Problem 49 - Thomas Calculus

Question

Find \(y^{\prime \prime}\) in Exercises \(49-52\) $$ y=\left(1+\frac{1}{x}\right)^{3} $$

Step-by-Step Solution

Answer

The second derivative is \( y'' = 6x^{-3}(1 + x^{-1})^2 + 6x^{-4}(1 + x^{-1}) \).

Step 1: Identify the Function and Goal

The function given is \( y = \left(1+\frac{1}{x}\right)^{3} \). We need to find the second derivative, \( y'' \), of this function.

Step 2: Simplify the Inner Function

Rewrite the function by expressing \(\frac{1}{x}\) as \(x^{-1}\), so that \( y = (1 + x^{-1})^3 \). This makes it easier to differentiate using the chain rule and power rule.

Step 3: Find the First Derivative (\(y'\))

Use the chain rule to differentiate. For \( y = (1 + x^{-1})^3 \), set \( u = 1 + x^{-1} \). Then, \( y = u^3 \). By the chain rule, \[ y' = 3u^2 \cdot \frac{du}{dx} \]Compute \(\frac{du}{dx} = -x^{-2}\). Thus, \[ y' = 3(1 + x^{-1})^2 \cdot (-x^{-2}) = -3x^{-2}(1 + x^{-1})^2 \].

Step 4: Simplify the First Derivative

Rewrite \(y' = -3x^{-2}(1 + x^{-1})^2 = -3x^{-2}\left(1 + \frac{1}{x}\right)^2 \), which expresses \(y'\) using the original form for easier differentiation in the next steps.

Step 5: Find the Second Derivative (\(y''\))

Differentiate \(y' = -3x^{-2}(1 + x^{-1})^2 \) again. Use the product rule \[ \frac{d}{dx}[uv] = u'v + uv' \]Let \(u = -3x^{-2}, v = (1 + x^{-1})^2 \).Thus, \( u' = 6x^{-3} \) and using the chain rule again, \( v' = 2(1 + x^{-1})(-x^{-2}) \).The second derivative is \[ y'' = u'v + uv' \]Substitute to get\[ y'' = 6x^{-3}(1 + x^{-1})^2 + (-3x^{-2})[2(1 + x^{-1})(-x^{-2})] \].

Step 6: Simplify the Second Derivative Expression

Simplify the expression obtained for \(y''\):\[ y'' = 6x^{-3}(1 + x^{-1})^2 + 6x^{-4}(1 + x^{-1}) \].Simplify further by factoring and combining similar terms if possible to present the second derivative in a neat form.

Key Concepts

Chain Rule

The chain rule is a crucial concept when differentiating composite functions. It allows us to find derivatives of functions that are composed of other functions. Consider a function of the form \( y = f(g(x)) \). To find the derivative \( y' \), we apply the chain rule, which states:

• Differentiate the outer function \( f \) with respect to the inner function \( g \). In mathematical terms, compute \( f'(g(x)) \).
• Then, multiply that result by the derivative of the inner function, \( g'(x) \).

In our problem, we have \( y = (1 + x^{-1})^3 \). Here, the outer function \( f(u) = u^3 \) and the inner function \( u = 1 + x^{-1} \). Applying the chain rule:

• Differentiate \( u^3 \) to get \( 3u^2 \).
• Then differentiate \( u = 1 + x^{-1} \) to get \( -x^{-2} \).
• Finally, multiply these results to obtain \( y' = 3(1 + x^{-1})^2(-x^{-2}) \).

This process efficiently handles the complexity added by the composition of functions in differentiation. When we move on to the product rule for the second derivative, the chain rule will again play an important part.

Product Rule

The product rule is essential when differentiating products of two functions. If you have a function \( y = u(x) \, v(x) \), the product rule states that:

• Differentiate \( u(x) \) to get \( u'(x) \).
• Leave \( v(x) \) as is.
• Then, differentiate \( v(x) \) to get \( v'(x) \).
• Leave \( u(x) \) as is.
• The derivative, \( y' \), is given by \( u'(x) \, v(x) + u(x) \, v'(x) \).

In the context of finding the second derivative in this exercise, the product rule helps us deal with the expression \( y' = -3x^{-2}(1 + x^{-1})^2 \). We treat \( u = -3x^{-2} \) and \( v = (1 + x^{-1})^2 \). For the second derivative,

• Find \( u' = 6x^{-3} \).
• Use the chain rule inside the product rule to get \( v' = 2(1 + x^{-1})(-x^{-2}) \).
• Substitute in: \( y'' = u'v + uv' = 6x^{-3}(1 + x^{-1})^2 + (-3x^{-2})[2(1 + x^{-1})(-x^{-2})] \).

This process breaks down the complexity of differentiating products, especially when paired with the chain rule.

Power Rule

The power rule is one of the fundamental rules in calculus for differentiating expressions involving powers of variables. It provides a simple formula to find the derivative of \( x^n \), where \( n \) is any real number. The rule states:

• If \( y = x^n \), then the derivative \( y' = nx^{n-1} \).

In our exercise, the power rule appears when differentiating \( (1 + x^{-1})^3 \), particularly useful after applying the chain rule.

• The outer function \( u^3 \) is differentiated using the power rule: \( 3u^2 \).
• For the first derivative \( y' \), this gives part of the expression: \( 3(1 + x^{-1})^2 \).
• In the second derivative calculation through the product rule, the power rule appears yet again: differentiating \( -3x^{-2} \) gives \( 6x^{-3} \), derived from \( -3 \times (-2) \times x^{-3} \).

The power rule simplifies the operation of finding derivatives significantly, especially when combined with other rules like the chain and product rules. By understanding how to apply these rules separately and together, one can efficiently handle differentiation tasks.