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Calculate \(\frac{d^{2} y}{d x^{2}}\). \(y=\frac{1}{x}-\ln x\)

Short Answer

Expert verified
The second derivative of \(y = \frac{1}{x} - \ln x\) with respect to \(x\) is \(\frac{d^2 y}{d x^2} = \frac{x^2 + 2x^3}{x^4}\).

Step by step solution

01

Calculate the first derivative \(y'\)

To find the first derivative of \(y\) with respect to \(x\), we apply the rules of differentiation: for \(\frac{d}{dx}(\frac{1}{x}) \) and \(\frac{d}{dx}(\ln x)\) separately. Recall that \(\frac{d}{dx} (\frac{1}{x}) = -\frac{1}{x^2}\) and \(\frac{d}{dx}(\ln x) = \frac{1}{x}\). Thus, \[y' = -\frac{1}{x^2} - \frac{1}{x}.\]
02

Simplify \(y'\)

To make the next steps easier, we can simplify the first derivative by writing both terms with the same denominator: \[y' = -\frac{1}{x^2} - \frac{1}{x} = -\frac{1}{x^2} - \frac{x}{x^2} = -\frac{1+x}{x^2}.\]
03

Calculate the second derivative \(\frac{d^2 y}{d x^2}\)

Now, we find the second derivative by taking the derivative of the simplified first derivative with respect to \(x\). We do this by using the quotient rule, which states that: \[\frac{d}{dx} (\frac{u}{v}) = \frac{v u' - u v'}{v^2},\] where \(u\) and \(v\) are functions of \(x\). In our case, we have \(u = -(1+x)\) and \(v = x^2\). The derivatives of \(u\) and \(v\) are: \( u' = -\frac{d(1+x)}{dx} = -1\) \( v' = \frac{d}{dx}(x^2) = 2x\) Now we apply the quotient rule: \[\frac{d^2 y}{d x^2} = \frac{x^2 (-1) - (-(1+x)(2x)}{(x^2)^2}.\]
04

Simplify the second derivative

Finally, we simplify the expression for the second derivative: \[\frac{d^2 y}{d x^2} = \frac{-x^2 + (2x^2 + 2x^3)}{x^4} = \frac{-x^2 + 2x^2 + 2x^3}{x^4} = \frac{x^2 + 2x^3}{x^4}.\] Now the expression for the second derivative is simplified and we have: \[\frac{d^2 y}{d x^2} = \frac{x^2 + 2x^3}{x^4}.\]

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Key Concepts

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

Differentiation Rules
Differentiation is a fundamental tool in calculus that allows us to study how a function changes as its input changes. There are several key rules that help us find derivatives efficiently, and these are known as differentiation rules. Some basic ones include the power rule, product rule, and chain rule.

In our exercise, we're using some of these rules to differentiate the function \( y = \frac{1}{x} - \ln x \). To find the first derivative, we apply the following rules:
  • The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \) by the power rule.
  • The derivative of \( \ln x \) is \( \frac{1}{x} \), derived from the properties of logarithms.
Combining these results, we have the first derivative of \( y \): \( y' = -\frac{1}{x^2} - \frac{1}{x} \). Understanding these foundational rules sets the stage for more complex differentiation processes, such as higher order derivatives.
Quotient Rule
When taking derivatives of expressions involving division, we use the quotient rule. This rule is vital in our exercise when we find the second derivative of the function. The quotient rule states:

\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2} \]

where \( u \) and \( v \) are differentiable functions of \( x \). In this case, when we calculate \( \frac{d^2 y}{d x^2} \), we set \( u = -(1+x) \) and \( v = x^2 \).
  • The derivative of \( u \), \( u' \), is \(-1\).
  • The derivative of \( v \), \( v' \), is \( 2x \).
Plugging these into the quotient rule formula helps us find the second derivative, which is then simplified in subsequent steps. Mastery of the quotient rule is crucial for tackling complex algebraic fractions and ensuring precision in higher-order differentiation.
Simplifying Derivatives
Simplification is an often overlooked but essential part of working with derivatives. To make algebraic expressions more manageable and the calculations clearer, we simplify the derivatives whenever possible.

In our exercise, once we found the first derivative \( y' = -\frac{1}{x^2} - \frac{1}{x} \), we simplified it to \( -\frac{1 + x}{x^2} \). This step sets us up for a cleaner application of the quotient rule.

After calculating the second derivative while using the quotient rule, we simplify further as follows:
  • The expression is \( \frac{-x^2 + 2x^2 + 2x^3}{x^4} \).
  • Combine like terms to get \( \frac{x^2 + 2x^3}{x^4} \).
This final simplification makes the derivative expression neater and more comprehensible. Simplifying derivatives not only reduces computational work but also enhances understanding by revealing the essential parts of the function.

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Most popular questions from this chapter

The consumer demand equation for tissues is given by \(q=(100-p)^{2}\), where \(p\) is the price per case of tissues and \(q\) is the demand in weekly sales. a. Determine the price elasticity of demand \(E\) when the price is set at \(\$ 30\), and interpret your answer. b. At what price should tissues be sold in order to maximize the revenue? c. Approximately how many cases of tissues would be demanded at that price?

The demand and unit price for your store's checkered T-shirts are changing with time. Show that the percentage rate of change of revenue equals the sum of the percentage rates of change of price and demand. (The percentage rate of change of a quantity \(Q\) is \(\left.Q^{\prime}(t) / Q(t) .\right)\)

Complete the following: If the graph of a function is concave up on its entire domain, then its second derivative is ________ on the domain.

The average cost function for the weekly manufacture of portable CD players is given by \(\bar{C}(x)=150,000 x^{-1}+20+0.0001 x\) dollars per player, where \(x\) is the number of CD players manufactured that week. Weekly production is currently 3,000 players and is increasing at a rate of 100 players per week. What is happening to the average cost? HINT [See Example 3.]

Demand for your tie-dyed T-shirts is given by the formula $$q=500-100 p^{0.5}$$ where \(q\) is the number of T-shirts you can sell each month at a price of \(p\) dollars. If you currently sell T-shirts for \(\$ 15\) each and you raise your price by \(\$ 2\) per month, how fast will the demand drop? (Round your answer to the nearest whole number.)

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