Given \(w=\sqrt{u^{2}+v^{2}}, u=\cos \left[\ln \tan \left(p+\frac{1}{4} \pi\right)\right], v=\sin \left[\ln \tan \left(p+\frac{1}{4} \pi\right)\right],\) find \(d w / d p.\)

Partial derivatives help us find the rate of change of a function with respect to one variable while keeping others constant. This becomes essential when dealing with functions of multiple variables.
For example, consider a function \(f(x, y)\). Its partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\), while keeping \(y\) constant.
In our exercise, we have variables \(u\) and \(v\) dependent on another variable \(p\). To find \(\frac{d w}{d p}\), we first need expressions for \(\frac{\partial w}{\partial u}\), \(\frac{\partial w}{\partial v}\), and the partial derivatives of \(u\) and \(v\) with respect to \(p\). This approach often employs the chain rule, which we'll explore next.

Trigonometric functions like \(\sin\), \(\cos\), and \(\tan\) are significant in various fields, including calculus. These functions are periodic and often come with useful identities.
In our problem, \(u\) and \(v\) are defined using the \(\cos\) and \(\sin\) functions respectively:

• \(u = \cos \left[ \ln \tan \left( p + \frac{1}{4} \pi \right) \right]\)
• \(v = \sin \left[ \ln \tan \left( p + \frac{1}{4} \pi \right) \right]\)

These expressions are part of the input for \(w = \sqrt{u^2 + v^2}\).
Important trigonometric identities, such as \( \cos^2 \theta + \sin^2 \theta = 1 \), help simplify our calculations. Here, this identity reveals that \(w\) simplifies to a constant as \(u^2 + v^2 = 1\).

The chain rule allows us to differentiate composite functions. It's crucial when dealing with functions dependent on other functions.
For a composite function like \(f(g(x))\), the derivative is found using:
\( (f(g(x)))' = f'(g(x)) \, g'(x)\)
This means we differentiate the outer function and then multiply it by the derivative of the inner function.
Applying this concept to our problem, we can express \(w\) in terms of intermediate functions \(u\) and \(v\), which themselves depend on \(p\).
First, determine \( \frac{\partial w}{\partial u}\) and \(\frac{\partial w}{\partial v}\):

• \(\frac{\partial w}{\partial u} = \frac{u}{\sqrt{u^2 + v^2}}\)
• \(\frac{\partial w}{\partial v} = \frac{v}{\sqrt{u^2 + v^2}}\)

Next, use the chain rule to combine these partial derivatives with \(\frac{\partial u}{\partial p}\) and \(\frac{\partial v}{\partial p}\). However, as we simplified earlier using trigonometric identities, \(w\) turns out to be constant, making \(\frac{dw}{dp} = 0\).
Thus, the chain rule confirms that the derivative of a constant function with respect to any variable is zero.