Problem 19 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ y=x^{3}-3 \sqrt{x} $$

Step-by-Step Solution

Answer

\(dy = (3x^2 - \frac{3}{2}x^{-1/2}) \, dx\).

Step 1: Identify the Derivative

The first task in finding the differential \(dy\) is to find the derivative of the function \(y = x^3 - 3\sqrt{x}\) with respect to \(x\). Let's represent this derivative as \(\frac{dy}{dx}\).

Step 2: Differentiate Each Term Separately

To differentiate, break the function into two terms: \(x^3\) and \(-3\sqrt{x}\).1. The derivative of \(x^3\) is \(3x^2\).2. The derivative of \(-3\sqrt{x}\) is \(-3 \cdot \frac{1}{2}x^{-1/2}\), which simplifies to \(-\frac{3}{2}x^{-1/2}\).

Step 3: Combine Derivatives

Now combine the derivatives of each term:\[ \frac{dy}{dx} = 3x^2 - \frac{3}{2}x^{-1/2} \]

Step 4: Express \(dy\)

To find \(dy\), multiply \(\frac{dy}{dx}\) by \(dx\):\[ dy = \left(3x^2 - \frac{3}{2}x^{-1/2}\right) dx \]

Key Concepts

Understanding Derivatives

Derivatives are at the heart of differential calculus. They measure how a function changes as its input changes. Imagine driving a car: the derivative is like your speedometer, telling you how fast your speed (function) changes with time (input). In mathematical terms, it's the rate of change or the slope of the curve at any given point.
To find the derivative, we use a process called differentiation. Here, functions are often broken down into basic terms that can be easily handled using rules of differentiation. For example:

• The Power Rule tells us that for any term like \(x^n\), its derivative is \(nx^{n-1}\).
• Multiplying constants are carried along when finding the derivative, just like we did with \(-3\sqrt{x}\), which simplifies using the power rule to \(-\frac{3}{2}x^{-1/2}\).

Each of these rules makes finding derivatives as straightforward as applying the formula to each part of your function separately.

The Process of Differentiation

Differentiation is the mathematical process used to find a derivative. It involves applying derivative rules to each term of a function to determine its rate of change. When differentiating, it's essential to break down the function into simpler parts that you can tackle one at a time.
Let's look at the exercise where we differentiate \(y = x^3 - 3\sqrt{x}\):

• First, we recognize each part: \(x^3\) and \(-3\sqrt{x}\).
• Next, apply differentiation rules to these terms: the Power Rule for \(x^3\) gives \(3x^2\), and for \(-3\sqrt{x}\), rewriting it as \(-3x^{1/2}\), it gives \(-\frac{3}{2}x^{-1/2}\).

Each term is handled separately and then combined back, resulting in the derivative \(\frac{dy}{dx} = 3x^2 - \frac{3}{2}x^{-1/2}\). This showcases the simplicity and power of differentiation when applied step by step.

Introduction to the Differential

In calculus, the differential (often noted as \(dy\)) is used to signify an infinitesimally small change in a function's output corresponding to a small change in its input (\(dx\)). It represents a slight modification in \(y\) due to a slight alteration in \(x\).
For practical purposes, when you find \(dy\), you multiply the derivative \(\frac{dy}{dx}\) by \(dx\) itself. In our exercise, after differentiating to get \(3x^2 - \frac{3}{2}x^{-1/2}\), we multiply by \(dx\) to find \(dy\):

• \(dy = \left(3x^2 - \frac{3}{2}x^{-1/2}\right) dx\)

The differential \(dy\) thus represents how even the tiniest shifts in \(x\) affect \(y\). It's a key concept for understanding how changes propagate through functions, leading to deeper applications like linear approximations and error estimations.