Problem 61 - Calculus and its applications

Question

Find \(y^{\prime \prime \prime}\) for each function. $$ y=\frac{1}{1-x} $$

Step-by-Step Solution

Answer

The third derivative, \( y''' \), is \( 6(1-x)^{-4} \).

Step 1: Understand the Function

We are given the function \( y = \frac{1}{1-x} \). This function can be rewritten for differentiation by using the form \( y = (1-x)^{-1} \).

Step 2: First Derivative

To find \( y' \), apply the power rule. Using \( u = 1-x \) with \( u' = -1 \) and \( y = u^{-1} \), the derivative is \( y' = -1 \times (-1)(1-x)^{-2} = (1-x)^{-2} \).

Step 3: Second Derivative

Differentiate the first derivative \( y' = (1-x)^{-2} \). Again using the power rule, \( y'' = -2(1-x)^{-3} \times (-1) = 2(1-x)^{-3} \).

Step 4: Third Derivative

Differentiate the second derivative \( y'' = 2(1-x)^{-3} \). Using the power rule, \( y''' = -3 \times 2(1-x)^{-4} \times (-1) = 6(1-x)^{-4} \).

Step 5: Final Expression for the Third Derivative

Therefore, the third derivative of \( y = \frac{1}{1-x} \) is \( y''' = 6(1-x)^{-4} \).

Key Concepts

Third Derivative

Understanding the third derivative is an important aspect of calculus, especially when analyzing the motion of objects or the behavior of functions. The third derivative, often denoted as \( y''' \), provides insight into the rate at which the rate of change of a function is changing. It helps us comprehend the function's curvature and how it evolves over time.

To find the third derivative, you rely on calculating the first and second derivatives first. For example, in our exercise, the function \( y = \frac{1}{1-x} \) is initially rewritten in a form suitable for differentiation: \( y = (1-x)^{-1} \). This reformulation helps apply the power rule efficiently.

• The first derivative, \( y' \), is found by applying the power rule to get \( (1-x)^{-2} \).
• The second derivative, \( y'' \), is similarly calculated by differentiating \( y' \), resulting in \( 2(1-x)^{-3} \).
• Finally, the third derivative, \( y''' \), occurs from differentiating the second derivative, which gives us \( 6(1-x)^{-4} \).

Each step involves using previous results, showing the interdependence of derivatives in calculus.

Power Rule

The power rule is a key tool in differentiation, simplifying the process of finding derivatives of functions. It states that for any function \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \). This rule is especially useful for polynomial functions or when dealing with expressions that can be rewritten as powers.

In our exercise, we start with \( y = \frac{1}{1-x} \), which is adjusted for the power rule into \( y = (1-x)^{-1} \). Here's how the power rule applies:

• When calculating the first derivative, \( y' \), the rule turns \( (1-x)^{-1} \) into \( -(1)(1-x)^{-2} \), which simplifies to \( (1-x)^{-2} \) because of the negative sign multiplication.
• For the second derivative, the same method leads to \( 2(1-x)^{-3} \).
• The third derivative follows similarly using the power rule to become \( 6(1-x)^{-4} \).

The power rule is efficient and eliminates the need for more complex division or multi-step processes, enabling swift and straightforward derivations.

Calculus Problem Solving

When solving calculus problems involving differentiation, understanding the concepts and techniques is crucial for finding accurate solutions. Each problem, like finding derivatives, requires a clear plan and methodical execution.

For the given exercise of calculating the third derivative, one must:

• Transform the given function into a differentiable form, such as changing \( \frac{1}{1-x} \) into \( (1-x)^{-1} \).
• Apply differentiation rules, such as the power rule, consistently across all steps.
• Check each step to ensure accuracy, since differentiation involves cumulative calculations.

Successfully finding derivatives demands a blend of methodical application of rules and careful simplification at each stage, as was demonstrated in finding \( y''' = 6(1-x)^{-4} \). Calculus challenges are best approached with a step-by-step mindset, ensuring every transformation and calculation paves the way for the next, leading to a correct and satisfying conclusion.