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Chapter 2: Problem 114

Use the result of Exercise 112 to find the derivative of the function. $$ g(x)=x\left|x^{2}+x\right| $$

Short Answer

Expert verified
The derivative of the function \(g(x) = x|x^2 + x|\) is given by: \[ g'(x) = \begin{cases} 3x^2 + 2x, & \text{if } x^2 + x \geq 0 \\ -3x^2 - 2x, & \text{if } x^2 + x < 0 \end{cases} \]

Step by step solution

01

Define the two cases based on absolute value expression

Firstly, we need to distinguish between the two cases based on the absolute value expression: Case 1: When \(x^2 + x \geq 0\), the function is \(g(x) = x(x^2 + x)\). Case 2: When \(x^2 + x < 0\), the function is \(g(x) = x(-x^2 - x)\).
02

Find the derivative for Case 1

We will now find the derivative of the function for Case 1 (\(x^2 + x \geq 0\)): \(g(x) = x(x^2 + x)\) Using the product rule, we have: \(g'(x) = (1).(x^2 + x) + x(2x + 1)\) Simplifying, we get: \(g'(x) = x^2 + x + 2x^2 + x\) Combining like terms, we have: \(g'(x) = 3x^2 + 2x\) So, for Case 1, the derivative is \(g'(x) = 3x^2 + 2x\).
03

Find the derivative for Case 2

We will now find the derivative of the function for Case 2 (\(x^2 + x < 0\)): \(g(x) = x(-x^2 - x)\) Using the product rule, we have: \(g'(x) = (1).(-x^2 - x) + x(-2x - 1)\) Simplifying, we get: \(g'(x) = -x^2 - x - 2x^2 - x\) Combining like terms, we have: \(g'(x) = -3x^2 - 2x\) So, for Case 2, the derivative is \(g'(x) = -3x^2 - 2x\).
04

Combine the results of Step 2 and Step 3

Now that we have found the derivatives for both cases, we can combine them to get the final derivative: If \(x^2 + x \geq 0\), then \(g'(x) = 3x^2 + 2x\), If \(x^2 + x < 0\), then \(g'(x) = -3x^2 - 2x\). The final answer is: \[ g'(x) = \begin{cases} 3x^2 + 2x, & \text{if } x^2 + x \geq 0 \\ -3x^2 - 2x, & \text{if } x^2 + x < 0 \end{cases} \]

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Key Concepts

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

Product Rule
The product rule is a fundamental concept in calculus used when you need to find the derivative of a product of two functions. It states that if you have two differentiable functions, say \(u(x)\) and \(v(x)\), the derivative of their product is given by:\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]In simpler terms, to differentiate a product of two functions, follow these steps:
  • Differentiate the first function (\(u(x)\)) and multiply by the second function (\(v(x)\)) as it is.
  • Then, differentiate the second function and multiply by the first function as it is.
  • Add the two results together.
In the context of the given exercise, for Case 1, the function was \(g(x) = x(x^2 + x)\). By applying the product rule:
  • Differentiate \(x\), which is 1, and multiply by \((x^2 + x)\).
  • Then, \((2x + 1)\) is differentiated by multiplying by \(x\).
  • Add these to get \(g'(x) = x^2 + x + 2x^2 + x\) which simplifies to \(3x^2 + 2x\).
This method was also used for the Case 2, yielding similar steps but different results as per the negative presence in the expression.
Absolute Value
The concept of absolute value is crucial for dealing with expressions that can yield positive or negative results.The absolute value \( |x| \) expresses the magnitude of \(x\) without regard to its sign, meaning:
  • If \(x\) is positive, \(|x| = x\).
  • If \(x\) is negative, \(|x| = -x\).
In calculus, dealing with absolute values often means evaluating functions differently based on the sign of a specific expression or variable.In the exercise above, \(|x^2 + x|\) creates two different cases:
  • Case 1 where \(x^2 + x \geq 0\), the absolute value does not alter the expression: \(g(x) = x(x^2 + x)\).
  • In Case 2 where \(x^2 + x < 0\), the absolute value does, resulting in \(g(x) = x(-x^2 - x)\).
Recognizing when to switch between these cases is key, especially when applying derivative rules like the product rule.
Piecewise Functions
Piecewise functions are functions that have different expressions based on the value of \(x\).These are very common when working with functions that involve absolute value, as they naturally divide the domain into separate regions.In our exercise, the piecewise function is represented by:\[g'(x) =\begin{cases}3x^2 + 2x, & \text{if } x^2 + x \geq 0 \-3x^2 - 2x, & \text{if } x^2 + x < 0\end{cases}\]This format allows us to detail which mathematical expression applies to which segment of the domain. Here's how it works:
  • For \(x\) values that make \(x^2 + x \geq 0\), substitutes into the first expression.
  • For \(x\) values where \(x^2 + x < 0\), the second expression takes over.
Understanding how to set up and solve piecewise functions is crucial, especially in calculus problems involving more complex scenarios like this one.

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Most popular questions from this chapter

A horizontal uniform beam of length \(L\) is supported at both ends and bends under its own weight \(w\) per unit length. Because of its elasticity, the beam is distorted in shape, and the resulting distorted axis of symmetry (shown dashed in the figure) is called the elastic curve. It can be shown that an equation for the elastic curve is $$y=\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right)$$ where the product \(E I\) is a constant called the flexural rigidity. (a) The distorted beam (b) The elastic curve in the \(x y\) -plane (The positive direction of the \(y\) -axis is directed downward.) a. Find the angle that the elastic curve makes with the positive \(x\) -axis at each end of the beam in terms of \(w, E\), and \(I .\) b. Show that the angle that the elastic curve makes with the horizontal at \(x=L / 2\) is zero. c. Find the deflection of the beam at \(x=L / 2\). (We will show that the deflection is maximal in Section 3.1, Exercise 74.)

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