Problem 71 - University Calculus: Early Transcendentals

Question

Find \(y^{\prime \prime}\) in Exercises \(71-78\). $$y=\left(1+\frac{1}{x}\right)^{3}$$

Step-by-Step Solution

Answer

After differentiating twice and simplifying, the second derivative is determined.

Step 1: Expand the function

First, write the function in its current form before applying any derivatives. Our function is:\[y = \left(1 + \frac{1}{x}\right)^3\].

Step 2: Differentiate to find the first derivative

Apply the chain rule to differentiate. Let \(u = 1 + \frac{1}{x}\). Then, \(y = u^3\) and \(y' = 3u^2 \cdot u'\). Now find \(u'\):Since \(u = 1 + \frac{1}{x}\), then \(u' = -\frac{1}{x^2}\). So:\[y' = 3\left(1+\frac{1}{x}\right)^2 \cdot \left(-\frac{1}{x^2}\right) = -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2\].

Step 3: Simplify the first derivative

Simplify the expression for the first derivative:\[y' = -\frac{3\left(1+\frac{1}{x}\right)^2}{x^2}\].

Step 4: Differentiate to find the second derivative

Differentiate the first derivative to find the second derivative. Use the quotient rule, where if \(v = u/v\) then \(v' = \frac{u'v - uv'}{v^2}\).Let \(u = -3\left(1+\frac{1}{x}\right)^2\) and \(v = x^2\).Firstly, find \(u'\):\(u' = -3 \cdot 2\left(1+\frac{1}{x}\right)\cdot -\frac{1}{x^2} = \frac{6\left(1+\frac{1}{x}\right)}{x^2}\).Then, \(v' = 2x\).Applying the quotient rule:\[y'' = \frac{\left(\frac{6\left(1+\frac{1}{x}\right)}{x^2}\right)x^2 - (-3\left(1+\frac{1}{x}\right)^2)2x}{(x^2)^2}\].

Step 5: Simplify the second derivative

In the second derivative expression from Step 4, simplify:Calculate each part:The numerator becomes \[ 6\left(1+\frac{1}{x}\right)x^2 + 6x\left(1+\frac{1}{x}\right)^2\].Thus, the expression for \(y''\) is simplified to:\[ y'' = \frac{6\left(1+\frac{1}{x}\right)x^2 + 6x\left(1+\frac{1}{x}\right)^2}{x^4} = \frac{6x + 12 + 6x \left(\frac{1}{x} + \frac{1}{x^2}\right)}{x^4}\].Continue simplifying to get the final form of \( y'' \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. When dealing with functions within functions, like our example of \(y = \left(1 + \frac{1}{x}\right)^3\), the chain rule helps to differentiate them effectively.

The chain rule states that if you have a function \(y = f(g(x))\), then the derivative is \(y' = f'(g(x)) \cdot g'(x)\).

• For our exercise, we first defined \(u = 1 + \frac{1}{x}\) and then differentiated \(u^3\) using the chain rule.
• This gives us \(y' = 3u^2 \cdot u'\).
• Finding \(u'\) involved differentiating \(u = 1 + \frac{1}{x}\), which indeed leads to \(-\frac{1}{x^2}\).

This shows how the chain rule simplifies the process of differentiation for nested functions.

Quotient Rule

The quotient rule is essential when you are working with derivatives of functions that are divided by one another. In our task, the first derivative \(y' = -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2\) required us to employ the quotient rule to find the second derivative.

The quotient rule states that for a quotient \(y = \frac{u}{v}\), the derivative is \(y' = \frac{u'v - uv'}{v^2}\). Applying it:

• We choose \(u = -3\left(1+\frac{1}{x}\right)^2\) and \(v = x^2\).
• We found \(u'\) and \(v'\), respectively, as \(\frac{6\left(1+\frac{1}{x}\right)}{x^2}\) and \(2x\).

Inserting these into the quotient rule helped simplify to get the second derivative, enforcing the clarity and utility of the quotient rule in differentiation.

Differentiation

Differentiation is the process of finding the derivative of a function, which describes the rate at which a function's value changes at any given point. In calculus, it is a basic and vital operation.

In our specific example, differentiation was applied twice:

• The first differentiation used the chain rule to manage the composite function, helping us get first \(y'\).
• The second application employed the quotient rule to differentiate that result further to find Second Derivative \(y''\).

The beauty of differentiation lies in its ability to depict changes, turning complex expressions into understandable derivatives, as seen in our step-by-step approach to solving the exercise.

Calculus

Calculus, a branch of mathematics focusing on the study of change, underpins many of the concepts covered in this exercise. It includes differential calculus, which deals with derivatives, like the ones we've computed here.

The goal in calculus is often to understand how functions behave and change. Whether through using the chain rule for composite functions, or the quotient rule for function divisions, calculus provides the necessary tools.

• It lets us compute derivatives, which reveal the rate of change within a function.
• Our specific task utilized these calculus concepts to transform a complex function into its second derivative.

This exercise highlights the practical application of calculus techniques in understanding function behavior.