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Partial derivatives are like regular derivatives, but they show how a function changes as just one variable changes, while keeping other variables constant. Imagine a multivariable function defined as \( z = f(x, y) \). Here, \( x \) and \( y \) are variables that affect \( z \). The partial derivative of \( z \) with respect to \( x \), denoted as \( \partial z/\partial x \), tells us how \( z \) changes as \( x \) changes, holding \( y \) constant.

When solving real-world problems, such as the one in the exercise, you are often given a function that depends on more than one variable. Therefore, you need to find the rate at which this function changes with respect to one of these variables, which is the whole point of finding partial derivatives. In the given exercise, we find \( \partial z/\partial u \) and \( \partial z/\partial v \) using two approaches: applying the chain rule and by expressing \( z \) in terms of \( u \) and \( v \) before differentiating.

• The first step always involves expressing the function in terms of the variables of interest.
• Then, calculate the derivative while keeping other variables fixed to see how the change in one variable affects the entire function.

The chain rule is a fundamental tool in calculus that helps differentiate composite functions. It tells us how to differentiate a function that is nested within another function. Think of it like peeling an onion—you're differentiating layer by layer!

In the context of multivariable calculus, where you have functions like \( z = \operatorname{tan}^{-1}(x/y) \) and \( x \) and \( y \) are themselves functions of \( u \) and \( v \), the chain rule becomes very handy. It helps express partial derivatives in terms of other partial derivatives.

• To apply the chain rule, first identify the "outside" function, here \( \operatorname{tan}^{-1} \), and the "inside" functions \( x/y \).
• Next, multiply the derivative with respect to the "inside" functions by the derivative of these functions with respect to the new variables (e.g., \( u \) and \( v \)).

For example, \( \partial z/\partial u \) and \( \partial z/\partial v \) are computed using:
\[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \]
\[ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v} \]
This layered approach makes complex differentiations manageable by breaking them into simpler, sequential steps.

Trigonometric functions are mathematical functions based on angles, commonly used in calculus to solve problems involving triangles and oscillations. The main functions are sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)).

These functions often appear in multivariable calculus problems as they are key to modeling cyclical phenomena like waves. In the exercise, functions like \( x = u \cos v \) and \( y = u \sin v \) show how these trigonometric functions interplay with calculus concepts.

• The function \( \tan^{-1}(x/y) \) is an inverse trigonometric function that represents the angle whose tangent is \( x/y \).
• In solving the exercise, understanding these trigonometric relationships helps in properly setting up equations and finding derivatives.

Differentiating these functions requires understanding their derivatives:

• For \( \sin v \), the derivative is \( \cos v \).
• For \( \cos v \), the derivative is \(-\sin v \).
• Understanding these derivatives is crucial for calculating the partial derivatives of \( x \) and \( y \) when using the chain rule.

With this knowledge, you can confidently tackle calculus problems that involve trigonometric expressions.