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Chapter 3: Problem 26

Find the derivative of the following functions. $$h(w)=\frac{w^{2}-1}{w^{2}+1}$$

Short Answer

Expert verified
**Question**: Find the derivative of the function \(h(w) = \frac{w^2 - 1}{w^2 + 1}\). **Answer**: The derivative of the given function is \(h'(w) = \frac{4w}{(w^2 + 1)^2}\).

Step by step solution

01

Identify the functions f(w) and g(w)

In the given function, \(h(w) = \frac{w^2 - 1}{w^2 + 1}\), we have: $$f(w) = w^2 - 1 \quad \text{and} \quad g(w) = w^2 + 1$$
02

Find the derivatives of f(w) and g(w)

Using basic power rule for derivatives: $$f'(w) = \frac{d}{dw}(w^2 - 1) = 2w$$ $$g'(w) = \frac{d}{dw}(w^2 + 1) = 2w$$
03

Apply the quotient rule

Now that we have the derivatives \(f'(w)\) and \(g'(w)\), we can apply the quotient rule to find the derivative \(h'(w)\): $$h'(w) = \frac{f'(w)g(w) - f(w)g'(w)}{(g(w))^2}$$ Substitute the expressions for \(f(w)\), \(f'(w)\), \(g(w)\), and \(g'(w)\), into the formula: $$h'(w) = \frac{(2w)(w^2 + 1) - (w^2 - 1)(2w)}{((w^2 + 1))^2} $$
04

Simplify the expression

Now, let's simplify the expression by expanding and combining like terms: \begin{align*} h'(w) &= \frac{2w^3 + 2w - 2w^3 + 2w}{(w^2 + 1)^2}\\ h'(w) &= \frac{4w}{(w^2 + 1)^2} \end{align*} So, the derivative of the given function is: $$h'(w) = \frac{4w}{(w^2 + 1)^2}$$

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