Question

Find all first partial derivatives of each function. \(F(w, z)=w \sin ^{-1}\left(\frac{w}{z}\right)\)

Short answer

\(\frac{\partial F}{\partial w} = \sin^{-1}\left(\frac{w}{z}\right) + \frac{w}{z\sqrt{z^2 - w^2}}\), \(\frac{\partial F}{\partial z} = -\frac{w^2}{z^2\sqrt{z^2 - w^2}}\).

Step-by-step solution

Identify the Function to Differentiate

We are given the function \( F(w, z) = w \sin^{-1}\left(\frac{w}{z}\right) \). Our goal is to find the first partial derivatives \( \frac{\partial F}{\partial w} \) and \( \frac{\partial F}{\partial z} \).

Differentiate with Respect to \(w\)

To find \( \frac{\partial F}{\partial w} \), treat \( z \) as a constant:\[\frac{\partial F}{\partial w} = \frac{d}{dw}\left( w \sin^{-1}\left(\frac{w}{z}\right) \right)\]Apply the product rule:\[= \sin^{-1}\left(\frac{w}{z}\right) + w \cdot \frac{d}{dw}\left(\sin^{-1}\left(\frac{w}{z}\right)\right)\]The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dw} \), where \( u = \frac{w}{z} \):\[= \sin^{-1}\left(\frac{w}{z}\right) + w \cdot \frac{1}{\sqrt{1-\left(\frac{w}{z}\right)^2}} \cdot \left(\frac{1}{z}\right)\]Thus,\[\frac{\partial F}{\partial w} = \sin^{-1}\left(\frac{w}{z}\right) + \frac{w}{z\sqrt{z^2 - w^2}}\]

Differentiate with Respect to \(z\)

To find \( \frac{\partial F}{\partial z} \), treat \( w \) as a constant:\[\frac{\partial F}{\partial z} = \frac{d}{dz}\left( w \sin^{-1}\left(\frac{w}{z}\right) \right)\]This becomes the product of a constant with the derivative of the inverse sine function:\[= w \cdot \frac{d}{dz}\left(\sin^{-1}\left(\frac{w}{z}\right)\right)\]\[= w \cdot \frac{1}{\sqrt{1-\left(\frac{w}{z}\right)^2}} \cdot \left(-\frac{w}{z^2}\right)\]Simplifying this gives:\[\frac{\partial F}{\partial z} = -\frac{w^2}{z^2\sqrt{z^2 - w^2}}\]

Verify the Results

Reassess the calculations to ensure all derivative rules were applied correctly and verify that none of the variables were treated incorrectly. Both partial derivatives now reflect how the function \( F(w, z) \) changes with respect to \( w \) and \( z \).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are incredibly useful in calculus and beyond. These functions help us determine angles when trigonometric function values are known. In this exercise, we see the inverse sine function, denoted as \( \sin^{-1}(x) \). It's commonly known as arcsine.
It's important to understand how the derivative of inverse trigonometric functions works. For the arcsine function, \( \sin^{-1}(x) \), the derivative is \( \frac{1}{\sqrt{1-x^2}} \). This derivative, however, may need further chain rule applications.

• When differentiating \( \sin^{-1}(u) \), where \( u \) is a function of another variable, you use the chain rule.
• This involves multiplying by \( \frac{du}{dx} \) where \( u = f(x) \).
• In our solution, \( u = \frac{w}{z} \) turning our final derivative into \( \frac{1}{\sqrt{1-(\frac{w}{z})^2}} \cdot \frac{1}{z} \).

Understanding and simplifying these derivatives correctly ensures that we grasp how functions change, even within compositions involving inverse trigonometry.

Product Rule

The product rule is essential in calculus when differentiating products of two or more functions. If you have a function in the form \( u \cdot v \), its derivative with respect to \( x \) is \( \frac{du}{dx} \, v + u \, \frac{dv}{dx} \). In simple terms, you differentiate one function while keeping the other constant, then switch and add both results together.
This is exactly what we use in the problem with \( F(w,z) = w \cdot \sin^{-1}\left(\frac{w}{z}\right) \).

• Differentiating this with respect to \( w \) involves applying the product rule.
• The first function, \( w \), differentiates to 1, while the second function is treated as constant, leaving us with \( \sin^{-1}\left(\frac{w}{z}\right) \).
• In the second part, the inverse sine derivative is calculated while \( w \) is constant, and multiplied to give the additional term.

The product rule also applies similarly when treating \( w \) as constant, differentiating with respect to \( z \). Mastery of the product rule leads to accurate problem-solving in scenarios involving products of functions.

Calculus

Calculus is the mathematical study of continuous change and forms the backbone of many scientific disciplines. It's often divided into differential and integral calculus. In the given problem, we're mainly dealing with differential calculus. Here, the focus is on how functions change with respect to their variables, through differentiation.
Partial derivatives, like the ones we calculated here, are fundamental in multivariable calculus. They show how a function changes as one variable changes while others are held constant. This is immensely useful in real-world problems where multiple factors are at play. Knowing the partial derivative gives insight into the rate of change in multidimensional spaces.

• If you're navigating a multi-faceted problem, partial derivatives help determine impacts of single factors.
• They form the basis for gradient vectors and are used extensively in optimization problems.

Understanding these core concepts of calculus is critical, as they provide deep insights and solutions in diverse fields, from engineering to economics. It's not just about solving for values, but understanding the terrain of changes each variable induces.