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Chapter 12: Problem 4

Calculate \(\frac{d^{2} y}{d x^{2}}\). \(y=-\frac{2}{x^{2}}\)

Short Answer

Expert verified
The second derivative of the given function is \(\boxed{\frac{d^2 y}{d x^2} = -12x^{-4}}\).

Step by step solution

01

Calculate the first derivative.

To calculate the first derivative, we will apply the power rule for differentiation, which states that if \(y=kx^n\) then \(\frac{dy}{dx}=knx^{n-1}\). Here, \(k=-2\) and \(n=-2\). So, applying the power rule, we get: \[\frac{dy}{dx} = (-2)(-2)x^{-2-1} = 4x^{-3}\] Now that we have the first derivative, we can move on to calculating the second derivative.
02

Calculate the second derivative.

To calculate the second derivative, we will differentiate the first derivative, \(\frac{dy}{dx} = 4x^{-3}\), with respect to \(x\). Again, we will apply the power rule for differentiation. In this case, \(k=4\) and \(n=-3\). So, applying the power rule, we get: \[\frac{d^2y}{dx^2} = (4)(-3)x^{-3-1} = -12x^{-4}\]
03

Final answer.

The second derivative of the given function is: \[\boxed{\frac{d^2 y}{d x^2} = -12x^{-4}}\]

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Key Concepts

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

Understanding the Power Rule
The power rule is a fundamental tool in differentiation that allows us to find the derivative of a function that involves powers of a variable, such as a polynomial. It simplifies the process of finding derivatives, making it quicker and more straightforward. The rule states that if you have a function of the form \(y = kx^n\), where \(k\) is a constant and \(n\) is a real number, the derivative of this function with respect to \(x\) is \(\frac{dy}{dx} = knx^{n-1}\). Envision this as reducing the power by one and multiplying by the original exponent and constant.
  • In our exercise, we first have \(y = -\frac{2}{x^2}\), which can be rewritten as \(y = -2x^{-2}\) for easier application of the power rule.
  • Applying the power rule to \(y = -2x^{-2}\), we calculate the first derivative: \(\frac{dy}{dx} = 4x^{-3}\).
  • Repeating the application of the power rule to the first derivative gives us the second derivative: \(\frac{d^2y}{dx^2} = -12x^{-4}\).
Using this rule helps in efficiently handling derivatives, especially when dealing with polynomial expressions.
The Process of Differentiation
Differentiation is the process used in calculus to find the rate at which a function is changing at any given point. It provides the gradient or slope of a function by calculating its derivative. This is a crucial concept, as it reveals how output values change with respect to changes in input values. When differentiating functions, especially polynomial functions, the power rule becomes incredibly handy.
  • Differentiating involves two main steps for our example:
  • First, we differentiate \(y = -2x^{-2}\) using the power rule, resulting in \(\frac{dy}{dx} = 4x^{-3}\).
  • Next, we differentiate the derivative \(\frac{dy}{dx} = 4x^{-3}\) to find the second derivative \(\frac{d^2y}{dx^2} = -12x^{-4}\).
This method highlights the systematic approach of differentiation, which involves taking derivatives step-by-step.
An Introduction to Calculus
Calculus is a branch of mathematics focused on the concepts of limits, functions, derivatives, integrals, and infinite series. It is split into two main parts: differential calculus and integral calculus. Differential calculus concerns itself with the rate of change and slopes of curves, which is where differentiation comes in. It's used extensively in fields like physics, engineering, economics, statistics, and more.
  • The second derivative, \(\frac{d^2y}{dx^2}\), is a part of differential calculus and tells us about the concavity of the function and points of inflection.
  • In our problem, by reaching the second derivative, we understand not just the slope but how the rate of change itself is changing.
  • This ability to quantify changes and predict future values is why calculus, particularly differentiation, is so powerful.
Understanding calculus opens numerous doors to solving complex problems in various domains by providing the tools to model and analyze dynamic systems.

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Most popular questions from this chapter

A fried chicken franchise finds that the demand equation for its new roast chicken product, "Roasted Rooster," is given by $$p=\frac{40}{q^{1.5}}$$ where \(p\) is the price (in dollars) per quarter-chicken serving and \(q\) is the number of quarter-chicken servings that can be sold per hour at this price. Express \(q\) as a function of \(p\) and find the price elasticity of demand when the price is set at \(\$ 4\) per serving. Interpret the result.

A production formula for a student's performance on a difficult English examination is given by $$g=4 h x-0.2 h^{2}-10 x^{2}$$ where \(g\) is the grade the student can expect to obtain, \(h\) is the number of hours of study for the examination, and \(x\) is the student's grade point average. The instructor finds that students' grade point averages have remained constant at \(3.0\) over the years, and that students currently spend an average of 15 hours studying for the examination. However, scores on the examination are dropping at a rate of 10 points per year. At what rate is the average study time decreasing?

Given that the demand \(q\) is a differentiable function of income \(x\), show that the quantity \(R=q / x\) has a stationary point when $$q-x \frac{d q}{d x}=0$$ Deduce that stationary points of \(R\) are the same as the points of unit income elasticity of demand. HINT [Differentiate \(R\) with respect to \(x\).]

The area of a circular sun spot is growing at a rate of \(1,200 \mathrm{~km}^{2} / \mathrm{s}\) a. How fast is the radius growing at the instant when it equals \(10,000 \mathrm{~km}\) ? HINT [See Example 1.] b. How fast is the radius growing at the instant when the sun spot has an area of \(640,000 \mathrm{~km}^{2} ?\) HINT [Use the area formula to determine the radius at that instant.]

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where \(p\) is the price per book, \(q\) is the demand in annual sales, and \(m\) and \(b\) are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at \(\$ 50.00\) per book, the sales amount to 10,000 per year; when the price is set at \(\$ 80.00\) per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.
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