Problem 20 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ y=x \sqrt{1-x^{2}} $$

Step-by-Step Solution

Answer

\(dy = \frac{1-2x^2}{\sqrt{1-x^2}} dx\)

Step 1: Apply the Product Rule

The given function is a product of two separate functions: \(y = x \cdot \sqrt{1 - x^2}\). To differentiate this, we will apply the product rule, which states that if you have two functions multiplied together, \(f(x) = u(x) \cdot v(x)\), then the derivative \(dy/dx = u'v + uv'\). Here, \(u = x\) and \(v = \sqrt{1 - x^2}\). First, we need to find the derivatives of \(u\) and \(v\).

Step 2: Differentiate \(u = x\)

The derivative of \(u = x\) with respect to \(x\) is straightforward: \(u' = 1\).

Step 3: Differentiate \(v = \sqrt{1-x^2}\) using the Chain Rule

For \(v = \sqrt{1-x^2}\), rewrite it as \((1-x^2)^{1/2}\). Now apply the chain rule. The outer function is \((1-x^2)^{1/2}\), and its derivative is \(\frac{1}{2}(1-x^2)^{-1/2}\). The inner function is \(1-x^2\) and its derivative is \(-2x\). Thus, \(v' = \frac{1}{2}(1-x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}\).

Step 4: Apply the Product Rule to Find \(dy/dx\)

Now substitute \(u'\), \(v\), \(u\), and \(v'\) into the product rule formula: \(dy/dx = u'v + uv'\). So, \(dy/dx = (1)(\sqrt{1-x^2}) + x \left(\frac{-x}{\sqrt{1-x^2}}\right)\). Simplify this to \(dy/dx = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\).

Step 5: Simplify the Expression

To simplify \(dy/dx = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\), find a common denominator. The common denominator is \(\sqrt{1-x^2}\), so rewrite it as \(dy/dx = \frac{(1-x^2) - x^2}{\sqrt{1-x^2}} = \frac{1-2x^2}{\sqrt{1-x^2}}\).

Step 6: Multiply by \(dx\) to Find \(dy\)

Since we need to find \(dy\), multiply \(dy/dx\) by \(dx\): \(dy = \frac{1-2x^2}{\sqrt{1-x^2}} dx\). This is the expression for \(dy\).

Key Concepts

Product Rule

In the world of differentiation, the product rule is a vital tool for finding the derivative of the product of two functions. Consider a function that is expressed as the product of two separate functions, like our original function, \( y = x \cdot \sqrt{1-x^2} \). Here, it's important to treat each part of the product as distinct.

• Definition: If you have a product of two functions, say \( f(x) = u(x) \cdot v(x) \), the product rule allows you to find its derivative using: \( \frac{dy}{dx} = u'v + uv' \).
• Applying the Rule: In the exercise, \( u(x) = x \) and \( v(x) = \sqrt{1-x^2} \). To differentiate, compute each of their derivatives separately and then apply the formula.

This approach helps break down more complex expressions into manageable parts, ultimately making differentiation feasible.

Chain Rule

When differentiating composite functions, the chain rule becomes a key strategy. This rule is specifically helpful for functions nested within other functions, such as \( v = \sqrt{1 - x^2} \) in our exercise.

• Understanding the Rule: The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). This means you first differentiate the outer function and then multiply by the derivative of the inner function.
• Applying to \( v = \sqrt{1-x^2} \): Rewrite it as \((1-x^2)^{1/2}\). First, differentiate \((1-x^2)^{1/2}\) with respect to \(1-x^2\), and then multiply by the derivative of \(1-x^2\) with respect to \(x\). This gives us \(v' = \frac{-x}{\sqrt{1-x^2}}\), effectively combining the derivatives of the layers of functions.

By mastering the chain rule, students can handle multi-layered functions with greater ease, turning what seems complex into a series of simpler differentiations.

Derivative Simplification

After applying differentiation rules, expressions can be quite complex, so simplification is crucial. In the given exercise, we see this in the step where \(dy/dx\) is simplified.

• Combining Terms: The initial derivative obtained was \(dy/dx = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\). The goal here is to simplify this expression by resulting in a single fraction.
• Finding a Common Denominator: Because both terms had \(\sqrt{1-x^2}\) in them, gathering them under this common denominator results in \(dy/dx = \frac{(1-x^2) - x^2}{\sqrt{1-x^2}}\), which simplifies to \(\frac{1-2x^2}{\sqrt{1-x^2}}\).

Simplification not only makes the expression manageable but also prepares it for further calculations such as integration or evaluations. It’s an essential step in obtaining a practical and usable derivative.

Differentiation Techniques

Differentiation involves a variety of techniques suited to different types of functions and expressions. Understanding and correctly applying these ensures accurate results in calculus.

• Basic Derivatives: Start with basic derivatives such as \(\frac{d}{dx}(x) = 1\) or \(\frac{d}{dx}(x^n) = nx^{n-1}\). These are the building blocks for more advanced methods.
• Product Rule: As discussed, applies when a function is a product of two or more sub-functions. It allows differentiation of each component separately before recombining results.
• Chain Rule: Essential for nested functions where one function is defined within another. This technique is particularly useful for trigonometric and exponential functions as well.
• Practical Application: Often, real-world problems require employing multiple techniques simultaneously to find the derivative of complex expressions. Understanding the circumstances for each technique is crucial.

By familiarizing oneself with these tools, any student can confidently approach a wide array of differentiating problems, applying the right rule or combination for efficient solutions.