Problem 62 - Calculus and its applications

Question

Find \(y^{\prime \prime \prime}\) for each function. $$ y=\frac{1}{\sqrt{2 x+1}} $$

Step-by-Step Solution

Answer

The third derivative is \( y''' = -15(2x+1)^{-7/2} \).

Step 1: Rewrite the Function

First, rewrite the given function in a form that is easier to differentiate. Original function: \( y = \frac{1}{\sqrt{2x+1}} \). Rewrite it as \( y = (2x+1)^{-1/2} \).

Step 2: Differentiate First Time

Apply the power rule and chain rule to find the first derivative, \( y' \). Start with \( y = (2x+1)^{-1/2} \). Then \( y' = -\frac{1}{2} (2x+1)^{-3/2} \cdot 2 = -(2x+1)^{-3/2} \).

Step 3: Differentiate Second Time

Find the second derivative, \( y'' \), of \( y' = -(2x+1)^{-3/2} \). Again, apply the power rule and chain rule: \( y'' = -\left(-\frac{3}{2}(2x+1)^{-5/2}\right) \cdot 2 = 3(2x+1)^{-5/2} \).

Step 4: Differentiate Third Time

Find the third derivative, \( y''' \), by differentiating \( y'' = 3(2x+1)^{-5/2} \). Apply the power rule and chain rule: \( y''' = 3\left(-\frac{5}{2}(2x+1)^{-7/2}\right)\cdot 2 = -15(2x+1)^{-7/2} \).

Step 5: Simplify Final Result

The third derivative, \( y''' \), simplifies to: \( y''' = -15(2x+1)^{-7/2} \).

Key Concepts

Power Rule

The power rule is an essential tool in calculus for finding derivatives. It states that if you have a function of the form \( y = x^n \), where \( n \) is any real number, the derivative \( y' \) can be computed as \( nx^{n-1} \). This rule makes differentiation straightforward when dealing with power functions.
In the given exercise, the original function was rewritten as \( y = (2x+1)^{-1/2} \). Applying the power rule to differentiate this requires multiplying the exponent by the coefficient of the function and then reducing the exponent by 1.

• First differentiation: The power rule helped us get the derivative \( y' = -\frac{1}{2}(2x+1)^{-3/2} \).
• Second differentiation: For the derivative of \( y' \), the power rule was used again to find \( y'' = 3(2x+1)^{-5/2} \).
• Third differentiation: Similarly, it was applied to find the third derivative as \( y''' = -15(2x+1)^{-7/2} \).

Each differentiation step involved reducing the exponent, which is a core aspect of the power rule in action.

Chain Rule

The chain rule is crucial when differentiating composite functions. It allows us to differentiate a function that is nested inside another function. When you have a function within a function, like \( y = (2x+1)^{-1/2} \), you need to apply the chain rule.
The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative is \( f'(g(x)) \cdot g'(x) \). This means you differentiate the outer function and multiply it by the derivative of the inner function.

• First Differentiation: For \( y = (2x+1)^{-1/2} \), the outer function is \((u)^{-1/2}\) where \( u = 2x+1 \). Differentiating \( u \) gives us \( 2 \), and multiplying by the derivative of the outer function gives \( y' = -(2x+1)^{-3/2} \).
• Subsequent Differentiations: This process is repeated for each derivative step, applying the chain rule to correctly handle the inner structure of the composite function. For example, in the second differentiation, the outer function becomes \( (u)^{-5/2} \) leading to the calculation of \( y'' = 3(2x+1)^{-5/2} \).

The chain rule is consistently used with the power rule to tackle the nested nature of the function for each derivative.

Differentiation Steps

Differentiation is the process of finding the derivative of a function. In this exercise, higher order derivatives were needed, which means continuing the process multiple times.
The steps to finding the third derivative of the function \( y = \frac{1}{\sqrt{2x+1}} \) are as follows:

• Step 1: Rewrite the function from its original form to a more manageable, differentiable form: \( y = (2x+1)^{-1/2} \). This allows easier application of differentiation rules.
• Step 2: Find the first derivative \( y' \) using the power rule and chain rule. This gives \( y' = -(2x+1)^{-3/2} \).
• Step 3: Continue to the second derivative \( y'' \). Again apply the power and chain rules to find \( y'' = 3(2x+1)^{-5/2} \).
• Step 4: Reach the third derivative \( y''' \), obtaining \( y''' = -15(2x+1)^{-7/2} \) by repeating the same techniques.

Each step methodically applies differentiation rules, making sure each derivative is calculated based on the previous one, illustrating the systematic nature of calculus.