Question

In Exercises, find the second derivative of the function. $$ f(x)=4\left(x^{2}-1\right)^{2} $$

Short answer

The second derivative of the function \(f(x) = 4(x^2 -1)^2\) is \(f''(x) = 32x^3 + 32x^2 - 32x\)

Step-by-step solution

Finding the first derivative

First, find the derivative of the given function, using the chain rule and the power rule. The derivative \(f'(x)\) of \(f(x)\) is computed as follows: \(f'(x)= 2 \times 4 (x^2 - 1) \times 2x = 16x(x^2 - 1)\).

Finding the second derivative

Next, differentiate \(f'((x)) = 16x(x^2 - 1)\) to find the second derivative \(f''(x)\). Use the product rule for differentiation, which states that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second. Here, \(f''(x)= 16 \times 2x \times (x^2 - 1) + 16x \times 2x = 32x^3 - 32x + 32x^2 = 32x^3 + 32x^2 - 32x\)

Key concepts

Chain Rule

When dealing with derivatives of functions, we often encounter composite functions. The chain rule is a fundamental tool used to differentiate these composite functions. A composite function is one where you have a function inside another, like in \(f(x) = 4(x^2 - 1)^2\).

Think of the chain rule as peeling different layers of an onion. We differentiate the outer layer, then the inner.

• The outer layer here is \((u^2)\), where \(u = x^2 - 1\).
• The inner layer is the function inside, \(x^2 - 1\).

To apply the chain rule, differentiate the outer function with respect to the inner function first (using the power rule), \(2(u)\).Then, multiply that result by the derivative of the inner function, \(2x\). Combining these results gives \(4 imes 2(x^2 - 1) imes 2x = 16x(x^2 - 1)\). This makes finding derivatives of complex functions much more manageable. Understanding how to peel away these layers using the chain rule will serve you well in calculus.

Power Rule

The power rule is one of the simplest rules in calculus for finding derivatives. It is particularly useful when differentiating terms like \(x^n\), where \(n\) is any real number.

The power rule states:

• If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).

Applying this rule to the function in our exercise, we take each term of \(x^2 - 1\) inside the bracket and differentiate them using the power rule. This results in \(2x\). When paired with the chain rule in the differentiation process, it allows us to address more complex functions, such as the composite function \((x^2 - 1)^2\). By differentiating with respect to \((x^2 - 1)\) first, then the entire inner function, we smoothly apply the power rule in concert with the chain rule. Together, they create a straightforward technique to tackle even the trickiest functions.

Product Rule

The product rule is essential when you're dealing with the derivatives of products of functions. In our solution, once we find the first derivative using the chain and power rules, we obtain a product of two functions: \(16x\) and \((x^2 - 1)\).

The product rule gets applied here allowing us to differentiate each component separately. The rule states:

• For two functions \(u(x)\) and \(v(x)\), the derivative \((uv)'\) is \(u'v + uv'\).

In this context:

• \(u = 16x\) and its derivative \(u' = 16\).
• \(v = (x^2 - 1)\) and \(v' = 2x\).

Plug these into the product rule to find \(f''(x) = 16 imes 2x(x^2 - 1) + 16x imes 2x = 32x^3 + 32x^2 - 32x\).This efficiently simplifies calculating second derivatives when functions come in multiplicative pairs.