Problem 43 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=x^{2} \sqrt{1+x^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{2x + x^3}{(1+x^2)^{1/2}} \).

Step 1: Identify the product structure

The function is written as a product: \[ f(x) = x^2 \cdot (1 + x^2)^{1/2}. \]Here we recognize two functions: \( u(x) = x^2 \) and \( v(x) = (1+x^2)^{1/2} \). We will use the product rule to differentiate, which states \[ (uv)' = u'v + uv'. \]

Step 2: Differentiate the first function, \( u(x) \)

Differentiate \( u(x) = x^2 \). The derivative is found using the power rule: \[ u'(x) = \frac{d}{dx}(x^2) = 2x. \]

Step 3: Differentiate the second function, \( v(x) \)

Differentiate \( v(x) = (1+x^2)^{1/2} \) using the Generalized Power Rule. Recall the Generalized Power Rule states \[ \frac{d}{dx}(g(x)^n) = n \cdot g(x)^{n-1} \cdot g'(x). \]Here, \( n = \frac{1}{2} \) and \( g(x) = 1 + x^2 \), thus \[ g'(x) = 2x. \]So, \[ v'(x) = \frac{1}{2}(1 + x^2)^{-1/2} \cdot 2x = \frac{x}{(1+x^2)^{1/2}}. \]

Step 4: Apply the product rule

Use the product rule to find \( f'(x) \):\[ f'(x) = u'v + uv'. \]Substituting the differentiated parts:\[ f'(x) = (2x)(1 + x^2)^{1/2} + (x^2)\left(\frac{x}{(1+x^2)^{1/2}}\right). \]

Step 5: Simplify the expression

Simplify the expression:1. For the first term: \( 2x(1 + x^2)^{1/2} \).2. For the second term: \( x^3/(1+x^2)^{1/2} \).Combine over a common denominator:\[ f'(x) = \frac{2x(1+x^2) + x^3}{(1+x^2)^{1/2}} = \frac{2x + x^3}{(1+x^2)^{1/2}}. \]

Key Concepts

Product Rule in Calculus

When dealing with derivatives, the product rule is an essential tool. It helps us differentiate functions that are multiplied together. In our example, the function was given as a product of two parts: \( u(x) = x^2 \) and \( v(x) = (1+x^2)^{1/2} \). The product rule formula is \( (uv)' = u'v + uv' \), which says the derivative of a product is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

This approach ensures each part of the product is taken into account, allowing us to find a complete derivative.

• The first step is identifying each function involved in the product.
• Next, differentiate each function separately.
• Finally, plug the results into the product rule formula.

This method simplifies the process of differentiation when dealing with multiple functions that multiply together.

Understanding Derivatives

A derivative is a key concept in calculus that represents the rate at which a function changes at a given point. It provides a measure of how a function responds to changes in its input.

In simple terms, if you imagine a curve on a graph, the derivative at any point tells you the slope of the tangent line at that point. In our exercise, we first found derivatives by using basic differentiation rules.

• Use the power rule for basic powers of \( x \).
• Apply the product rule for products of functions.
• Apply generalized rules for more complex forms.

By understanding these initial rules, you can tackle more complicated functions using combinations of these basic derivatives.

The Role of Calculus

Calculus is a branch of mathematics focused on change. It's the mathematics of motion and change, and it paves the way for advanced studies in engineering, physics, economics, and beyond.

Two of the main concepts in calculus are differentiation and integration, which are inverse operations. Differentiation breaks down complex changes into simpler, more manageable pieces—just like the derivative we calculated.

• Use calculus to understand and predict change.
• Differentiate to find instantaneous rates of change.
• Perform integration to accumulate small increments.

This exercise and its solution show how we can use calculus, especially differentiation, to analyze how functions behave by finding their derivatives.