Problem 76 - Calculus for Biology and Medicine

Question

Find the first and the second derivatives of each function. \(h(s)=\frac{1}{s^{2}+2}\)

Step-by-Step Solution

Answer

First derivative: \( h'(s) = -\frac{2s}{(s^2 + 2)^2} \); Second derivative: \( h''(s) = \frac{6s^4 + 8s^2 - 8}{(s^2 + 2)^4} \).

Step 1: Recognize the Function Form

The function given is \( h(s) = \frac{1}{s^2 + 2} \). This can be rewritten as \( h(s) = (s^2 + 2)^{-1} \) by using the property of exponents for reciprocal terms.

Step 2: Apply the Power Rule for the First Derivative

To find \( h'(s) \), first use the power rule: if \( f(x) = (g(x))^n \), then \( f'(x) = n(g(x))^{n-1}g'(x) \). For \( h(s) = (s^2 + 2)^{-1} \), we have \( n = -1 \) and \( g(s) = s^2 + 2 \). Thus, \( g'(s) = 2s \). Now, apply the power rule: \[ h'(s) = -1 imes (s^2 + 2)^{-2} imes 2s = -\frac{2s}{(s^2 + 2)^2} \].

Step 3: Apply the Quotient Rule for the Second Derivative

The derivative \( h'(s) = -\frac{2s}{(s^2 + 2)^2} \) can be differentiated using the quotient rule. For \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \) where \( u(s) = -2s \) and \( v(s) = (s^2 + 2)^2 \). Calculate these derivatives: \( u'(s) = -2 \), and \( v'(s) = 2(s^2 + 2)(2s) = 4s(s^2 + 2) \).

Step 4: Compute the Second Derivative

Substitute our derivatives into the quotient rule: \( h''(s) = \frac{-2(s^2 + 2)^2 - (-2s)(4s)(s^2 + 2)}{(s^2 + 2)^4} \). Simplify the expression: \[ h''(s) = \frac{-2(s^2 + 2)^2 + 8s^2(s^2 + 2)}{(s^2 + 2)^4} = \frac{-2(s^4 + 4s^2 + 4) + 8s^4 + 16s^2}{(s^2 + 2)^4} \]. Simplifying further: \( h''(s) = \frac{6s^4 + 8s^2 - 8}{(s^2 + 2)^4} \).

Key Concepts

Power Rule

The power rule is a fundamental part of calculus, especially when it comes to differentiation. It provides a simple formula to find the derivative of a power function. Suppose you have a function of the form \( f(x) = x^n \), where \( n \) is any real number.
To find its derivative, the power rule states that \( f'(x) = nx^{n-1} \). This rule simplifies the differentiation process by reducing the power by one and multiplying the result by the original power.
For instance, if our function is \( (s^2 + 2)^{-1} \), the power rule helps us calculate the derivative efficiently. Here, \( n = -1 \), and we need to also find the derivative of the inside function, \( g(s) = s^2 + 2 \), using the chain rule. This step involves differentiating the inner function, \( g'(s) = 2s \). Mixing both the power and chain rules gives us:

• Step 1: Multiply by the power: \(-1 \times (s^2 + 2)^{-2}\)
• Step 2: Multiply by the derivative of the inner function: \(2s\)
• Result: \( h'(s) = -\frac{2s}{(s^2 + 2)^2} \)

Quotient Rule

The quotient rule is a crucial technique in calculus when dealing with derivatives of functions that are expressed as a ratio of two functions. It is used when you have a function of the form \( f(x) = \frac{u(x)}{v(x)} \).
The quotient rule formula for the derivative is:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \] This means you'll differentiate the top function \( u(x) \) and the bottom function \( v(x) \), then substitute them into the formula.
In our exercise, we're given \( h'(s) = -\frac{2s}{(s^2 + 2)^2} \). To find the second derivative, we identify:

• \( u(s) = -2s \)
• \( v(s) = (s^2 + 2)^2 \)
• Derivatives are \( u'(s) = -2 \) and \( v'(s) = 4s(s^2+2) \)

Substituting these into the quotient rule gives:

• Numerator: \(-2(s^2+2)^2 - (-2s)(4s)(s^2+2)\)
• Denominator: \((s^2+2)^4\)

Finally, by simplifying the result, we find the second derivative:
\( h''(s) = \frac{6s^4 + 8s^2 - 8}{(s^2 + 2)^4} \).

Calculus Problem Solving

When tackling calculus problems, especially derivatives, it is important to break the problem into smaller manageable steps. This process involves identifying applicable rules, then applying them systematically. Here is a generalized approach for tackling such problems:

• Begin with understanding the structure of the function: Identify if it requires the power, product, or quotient rule.
• Determine the sub-functions involved and how they might affect the differentiation process. Consider their derivatives.
• Apply the chosen rule(s): Ensure you methodically follow the steps related to that rule.
• Simplify: Once derivatives are found, algebraic simplification is often necessary to achieve the cleanest form.

This approach is particularly useful for complex expressions, as demonstrated in our exercise. Using the power and quotient rules in tandem facilitated solving the derivatives efficiently, highlighting the importance of multiple strategies in calculus problem-solving.