Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(f^{\prime \prime}(2)\). $$ f(s)=s\left(1-s^{2}\right)^{3} $$

Step-by-Step Solution

Answer

\(f''(2) = -36\)

Step 1: Differentiate the Function

The original function is \(f(s) = s(1-s^2)^3\). To find the first derivative, \(f'(s)\), we need to apply the product rule and chain rule. Let \(u = s\) and \(v = (1-s^2)^3\). Then, \(f(s) = u imes v\).The product rule states \((uv)' = u'v + uv'\). So,1. Find \(u'\): Since \(u = s\), we have \(u' = 1\).2. Find \(v'\) using the chain rule: The derivative of \((1-s^2)^3\) is \(3(1-s^2)^2(-2s) = -6s(1-s^2)^2\).Apply the product rule:\[f'(s) = 1 imes (1-s^2)^3 + s imes (-6s(1-s^2)^2)\]\[= (1-s^2)^3 - 6s^2(1-s^2)^2\].

Step 2: Simplify First Derivative

Simplify the expression obtained for \(f'(s)\):\[f'(s) = (1-s^2)^3 - 6s^2(1-s^2)^2\]Factor out \((1-s^2)^2\):\[f'(s) = (1-s^2)^2((1-s^2) - 6s^2)\]Simplify inside the parentheses:\[= (1-s^2)^2(1 - s^2 - 6s^2) = (1-s^2)^2(1 - 7s^2)\].

Step 3: Differentiate Again to Find Second Derivative

Now differentiate \(f'(s) = (1-s^2)^2(1 - 7s^2)\) to find \(f''(s)\). We'll use the product rule again:Let \(u = (1-s^2)^2\) and \(v = 1 - 7s^2\):1. Find \(u'\): Use the chain rule: \(u = (1-s^2)^2\), so \(u' = 2(1-s^2)(-2s) = -4s(1-s^2)\).2. Find \(v'\): This is \(v' = -14s\).Now apply the product rule:\[f''(s) = (-4s(1-s^2))(1 - 7s^2) + (1-s^2)^2(-14s)\].

Step 4: Simplify the Second Derivative

Simplify \(f''(s)\):Distribute in each term:\[f''(s) = -4s(1-7s^2)(1-s^2) + (-14s(1-s^2)^2)\]Combine like terms:First term: \[-4s(1-s^2 - 7s^4)\]Second term: \[-14s(1 - 2s^2 + s^4)\].Expand and combine:\[= -4s + 28s^3 - 4s + (-14s + 28s^3 - 14s^5)\]Combine all like terms:\[-18s + 56s^3 - 14s^5\].

Step 5: Evaluate Second Derivative at \(s=2\)

Now that we have \(f''(s) = -18s + 56s^3 - 14s^5\), evaluate it at \(s = 2\). Substitute \(s = 2\) into the expression:\[f''(2) = -18(2) + 56(2)^3 - 14(2)^5\]Calculate each term:\[-18(2) = -36\]\[56(2)^3 = 56(8) = 448\]\[14(2)^5 = 14(32) = 448\]Combine these:\[-36 + 448 - 448 = -36\].

Key Concepts

Product Rule

When we want to find the derivative of a function that is the product of two other functions, we use the product rule. The product rule helps us differentiate a function of the form \(f(x) = u(x) \cdot v(x)\), where both \(u\) and \(v\) are functions of \(x\). The rule states that the derivative is given by

• \((uv)' = u'v + uv'\)

In our exercise, \(f(s)\) is given as \(s(1-s^2)^3\). Here, \(u = s\) and \(v = (1-s^2)^3\). The product rule requires us to differentiate \(u\) and \(v\) independently before applying the formula.

• First, we find \(u'\) which is \(1\) because the derivative of \(s\) with respect to \(s\) is \(1\)
• Next, differentiate \(v\) using other rules such as the chain rule.

Chain Rule

The chain rule is a critical tool for finding the derivative of a composite function. Specifically, if we have a function \(g(x)\) inside another function \(f(x)\), like \(f(g(x))\), we apply the chain rule.
In formula terms, the chain rule is expressed as:

• \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

In our problem, we encountered the expression \((1-s^2)^3\). Here we identify the inner function as \(g(s) = 1-s^2\) and the outer function as \(h(x) = x^3\). Differentiating using the chain rule means first taking the derivative of the outer function \((h'(x) = 3x^2)\) evaluated at the inner function, then multiply by the derivative of the inner function \((g'(s) = -2s)\). Putting it together, the derivative \(v' = 3(1-s^2)^2(-2s) = -6s(1-s^2)^2\). This is the derivative we use for the product rule.

Derivative Evaluation

Once we've found the derivative expressions, evaluating them at specific points can tell us much about the behavior of the function. In our exercise, we go through several derivative evaluations:

• First, evaluate the first derivative \(f'(s)\) to understand its behavior over different values of \(s\).
• Once the second derivative \(f''(s)\) is calculated, we can further analyze it at specific points such as \(s = 2\).

To find \(f''(2)\), we substitute \(s=2\) into our second derivative equation \(-18s + 56s^3 - 14s^5\) and compute each term:

• \(-18(2) = -36\)
• \(56(2)^3 = 448\)
• \(-14(2)^5 = -448\)
• Putting it together, \(-36 + 448 - 448 = -36\)

Simplifying Derivatives

After deriving expressions, it's crucial to simplify them. Simplification makes it easier to evaluate at certain points and understand their behavior. For our given function, after applying the product and chain rules, we got an initial complex form for \(f'(s)\):

• \(f'(s) = (1-s^2)^3 - 6s^2(1-s^2)^2\)

Simplifying it involves factoring out common elements, allowing us to rewrite it as:

• \((1-s^2)^2(1 - 7s^2)\)

This factorization shows a clear relationship between components, making further derivatives and evaluations more manageable. The process is repeated for \(f''(s)\), resulting in an expression like \(-18s + 56s^3 - 14s^5\), which is simpler to interpret and calculate.