Problem 35 - Calculus and its applications

Question

Find \(y^{\prime \prime}\) $$ y=\frac{3 x+1}{2 x-3} $$

Step-by-Step Solution

Answer

The second derivative is \( y'' = \frac{44}{(2x-3)^3} \).

Step 1: Identify the Function to Differentiate

The given function is \( y = \frac{3x + 1}{2x-3} \). We need to find the second derivative \( y^{\prime\prime} \).

Step 2: Differentiate the Function Using the Quotient Rule

First, apply the quotient rule to find the first derivative \( y' \). For a function \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule is: \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). Let \( u(x) = 3x + 1 \) and \( v(x) = 2x - 3 \). Then find their derivatives: \( u'(x) = 3 \) and \( v'(x) = 2 \). Substitute into the quotient rule:\[ y' = \frac{(3)(2x - 3) - (3x + 1)(2)}{(2x - 3)^2} \].Simplify the expression:\[ y' = \frac{6x - 9 - (6x + 2)}{(2x - 3)^2} = \frac{-11}{(2x - 3)^2} \].

Step 3: Differentiate Again Using the Chain and Product Rules

To find the second derivative \( y'' \), differentiate \( y' = \frac{-11}{(2x - 3)^2} \). This can also be written as \( -11(2x - 3)^{-2} \). Use the chain rule and power rule:\[ y'' = \, d/dx \, [-11(2x-3)^{-2}] = -11 \, \cdot \, -2 \, (2x-3)^{-3} \, \cdot \, \frac{d}{dx}(2x-3) \].\( \frac{d}{dx}(2x-3) = 2 \), thus:\[ y'' = 22 \, \cdot \, (2x-3)^{-3} \, \cdot \, 2 = \frac{44}{(2x-3)^3} \].

Step 4: Simplify the Second Derivative

The expression for the second derivative is \( y'' = \frac{44}{(2x-3)^3} \). This is already in a simplified form.

Key Concepts

Quotient Rule

When you need to differentiate a function that is composed of two other functions divided by each other, the Quotient Rule is your go-to tool. Envision it as a recipe for finding the derivative of a quotient of two functions. Let's break it down with some steps:

• If you have a function \( y = \frac{u}{v} \), the first derivative, using the quotient rule, is given by:
• \( y' = \frac{u'v - uv'}{v^2} \)
• Here, \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives.

Using this rule ensures you carefully manage the combination of derivatives and the original functions. It's essential when tackling problems where differentiation needs to be applied to a ratio. Remember:

• The derivative of the numerator \( u \) is multiplied by the denominator \( v \), and vice versa — but subtracted.
• All of this is over the square of the denominator function \( v(x) \).

Chain Rule

The Chain Rule comes into play when you deal with composite functions, functions within functions. Think of it as a method to "unwind" layers of functions when taking derivatives. If a function can be seen as \( f(g(x)) \), where \( f \) is one function applied to another function \( g \), you can use:

• Chain Rule formula: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
The key idea is to first find the derivative of the outer function \( f \) with respect to the inner function \( g \), and then multiply it by the derivative of the inner function \( g(x) \). Break it down in steps:
• Differentiating the outer function as if the inner function was a simple variable.
• Multiply that derivative by the inner function’s derivative.
This rule becomes particularly useful when you need to differentiate powers of functions or compositions found in a wider range of complex expressions.

Power Rule

The Power Rule is perhaps one of the simplest and most widely used differentiation rules. It provides a quick way to find the derivative of a function in the form \( x^n \), where \( n \) is any real number. When you take derivatives using the Power Rule, remember:

• The derivative of \( x^n \) is \( nx^{n-1} \).

A couple of things to keep in mind:

• It works for any real number \( n \)—including negatives and fractions.
• A crucial step when differentiating higher powers or when combined with other rules like the chain rule.

When applied, such as in the original problem where you have to differentiate expressions like \((2x-3)^{-2}\), the Power Rule helps you quickly reduce these exponents. By reducing the power by one and multiplying by the original power, it fits seamlessly when differentiating polynomials or parts of more complex functions.