Question

Find \(f^{\prime \prime}(2)\). $$ f(x)=\frac{(x+1)^{2}}{x-1} $$

Short answer

The value of \( f''(2) \) is 2.

Step-by-step solution

Find the First Derivative

To find the first derivative, we use the quotient rule. The quotient rule states that for a function \( g(x) = \frac{u(x)}{v(x)} \), the derivative \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(x) = (x+1)^2 \) and \( v(x) = x-1 \).First, find \( u'(x) \) and \( v'(x) \): - \( u'(x) = 2(x+1) \) because of the chain rule.- \( v'(x) = 1 \) since it's a simple linear function.Now, apply the quotient rule:\[ f'(x) = \frac{2(x+1)(x-1) - (x+1)^2}{(x-1)^2} \]Simplify the expression:\[ f'(x) = \frac{2x^2 - 2x + 2x - 2 - (x^2 + 2x + 1)}{(x-1)^2} \]\[ f'(x) = \frac{x^2 - 3}{(x-1)^2} \].

Find the Second Derivative

To find the second derivative, \( f''(x) \), we need to differentiate \( f'(x) = \frac{x^2 - 3}{(x-1)^2} \) again, using the quotient rule:For \( g(x) = \frac{p(x)}{q(x)} \), where \( p(x) = x^2 - 3 \) and \( q(x) = (x-1)^2 \), the derivative \( g'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{(q(x))^2} \).Find \( p'(x) \) and \( q'(x) \):- \( p'(x) = 2x \).- \( q'(x) = 2(x-1) \) using the chain rule.Apply the quotient rule:\[ f''(x) = \frac{2x(x-1)^2 - (x^2 - 3)(2(x-1))}{((x-1)^2)^2} \].Simplify the expression:First calculate each term, then expand and combine:\[ f''(x) = \frac{2x(x^2 - 2x + 1) - 2(x^2 - 3)(x-1)}{(x-1)^4} \]Calculate term by term, then simplify.

Evaluate the Second Derivative at x = 2

Substitute \( x = 2 \) into the expression of the second derivative.First, substitute and simplify step by step:\[ f''(2) = \frac{2(2)((2-1)^2) - (2^2 - 3)(2(2-1))}{(2-1)^4} \]Calculate each term:- \((2-1)^2 = 1\) so the first term becomes \(2 \times 2 \times 1 = 4\).- The second term becomes \((4 - 3) \times 2 \times 1 = 2\).Now combine:\[ f''(2) = \frac{4 - 2}{1} = 2 \].

Key concepts

Second Derivative

The second derivative of a function provides us with important information about its concavity and the rate at which the slope of the curve is changing. While the first derivative, often denoted as \( f'(x) \), tells us the rate of change of the function \( f(x) \), the second derivative \( f''(x) \) gives us the rate of change of that rate of change.

For instance, in physical terms, if the function represents the position of a particle over time, the first derivative indicates velocity, while the second derivative indicates acceleration. A positive second derivative suggests that the function is concave up, indicating a local minimum. Conversely, a negative second derivative suggests the function is concave down, indicating a local maximum. In our exercise, calculating \( f''(2) \) involves using previous derivative results to understand the behavior of the curve specifically at \( x = 2 \).

Quotient Rule

The quotient rule is a key technique in calculus for differentiating functions that are the ratio of two other functions. The rule is expressed as follows: if \( g(x) = \frac{u(x)}{v(x)} \), then the derivative \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).

This rule simplifies finding the derivative of complex rational functions by transforming the task into finding the derivatives of simpler constituent functions \( u(x) \) and \( v(x) \). In our problem, we used the quotient rule twice to find the first and second derivatives of \( f(x) = \frac{(x+1)^2}{x-1} \).

• Find the derivatives of the numerator and denominator separately.
• Apply the formula to combine these results into the overall derivative.
• Simplify the expression.

This process is crucial for efficiently handling the division of functions in differentiation.

Chain Rule

The chain rule allows us to differentiate composite functions, which is indispensable in the context of the quotient rule. It is expressed as: if a function \( h(x) = f(g(x)) \), then the derivative \( h'(x) = f'(g(x)) \times g'(x) \).

In contexts like our exercise, the chain rule helps us identify how to differentiate parts of the function within other rules, such as the quotient rule. For example, in finding the derivative of \( u(x) = (x+1)^2 \), the chain rule helps to recognize the outer and inner functions, allowing us to take the derivative more effectively.

By applying the chain rule within the quotient rule, we ascertain the derivative of these nested or composite structures easily and correctly.

Differentiation Techniques

Differentiation techniques are essential tools in calculus. They equip us with methods to find derivatives for various types of functions. This includes foundational rules like the power rule, product rule, and specific rules such as the chain and quotient rules. These techniques are not just theoretical but practical tools which streamline complex problems.

In our exercise, we applied the quotient rule and chain rule effectively.

• The quotient rule allowed us to differentiate the ratio in \( f(x) = \frac{(x+1)^2}{x-1} \).
• The chain rule facilitated handling the composite function in the numerator.

By integrating these techniques:

• We systematically solve problems by breaking down complex functions into more manageable parts.
• This helps in not only obtaining the first and second derivatives but also in simplifying the calculations significantly.

Mastery of these techniques is crucial for tackling a wide range of problems in calculus.