Problem 20 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(s)=\frac{2(3-s)^{2}}{s^{2}+(7 s-1)^{2}}\)

Step-by-Step Solution

Answer

Differentiate using the quotient rule, and the derivative is \( h'(s) = \frac{-4(3-s)(s^2 + (7s-1)^2) - 2(3-s)^2(100s - 14)}{(s^2+(7s-1)^2)^2} \).

Step 1: Identify the Differentiation Technique

Given the function \(h(s) = \frac{2(3-s)^{2}}{s^{2}+(7s-1)^{2}}\), recognize that it is a rational function (a fraction of two functions). We'll need to use the quotient rule for differentiation: if \( u(s) \) and \( v(s) \) are functions of \( s \), then \( \frac{d}{ds} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).

Step 2: Apply the Quotient Rule

Identify \( u(s) = 2(3-s)^2 \) and \( v(s) = s^2 + (7s-1)^2 \). Compute \( u'(s) \) and \( v'(s) \) next, and then substitute them into the quotient rule formula.

Step 3: Differentiate the Numerator

To find \( u'(s) = \frac{d}{ds}[2(3-s)^2] \), use the chain rule. Start by differentiating \((3-s)^2\) as \(-2(3-s)\) and multiply by 2: \[ u'(s) = 2 \cdot 2 \cdot (-1)(3-s) = -4(3-s) \].

Step 4: Differentiate the Denominator

To find \( v'(s) = \frac{d}{ds}[s^2 + (7s-1)^2] \), differentiate each term: \( \frac{d}{ds}[s^2] = 2s \). For \((7s-1)^2\), use the chain rule: \( 2(7s-1)(7) = 14(7s-1) \).So, \[ v'(s) = 2s + 14(7s - 1) = 2s + 98s - 14 = 100s - 14 \].

Step 5: Substitute into the Quotient Rule

Now substitute back into the quotient rule: \[ h'(s) = \frac{-4(3-s)(s^2 + (7s-1)^2) - 2(3-s)^2(100s - 14)}{(s^2+(7s-1)^2)^2} \].

Step 6: Simplify the Result

When simplifying, carefully expand each term and combine like terms to achieve a concise expression for the derivative. Keep in mind that cleaning complex algebraic expressions might require persistence and attention to detail but provides an elegant final result.

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used for differentiating composite functions. When you have a function nested within another function, the chain rule comes to the rescue. Imagine you have a function \( g(x) = (f(x))^2 \). If you want to differentiate \( g(x) \), you don't need to expand and simplify it first. Instead, you use the chain rule.

This rule states that the derivative of \( g(x) \) is the derivative of the outer function evaluated at the inner function \( f(x) \), then multiplied by the derivative of the inner function \( f(x) \). In mathematical terms, it's expressed as:

• \[ g'(x) = 2f(x) imes f'(x) \]

This is exactly the method applied when differentiating \( u(s) = 2(3-s)^{2} \) in the original exercise, where \( (3-s) \) is embedded within the squaring operation. The chain rule provides a systematic way of tackling derivatives of such composite structures with ease and accuracy.

Differentiation Techniques

Differentiation is a cornerstone concept in calculus, allowing us to understand how functions change. Several techniques are used in differentiating various types of functions, such as the quotient rule, product rule, and chain rule.

In the given function \( h(s) = \frac{2(3-s)^2}{s^2 + (7s-1)^2} \), the quotient rule is essential for differentiating the entire expression because of its rational function nature. However, within this process, other rules like the chain rule are used specifically for differentiating parts of the function.

• **Quotient Rule:** Appropriately applied when dealing with \( \frac{u(s)}{v(s)} \), ensuring appropriate handling of each part's derivative in relation to the other parts.
• **Chain Rule:** As discussed earlier, useful for differentiating composite expressions like \( (3-s)^2 \) and \( (7s-1)^2 \).

Different techniques often merge in complicated expressions, making their understanding key to successful calculus problem solving. Each method serves a purpose, fine-tuning the differentiation for specific forms of functions within larger expressions.

Rational Functions

Rational functions are quotients of two polynomials, appearing frequently in calculus. They are highly familiar to come across in differential calculus exercises.

• A rational function can be represented as \( \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials.

In differentiation, tools like the quotient rule become crucial. This rule permits the effective differentiation of rational expressions by focusing on upper and lower functions separately.

In the given example, differentiating \( h(s) = \frac{2(3-s)^{2}}{s^{2}+(7s-1)^{2}} \) requires focusing on the numerator and the denominator separately, computing their derivatives, and substituting them into the quotient rule.

Understanding the nature of rational functions and being able to handle them analytically enhances problem-solving skills and prepares you for complex calculus challenges.