial derivatives are a core concept in multivariable calculus. They measure how a function changes as one variable changes while keeping other variables constant.
For a function like \( u = \tan^{-1}\big(\frac{y}{x}\big) \), we need to find its partial derivatives with respect to \( x \) and \( y \). These are the first steps in applying the chain rule.
In the given exercise, we calculate the partial derivatives as:

• \( \frac{\text{\textpartial} u}{\text{\textpartial} x} = \frac{\text{\textpartial}}{\text{\textpartial} x} \tan^{-1}\big(\frac{y}{x}\big) = \frac{-y}{x^2 + y^2} \)
• \( \frac{\text{\textpartial} u}{\text{\textpartial} y} = \frac{\text{\textpartial}}{\text{\textpartial} y} \tan^{-1}\big(\frac{y}{x}\big) = \frac{x}{x^2 + y^2} \)

Calculating partial derivatives precisely is crucial for correctly applying the chain rule, as they provide the rate of change of \( u \) concerning \( x \) and \( y \).

Differentiation is the process of finding the rate at which a function changes at any given point. It is one of the most important concepts in calculus.
For this exercise, we need to differentiate the functions \( x = \text{ln} \thinspace t \) and \( y = e^t \) concerning \( t \). This will give us the necessary derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

• \( \frac{dx}{dt} = \frac{d}{dt}( \text{\textln} \thinspace t ) = \frac{1}{t} \)
• \( \frac{dy}{dt} = \frac{d}{dt}( e^t ) = e^t \)

Now, substituting these derivatives back into our chain rule formula allows us to find the total derivative \( \frac{du}{dt} \):

• Substitute \( \frac{\text{\textpartial} u}{\text{\textpartial} x}, \frac{\text{\textpartial} u}{\text{\textpartial} y}, \frac{dx}{dt} \), and \( \frac{dy}{dt} \) into \( \frac{du}{dt} = \frac{\text{\textpartial} u}{\text{\textpartial} x} \frac{dx}{dt} + \frac{\text{\textpartial} u}{\text{\textpartial} y} \frac{dy}{dt} \)
• Plug \( x = \text{\textln} \thinspace t \) and \( y = e^t \) to get the final expression

By carefully following the differentiation steps, we simplify the complex function into something manageable and solvable.