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Chapter 3: Problem 29

a. Use the Quotient Rule to find the derivative of the given function. Simplify your result. b. Find the derivative by first simplifying the function. Verify that your answer agrees with part \((a)\) $$f(w)=\frac{w^{3}-w}{w}$$

Short Answer

Expert verified
Question: Find the derivative of the function \(f(w) = \frac{w^3 - w}{w}\), and verify that the answer is the same when using the Quotient Rule and when simplifying the function first. Answer: The derivative of the function is \(f'(w) = 2w\). When using both the Quotient Rule and simplifying the function before taking the derivative, the result is the same.

Step by step solution

01

Use the Quotient Rule.

The Quotient Rule states that if \(h(w) = \frac{f(w)}{g(w)}\), then: $$h'(w) = \frac{f'(w)g(w) - f(w)g'(w)}{(g(w))^2}$$ Here, \(f(w) = w^3 - w\) and \(g(w) = w\). We need to find the derivatives \(f'(w)\) and \(g'(w)\) first, then plug them into the Quotient Rule formula.
02

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

Calculate the derivatives of f(w) and g(w) using the power rule. \(f'(w) = \frac{d(w^3 - w)}{dw} = 3w^2 - 1\) \(g'(w) = \frac{d(w)}{dw} = 1\)
03

Apply the Quotient Rule to find the derivative.

Now, we will plug the functions and their derivatives into the Quotient Rule formula: \(h'(w) = \frac{f'(w)g(w)-f(w)g'(w)}{(g(w))^2} = \frac{(3w^2-1)w-(w^3-w)(1)}{w^2} = \frac{3w^3-w^2-w^3+w}{w^2}\) Simplifying the numerator: \(h'(w) = \frac{2w^3- w^2 + w}{w^2} = \frac{w(2w^2-w+1)}{w^2}\) Now, we move to part (b) and find the derivative after simplifying the function first.
04

Simplify the function.

Simplify the function f(w) by dividing both terms in the numerator by w: \(f(w) = \frac{w^3 - w}{w} = w^2 - 1\)
05

Find the derivative of the simplified function.

Now, we will find the derivative of the simplified function using the power rule: \(f'(w) = \frac{d(w^2-1)}{dw} = 2w\)
06

Verify the agreement between both methods.

From part (a), we have: \(h'(w) = \frac{w(2w^2-w+1)}{w^2}\) Now notice that when \(w\neq 0\), we have \(f'(w)=2w\). And when \(w=0\), both \(h'(w)\) and the simplified function have an undefined derivative. Thus, we can conclude that the result from both methods agrees, and we have successfully found the derivative of the given function using both methods.

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