Problem 64 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The second derivative \( y'' \) is derived using product and quotient rules.

Step 1: Rewrite for Easier Differentiation

Rewrite the given function \( y = \frac{x}{\sqrt{x-1}} \) using exponent notation for ease of differentiation: \[ y = x (x-1)^{-0.5} \]

Step 2: Apply the Product Rule

Use the product rule to differentiate \( y = x (x-1)^{-0.5} \). The product rule states: \[ (uv)' = u'v + uv' \]In this case, let \( u = x \) and \( v = (x-1)^{-0.5} \), then \( u' = 1 \) and \( v' = -0.5(x-1)^{-1.5} \).So,\[ y' = 1 \cdot (x-1)^{-0.5} + x \cdot (-0.5)(x-1)^{-1.5} \] This simplifies to:\[ y' = (x-1)^{-0.5} - 0.5x(x-1)^{-1.5} \]

Step 3: Simplify the First Derivative

Combine the terms to simplify \( y' \):\[ y' = \frac{1}{\sqrt{x-1}} - \frac{0.5x}{(x-1)^{1.5}} \]Express both terms with a common denominator: \[ y' = \frac{(x-1)^{1.5} - 0.5x\sqrt{x-1}}{(x-1)^{1.5}} \]

Step 4: Find the Second Derivative

Differentiate \( y' \) to find \( y'' \). Apply the quotient rule: \[ \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \]Let \( f = (x-1)^{1.5} - 0.5x\sqrt{x-1} \) and \( g = (x-1)^{1.5} \).Find \( f' \) and \( g' \) separately.

Step 5: Differentiate Numerator \(f\)

Find the derivative \( f' \):Separate the two terms:- Derivative of \( (x-1)^{1.5} \) using chain rule: - Let \( u = x-1 \), then \( f_1 = u^{1.5} \), so: \[ f_1' = 1.5(x-1)^{0.5} \]- Derivative of \( -0.5x\sqrt{x-1} \): - \( f_2 = -0.5x(x-1)^{0.5} \), using product rule: \[ f_2' = -0.5\sqrt{x-1} - 0.25x(x-1)^{-0.5} \]Combine:\[ f' = 1.5(x-1)^{0.5} - 0.5\sqrt{x-1} - 0.25x(x-1)^{-0.5} \]

Step 6: Differentiate Denominator \(g\)

Find the derivative \( g' \):\[ g = (x-1)^{1.5} \]Differentiating gives:\[ g' = 1.5(x-1)^{0.5} \]

Step 7: Apply Quotient Rule

Finally, apply the quotient rule on \( y' \):\[ y'' = \frac{f'g - fg'}{g^2} \]Substitute \( f, g, f', g' \) as determined previously:\[ y'' = \frac{(1.5(x-1)^{0.5} - 0.5\sqrt{x-1} - 0.25x(x-1)^{-0.5})(x-1)^{1.5} - ((x-1)^{1.5} - 0.5x\sqrt{x-1})\cdot 1.5(x-1)^{0.5}}{((x-1)^{1.5})^2} \]Simplify as needed based on specific focus.

Key Concepts

Product Rule

Understanding the product rule is crucial when dealing with functions that are products of two or more expressions. It's a rule in calculus that allows us to differentiate such functions effectively. Say you have two functions, \( u \) and \( v \), which are multiplied together to form a new function \( y = uv \). The product rule states that the derivative of this product is:

• \( (uv)' = u'v + uv' \)

Here, \( u' \) is the derivative of \( u \), and \( v' \) is the derivative of \( v \). This rule simplifies the differentiation process significantly.
In practice, as in the original exercise, you identify parts of your function that together form a product. For instance, if you start with \( y = x (x-1)^{-0.5} \), you can treat \( x \) as \( u \) and \( (x-1)^{-0.5} \) as \( v \). Calculating their derivatives individually, you can apply the product rule formula to find the overall derivative \( y' \).
Remember, mastering the product rule helps break down complex functions into manageable chunks, making them easier to differentiate.

Quotient Rule

The quotient rule is a powerful technique in calculus when you have a function that is the division of two others. Unlike products, quotients involve dividing one function by another, expressed as \( y = \frac{f}{g} \) where \( f \) and \( g \) are differentiable functions.
To differentiate such a function, use the quotient rule:

• \( \left(\frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \)

In this formula, \( f' \) and \( g' \) are the derivatives of \( f \) and \( g \) respectively. The process involves finding each of these derivatives separately, substituting them back into the formula, and simplifying as needed.
In the example, when you find \( y' \) and need the second derivative \( y'' \), the function is expressed as a fraction, making the quotient rule appropriate. Practice with the quotient rule aids in handling functions that can be unwieldy due to division, simplifying calculations and preventing algebraic errors.

Differentiation Techniques

Differentiation is a fundamental concept in calculus used to determine how functions change. Various techniques help tackle different types of functions:

• Basic Derivatives: Begin with simple polynomial rules and then naturally extend to more complicated derivatives.
• Product Rule: Useful for multiplying functions. Differentiates by dissecting into parts.
• Quotient Rule: Applies to functions that involve division. It’s more complex, but it reveals the change in ratios.
• Chain Rule: For compositions of functions. Differentiates the outer and inner functions iteratively.

In the original exercise, differentiation techniques like the product rule and subsequently the quotient rule were utilized. Having a versatile skill set with these techniques allows differentiation in almost any scenario, aiding in understanding the rate and direction of change across diverse functions.
Confidence in differentiation techniques stems from practice and familiarity. They not only solve problems regarding rates of change but also lay down the logic for more advanced studies in calculus.

Chain Rule

The chain rule is indispensable for differentiating composite functions, where one function is nestled inside another. It enables us to address situations where each function alone is simple, but their combination through composition makes the differentiation process more intricate.
The chain rule can be stated as follows: if you have a function \( y = f(g(x)) \), then its derivative is:

• \( y' = f'(g(x)) \cdot g'(x) \)

First, you differentiate the outer function while leaving the inner function unchanged, then multiply by the derivative of the inner function. In practical terms, this often involves identifying layers within a function and systematically peeling them away through differentiation.
In the task involving second derivatives, the chain rule plays a vital role when functions like \( (x-1)^{1.5} \) arise. You treat \( (x-1) \) as an inner function, distinctly finding its derivative before attending to the outer component.
The chain rule is particularly useful in making the calculation of second derivatives cleaner and more manageable. With compositional functions, it handles layers with precision, contributing to deeper insights into function behavior.