Problem 28 - Calculus and its applications

Question

Find \(y^{\prime \prime}\) $$ y=\left(x^{4}+x\right)^{2 / 3} $$

Step-by-Step Solution

Answer

The second derivative \(y''\) is calculated using the product and chain rules, resulting in a lengthy expression involving \(x^4+x\) and \(x\).

Step 1: Identify the differentiation method

Given the function \( y = (x^4 + x)^{2/3} \), we need to apply the chain rule to find its derivative, as it involves a composition of functions.

Step 2: Apply the chain rule for first derivative

First, differentiate the outer function \((u)^{2/3}\) with respect to \(u\), where \(u = x^4 + x\). The derivative is:\[ \frac{d}{du}(u^{2/3}) = \frac{2}{3}u^{-1/3} \]. Next, multiply by the derivative of the inner function \(x^4 + x\) with respect to \(x\), which is:\[ \frac{d}{dx}(x^4 + x) = 4x^3 + 1 \]. Thus, \[ y' = \frac{2}{3}(x^4 + x)^{-1/3} (4x^3 + 1) \].

Step 3: Simplify the first derivative

Substitute back the expression for \(u\). The first derivative becomes:\[ y' = \frac{2}{3}(x^4 + x)^{-1/3} (4x^3 + 1) \].

Step 4: Apply the chain rule again for second derivative

Apply the product rule and chain rule on \[ y' = \frac{2}{3}(x^4 + x)^{-1/3} (4x^3 + 1) \]. Let \( v = (x^4 + x)^{-1/3} \) and \( w = 4x^3 + 1 \). Then \( y' = \frac{2}{3}vw \). The product rule gives \( y'' = \frac{2}{3}(v'w + vw') \).

Step 5: Differentiate components using chain rule

First, find \( v' \): \[ v = (x^4 + x)^{-1/3} \implies v' = -\frac{1}{3}(x^4 + x)^{-4/3}(4x^3 + 1) \]. Then, differentiate \( w \): \[ w' = \frac{d}{dx}(4x^3 + 1) = 12x^2 \].

Step 6: Substitute derivatives into second derivative formula

Substitute \( v' \) and \( w' \) into \( y'' = \frac{2}{3}(v'w + vw') \). Thus, \[ y'' = \frac{2}{3} \left( -\frac{1}{3}(x^4 + x)^{-4/3}(4x^3 + 1)^2 + (x^4 + x)^{-1/3}(12x^2) \right) \].

Step 7: Simplify the second derivative

Combine and simplify each term neatly, keeping in mind the common factor from the expressions. This will be a lengthy expression but expressed in terms of powers of \(x\) and \(x^4 + x\).

Key Concepts

Chain Rule

Differentiation can sometimes be tricky, especially when dealing with composite functions. This is where the **Chain Rule** comes into play. Imagine you have a function that is layered within another function, like a matryoshka doll. The Chain Rule helps to differentiate such functions easily.
For instance, consider a function like \( y = (x^4 + x)^{2/3} \). It's a combination of inner and outer functions where the Chain Rule is particularly useful. Let's break it down:

• Identify the outer function: here, it's \( u^{2/3} \), where \( u = x^4 + x \).
• Differentiate the outer function with respect to \( u \): \( \frac{d}{du}(u^{2/3}) = \frac{2}{3}u^{-1/3} \).
• Differentiating the inner function with respect to \( x \): \( \frac{d}{dx}(x^4 + x) = 4x^3 + 1 \).
• Combine these results: \( y' = \frac{2}{3}(x^4 + x)^{-1/3}(4x^3 + 1) \).

This method allows for effective differentiation of composite functions by breaking it into manageable parts.

Product Rule

Differentiating products of functions is another important concept in calculus known as the **Product Rule**. This rule applies when two functions are multiplied together, as in the case of calculating the second derivative of \( y \) where multiple derivative rules combine.
The Product Rule can be summarized as follows:

• Let \( y' = \frac{2}{3}vw \) where \( v = (x^4 + x)^{-1/3} \) and \( w = 4x^3 + 1 \).
• According to the Product Rule: \( y'' = \frac{2}{3}(v'w + vw') \).
• To find \( v' \), apply the Chain Rule to \( v \), giving \( v' = -\frac{1}{3}(x^4 + x)^{-4/3}(4x^3 + 1) \).
• The derivative of \( w \) with respect to \( x \) is straightforward: \( w' = 12x^2 \).
• Substitute these into the Product Rule formula: \( y'' = \frac{2}{3}(-\frac{1}{3}(x^4 + x)^{-4/3}(4x^3 + 1)^2 + (x^4 + x)^{-1/3}(12x^2)) \).

Understanding and applying the Product Rule allows you to tackle complex expressions with ease.

Second Derivative

The **Second Derivative** provides insight into the curvature or concavity of a function. It is simply the derivative of the first derivative. Differentiating twice, as seen in the given function \( y = (x^4 + x)^{2/3} \), gives us an expression that further reveals the nature of the graph.
Specifically, here's how the second derivative adds depth:

• The second derivative \( y'' \) tells us about the concavity of \( y \). If \( y'' > 0 \), the function is concave up; if \( y'' < 0 \), it's concave down.
• In the solution, we used both the Chain Rule and Product Rule to find \( y' \) and then \( y'' \).
• By examining \( y'' = \frac{2}{3}(-\frac{1}{3}(x^4 + x)^{-4/3}(4x^3 + 1)^2 + (x^4 + x)^{-1/3}(12x^2)) \), the interplay between the terms tells us how steep or flat the curve is at a given point.

Thus, understanding the second derivative is crucial for examining the behavior of functions, helping you predict their movement and general shape on a graph.