Problem 76 - University Calculus: Early Transcendentals

Question

Find \(y^{\prime \prime}\) in Exercises \(71-78\). $$y=x^{2}\left(x^{3}-1\right)^{5}$$

Step-by-Step Solution

Answer

The second derivative is a combination of terms from applying product and chain rules.

Step 1: Identify the Formula

Notice that the function \(y=x^{2}(x^{3}-1)^{5}\) is a product of two functions: \(u=x^2\) and \(v=(x^3 - 1)^5\). We'll need to utilize the product rule for differentiation.

Step 2: Differentiate Using the Product Rule

The product rule states that \((uv)' = u'v + uv'\). First, find \(u'\) which is the derivative of \(x^2\), so \(u' = 2x\). Next, hold \(u\) and differentiate \(v = (x^3 - 1)^5\) using the chain rule.

Step 3: Apply the Chain Rule

The differentiation of \(v = (x^3 - 1)^5\) requires the chain rule. Setting \(w = x^3 - 1\), then \(v = w^5\). Thus, \(v' = 5w^4 \cdot w'\). Since \(w = x^3 - 1\), then \(w' = 3x^2\). So, \(v' = 5(x^3 - 1)^4 \cdot 3x^2 = 15x^2(x^3 - 1)^4\).

Step 4: Substitute Back into the Product Rule

Substituting, we have \(y' = 2x(x^3 - 1)^5 + x^2 \cdot 15x^2(x^3 - 1)^4\). Simplifying, \(y' = 2x(x^3 - 1)^5 + 15x^4(x^3 - 1)^4\).

Step 5: Find the Second Derivative

Now, differentiate \(y'\) to find \(y''\). This again involves using the product rule on each term. Start with the first term, then the second term. Notice \((x^n)\)' is \(nx^{n-1}\).

Step 6: Differentiate Each Term Separately

For the term \(2x(x^3 - 1)^5\), let \(a = 2x\), \(a' = 2\), and \(b = (x^3 - 1)^5\), \(b' = 15x^2(x^3 - 1)^4\). Thus, the first part's derivative is \(a'b + ab' = 2(x^3 - 1)^5 + 30x^3(x^3 - 1)^4\).

Step 7: Differentiate the Second Term

For the term \(15x^4(x^3-1)^4\), \(c = 15x^4\), \(c' = 60x^3\), and already we know \((x^3-1)^4\)'s part. Applying the product rule again gives: \(c'd + cd' = 60x^3(x^3-1)^4 + 60x^6(x^3-1)^3\).

Step 8: Combine and Simplify

Combine all differentiated parts: \[ y'' = 2(x^3 - 1)^5 + 30x^3(x^3 - 1)^4 + 60x^3(x^3-1)^4 + 60x^6(x^3-1)^3 \].Combine like terms and organize for a simplified expression.

Key Concepts

Product Rule

The product rule is an essential technique in differentiation used whenever you encounter a function that is the product of two other functions. Here's a quick rundown to help you understand it.

• Consider two functions, say \( u \) and \( v \). If you wish to differentiate the product \( uv \), the product rule comes into play.
• The rule states: \((uv)' = u'v + uv'\).
• This means: Differentiate the first function \( u \) and multiply it by \( v \), then add \( u \) multiplied by the derivative of \( v \).

Imagine you have the function \( y = x^2(x^3 - 1)^5 \). Here, \( u = x^2 \) and \( v = (x^3 - 1)^5 \). Applying the product rule allows you to tackle each part separately before combining them. Always remember, when dealing with products, break it down into manageable pieces.

Chain Rule

The chain rule is a powerful tool in calculus to differentiate composite functions - that is, functions within other functions.

• In simple terms, the chain rule lets you "peel back" these function layers, differentiating them step by step.
• Mathematically, if you have a function \( y = f(g(x)) \), then the derivative \( y' \) is \( f'(g(x)) \cdot g'(x) \).

In our example, for differentiating \( v = (x^3 - 1)^5 \), note that this function is built from two functions: the outer \( f(u) = u^5 \) and inner \( u = x^3 - 1 \).

• First, differentiate the outer layer: \( f'(u) = 5u^4 \).
• Then, differentiate the inner function: \( u'(x) = 3x^2 \).
• Finally, multiply these derivatives: \( v' = 5(x^3 - 1)^4 \cdot 3x^2 = 15x^2(x^3 - 1)^4 \).

The chain rule requires attention to detail, but with practice, it becomes manageable and is crucial for handling complex differentiation tasks.

Second Derivative

Finding the second derivative, noted as \( y'' \), involves differentiating the derivative of a function, essentially repeating differentiation.

• It provides insights into the function's concavity or convexity and helps in determining points of inflection.
• Let's find the second derivative of \( y = x^2(x^3 - 1)^5 \).
• Our first derivative is \( y' = 2x(x^3 - 1)^5 + 15x^4(x^3 - 1)^4 \).

To compute \( y'' \), differentiate each term separately by applying the product rule again:

• For \( 2x(x^3 - 1)^5 \), you take the derivative of \( 2x \), which is \( 2 \), and go through the process of finding the derivative of \( (x^3 - 1)^5 \) as before, resulting in the first part's derivative.
• Similarly, differentiate \( 15x^4(x^3 - 1)^4 \) using the product rule to get the next segment.

Put together, you simplify and combine. Calculating the second derivative aids in a deeper understanding of the function's behavior, making it vital for analyzing graphs and optimizing problems.