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Chapter 3: Problem 16

Find the first derivative. $$ g(w)=\frac{(w-1)(w-3)}{(w+1)(w+3)} $$

Short Answer

Expert verified
The first derivative is \( g'(w) = \frac{4w^3 + 2w^2 - 24w}{((w+1)(w+3))^2} \).

Step by step solution

01

Identify Derivative Rule

To find the derivative of the function \(g(w) = \frac{(w-1)(w-3)}{(w+1)(w+3)}\), we'll use the quotient rule for derivatives. The quotient rule states that if you have a function \( \frac{u(w)}{v(w)} \), its derivative is \( \frac{u'(w)v(w) - u(w)v'(w)}{(v(w))^2} \). Identify \( u(w) \) as \((w-1)(w-3)\) and \( v(w) \) as \((w+1)(w+3)\).
02

Find Derivative of u(w)

The function \( u(w) = (w-1)(w-3) \) can be expanded to \( w^2 - 4w + 3 \). Find the derivative, \( u'(w) = \frac{d}{dw}(w^2 - 4w + 3) = 2w - 4 \).
03

Find Derivative of v(w)

Similarly, the function \( v(w) = (w+1)(w+3) \) can be expanded to \( w^2 + 4w + 3 \). Find the derivative, \( v'(w) = \frac{d}{dw}(w^2 + 4w + 3) = 2w + 4 \).
04

Apply Quotient Rule

Using the quotient rule, the derivative of \( g(w) \) is given by: \[g'(w) = \frac{(2w - 4)(w+1)(w+3) - (w-1)(w-3)(2w + 4)}{((w+1)(w+3))^2}\].
05

Simplify the Expression

First, expand each term in the numerator. Calculate \((2w - 4)(w+1)(w+3)\) and \((w-1)(w-3)(2w + 4)\), then simplify:- For \((2w - 4)(w+1)(w+3)\): Expand stepwise, resulting in \(2w^3 + 10w^2 + 12w - 12\).- For \((w-1)(w-3)(2w + 4)\): Expand stepwise, resulting in \(2w^3 - 8w^2 - 12w + 12\).Subtract these expressions to simplify.
06

Final Simplification

After simplification, combine like terms to obtain a simplified expression for the numerator. Check and simplify further if needed. Then provide the final expression for \( g'(w) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
The quotient rule is a powerful tool in calculus used to find derivatives of functions that are given as a ratio of two other functions. If you have a function in the form \( \frac{u(w)}{v(w)} \), the derivative \( \frac{d}{dw}\left( \frac{u(w)}{v(w)} \right) \) is calculated using the formula:
  • \( \frac{u'(w)v(w) - u(w)v'(w)}{(v(w))^2} \)
Here鈥檚 what the formula tells us:
  • Differentiate the numerator \( u(w) \) to get \( u'(w) \)
  • Differentiate the denominator \( v(w) \) to get \( v'(w) \)
  • Multiply \( u'(w) \) by \( v(w) \) and \( u(w) \) by \( v'(w) \)
  • Subtract the second product from the first
  • Divide by the square of the denominator \( (v(w))^2 \)
This approach allows us to handle functions that would be otherwise daunting, especially when both parts \( u(w) \) and \( v(w) \) are more complex expressions.
Derivative of a Polynomial
A polynomial is an algebraic expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. Finding the derivative of a polynomial is straightforward. You apply basic rules which simplify down to:
  • The power rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \)
  • Sum rule: The derivative of a sum is the sum of the derivatives
  • Constant rule: The derivative of a constant is zero
For instance, deriving \( u(w) = (w-1)(w-3) \) involves expanding it to \( w^2 - 4w + 3 \), then applying:
  • \( \frac{d}{dw}(w^2) = 2w \)
  • \( \frac{d}{dw}(-4w) = -4 \)
  • \( \frac{d}{dw}(3) = 0 \)
This gives \( u'(w) = 2w - 4 \). This clarity and process structure make working with polynomials highly approachable for finding derivatives.
Simplification of Expressions
Simplifying mathematical expressions is essential in calculus to make outcomes readable and comprehensible. After deriving or transforming expressions, complexity might arise, which means simplifying is a necessary step. Here's why simplification matters and how it鈥檚 done:
  • Reduce expressions by combining like terms and eliminating redundant components.
  • Avoid unwieldy numbers by canceling out terms wherever possible, provided the operations are valid.
  • Simplified expressions not only look neater but are also easier to interpret and apply in further mathematical operations.
For example, after applying the quotient rule to \( g(w) \), you get:\[g'(w) = \frac{(2w^3 + 10w^2 + 12w - 12) - (2w^3 - 8w^2 - 12w + 12)}{((w+1)(w+3))^2}\]By expanding and simplifying the numerator by canceling terms where possible, the expression becomes clearer. Always try to reach the simplest form to easily understand and solve mathematical problems.

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