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Chapter 23: Problem 33

Evaluate the second derivative of the given function for the given value of \(x\). $$y=3 x^{2 / 3}-\frac{2}{x}, x=-8$$

Short Answer

Expert verified
The second derivative evaluated at \( x = -8 \) is \(-\frac{13}{384}\).

Step by step solution

01

Find the First Derivative

First, we need to differentiate the function \( y = 3x^{2/3} - \frac{2}{x} \). Use the power rule and the derivative of a fraction for this. The derivative of \( x^{2/3} \) is \( \frac{2}{3}x^{-1/3} \) and of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Therefore, the first derivative is \( y' = 3 \cdot \frac{2}{3}x^{-1/3} + 2x^{-2} \), which simplifies to \( y' = 2x^{-1/3} + \frac{2}{x^2} \).
02

Simplify the First Derivative

Simplify the expression \( y' = 2x^{-1/3} + \frac{2}{x^2} \). This gives us \( y' = 2x^{-1/3} + 2x^{-2} \). It is already in a simplified, manageable form for further differentiation.
03

Find the Second Derivative

Differentiate \( y' = 2x^{-1/3} + 2x^{-2} \) to find the second derivative. For \( 2x^{-1/3} \), the derivative is \( -\frac{2}{3}x^{-4/3} \). For \( 2x^{-2} \), the derivative is \( -4x^{-3} \). Thus, the second derivative \( y'' = -\frac{2}{3}x^{-4/3} - 4x^{-3} \).
04

Evaluate the Second Derivative at \( x = -8 \)

Substitute \( x = -8 \) into the second derivative \( y'' = -\frac{2}{3}x^{-4/3} - 4x^{-3} \). Calculate \( (-8)^{-4/3} \) and \( (-8)^{-3} \). We have \( (-8)^{-4/3} = \left(\frac{1}{(-8)^{4/3}}\right) \) and as \( (-8)^{1/3} = -2 \), \((-8)^{4/3} = 16 \), so \( (-8)^{-4/3} = \frac{1}{16} \). For \( (-8)^{-3} = -\frac{1}{512} \). Substitute back to get \( y'' = -\frac{2}{3} \cdot \frac{1}{16} - 4 \cdot \left(-\frac{1}{512}\right) \). Simplify to \( y'' = -\frac{1}{24} + \frac{1}{128} \).
05

Simplify Final Expression

To combine \(-\frac{1}{24}\) and \(\frac{1}{128}\), find a common denominator, which is 384. Convert terms: \(-\frac{1}{24} = -\frac{16}{384}\) and \(\frac{1}{128} = \frac{3}{384}\). Combine them to get \(-\frac{16}{384} + \frac{3}{384} = -\frac{13}{384}\). Thus, \( y''(-8) = -\frac{13}{384} \).

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Key Concepts

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

Power Rule
The power rule is a fundamental tool in calculus used for differentiating functions with terms raised to a power. It's very useful when you have expressions of the form \( x^n \). The rule states that you bring down the exponent as a coefficient in front and then subtract one from the original exponent. Let's break it down into steps:
  • If you have \( x^n \), you multiply \( n \) by the term, resulting in \( n \cdot x^{n-1} \).
  • This simplifies the process of differentiation, allowing you to find the derivative easily without dealing with complex algebra.
In the given exercise, you applied the power rule to terms like \( 3x^{2/3} \), transforming it by multiplying the coefficient 3 by \( \frac{2}{3} \) and reducing the power by one, leading to a new term \( \frac{2}{3} x^{-1/3} \). This process of simplifying through the power rule makes handling derivatives straightforward!
Differentiation
Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to a variable. It鈥檚 like finding how fast or slow a function changes at any point.
  • To differentiate a function, you take each part of the expression and apply differentiation rules to transform it into a derivative.
  • In simple terms, differentiation converts a function into something that tells you about the slopes or changes of that function along the x-axis.
For example, in our exercise, we differentiated the function \( y = 3x^{2/3} - \frac{2}{x} \) to get the first derivative \( y' = 2x^{-1/3} + \frac{2}{x^2} \). This first derivative gives us a new function that describes the slope of the tangent line to the curve at any point \( x \). Differentiating again gives us the second derivative, which indicates how the slope itself is changing.
Function Evaluation
Function evaluation is about determining the value of a function at a specific point by substituting a number for the variable. This is especially useful after you have found a derivative and need to understand what it means at a particular x-value.
  • You substitute the given x-value into your function or derivative and simplify to find the result.
  • Evaluating the second derivative, as we did for \( y'' = -\frac{2}{3}x^{-4/3} - 4x^{-3} \) at \( x = -8 \), gives us insight into the curvature or concavity of the function at that point.
In our example, substituting \( x = -8 \) involves computing powers of negative numbers and manipulating fractions until you find a clear numerical result. The careful evaluation helps in understanding the nature of the graph at specific points and can reveal local maxima, minima, or points of inflection.
Common Denominators
Finding common denominators is a crucial step in combining fractions. When performing operations like addition or subtraction on fractions with different denominators, you need a common denominator to combine them into a single fraction.
  • To find a common denominator, identify the least common multiple (LCM) of the denominators involved.
  • Rewrite each fraction with this new denominator, adjusting the numerators accordingly.
For instance, in simplifying \( -\frac{1}{24} + \frac{1}{128} \), the common denominator is 384. Rewriting \( -\frac{1}{24} \) as \( -\frac{16}{384} \) and \( \frac{1}{128} \) as \( \frac{3}{384} \), allows these fractions to be combined easily. The final result, \( -\frac{13}{384} \), is a straightforward addition thanks to matching denominators, concluding our function evaluation process.

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Most popular questions from this chapter

Find the second derivative of each of the given functions. $$x^{2}-4 y^{2}=9$$

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