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Chapter 2: Problem 43

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

Short Answer

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

Step by step solution

01

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 \[ (uv)' = u'v + uv'. \]
02

Differentiate the first function, \( u(x) \)

Differentiate \( u(x) = x^2 \). The derivative is found using the power rule: \[ u'(x) = \frac{d}{dx}(x^2) = 2x. \]
03

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, \[ v'(x) = \frac{1}{2}(1 + x^2)^{-1/2} \cdot 2x = \frac{x}{(1+x^2)^{1/2}}. \]
04

Apply the product rule

Use the product rule to find \( f'(x) \):\[ f'(x) = u'v + uv'. \]Substituting the differentiated parts:\[ f'(x) = (2x)(1 + x^2)^{1/2} + (x^2)\left(\frac{x}{(1+x^2)^{1/2}}\right). \]
05

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:\[ f'(x) = \frac{2x(1+x^2) + x^3}{(1+x^2)^{1/2}} = \frac{2x + x^3}{(1+x^2)^{1/2}}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攋ust 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.

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Most popular questions from this chapter

For each function: a. Find \(f^{\prime}(x)\) using the definition of the derivative. b. Explain, by considering the original function, why the derivative is a constant. $$ f(x)=2 x-9 $$

Learning Theory In a psychology experiment, a person could memorize \(x\) words in \(f(x)=2 x^{2}-x\) seconds (for \(0 \leq x \leq 10)\). a. Find \(f^{\prime}(x)\) by using the definition of the derivative. b. Find \(f^{\prime}(5)\) and interpret it as an instantaneous rate of change in the proper units.

Derive the Quotient Rule from the Product Rule as follows. a. Define the quotient to be a single function, $$ Q(x)=\frac{f(x)}{g(x)} $$ b. Multiply both sides by \(g(x)\) to obtain the equation \(Q(x) \cdot g(x)=f(x) .\) c. Differentiate each side, using the Product Rule on the left side. d. Solve the resulting formula for the derivative \(Q^{\prime}(x) .\) e. Replace \(Q(x)\) by \(\frac{f(x)}{g(x)}\) and show that the resulting formula for \(Q^{\prime}(x)\) is the same as the Quotient Rule. Note that in this derivation when we differentiated \(Q(x)\) we assumed that the derivative of the quotient exists, whereas in the derivation on pages \(135-136\) we proved that the derivative exists.

Find the equation for the tangent line to the curve \(y=f(x)\) at the given \(x\) -value. $$ f(x)=(2 x+1)^{4} \text { at } x=-1 $$

The percentage of people in the United States who are immigrants (that is, were born elsewhere) for different decades is shown below. These percentages are approximated by the function \(f(x)=\frac{1}{2} x^{2}-3.7 x+12\), where \(x\) stands for the number of decades since 1930 (so that, for example, \(x=5\) would stand for 1980 ). a. Find \(f^{\prime}(x)\) using the definition of the derivative b. Evaluate the derivative at \(x=1\) and interpret the result. c. Find the rate of change of the immigrant percentage in the year 2010 .
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