Question

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

Short answer

The derivative dw/dp is 0.

Step-by-step solution

Understanding the given expressions

Identify the given variables and expressions for substitution. The variables are: $u=\cos [\ln \tan (p+\frac{1}{4}\pi )]$ and$v=\sin [\ln \tan (p+\frac{1}{4}\pi )]$.The goal is to find the derivative of $w=\sqrt{u^2+v^2}$ with respect to $p$.

Simplification of the expressions

Note that $\tan (\frac{1}{4}\pi +p)=1/p$ for certain values of $p$, and thus the expressions: $u=\cos [\theta ]$ where $\theta =ln(\tan (p+1/4\pi ))$ and $v=\sin [\theta ].$

Substitute $u$ and $v$

Since $w=\sqrt{u^2+v^2}$, and knowing identity $u^2+v^2={\cos }^2[\theta ]+{\sin }^2[\theta ]=1$ we get: $w=\sqrt{1}=1.$

Differentiate $w$

Since $w=1$ is a constant function with no dependency on p, taking the derivative $dw/dp$ yields zero. Therefore: $\frac{dw}{dp}=0$

Key concepts

Partial Derivatives

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 $\frac{dw}{dp}$, 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

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:

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

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 $u^2+v^2=1$.

Chain Rule

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:
$(f(g(x)){)}^{\prime }=f^{\prime }(g(x))g^{\prime }(x)$
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.