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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
Answer: The derivative of the function $$f(w) = \frac{w^3 - w}{w}$$ is $$f'(w) = 2w$$.

Step by step solution

01

In this step, we will apply the Quotient Rule to find the derivative of $$f(w) = \frac{w^3 - w}{w}$$. Let $$u(w) = w^3 - w$$ and $$v(w) = w$$, then we have: $$u'(w) = \frac{d}{dw}(w^3 - w) = 3w^2 - 1$$ $$v'(w) = \frac{d}{dw}(w) = 1$$ Now, we can apply the Quotient Rule formula: $$f'(w) = \frac{u'(w)v(w) - u(w)v'(w)}{v(w)^2}$$ #Step 2: Calculate and simplify f'(w)#

Substitute u(w), v(w), u'(w), and v'(w) into the formula and simplify: $$f'(w) = \frac{(3w^2 - 1)w - (w^3 - w)(1)}{w^2}$$ $$f'(w) = \frac{3w^3 - w - w^3 + w}{w^2}$$ $$f'(w) = \frac{2w^3}{w^2}$$ So, the derivative of the given function using the Quotient Rule is: $$f'(w) = 2w$$ Now, let's move on to part (b) and find the derivative by first simplifying the function. #Step 3: Simplify the function f(w)#
02

Simplify the given function by dividing the numerator and denominator: $$f(w) = \frac{w^3 - w}{w}$$ $$f(w) = \frac{w(w^2 - 1)}{w}$$ $$f(w) = w^2 - 1$$ #Step 4: Calculate the derivative of the simplified function f(w)#

Now, we can find the derivative of the simplified function $$f(w) = w^2 - 1$$ by differentiating with respect to w: $$f'(w) = \frac{d}{dw}(w^2 - 1)$$ $$f'(w) = 2w$$ Both methods give us the same result, $$f'(w) = 2w$$. Therefore, the final answer is the derivative of f(w) is $$f'(w) = 2w$$.

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Key Concepts

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

Quotient Rule
When we want to differentiate a function where one expression is divided by another, we use the Quotient Rule. This rule is crucial because it allows us to find the derivative without first having to simplify the expression. Here鈥檚 how it works:
  • Consider a function \( f(x) = \frac{u(x)}{v(x)} \). Here, \( u(x) \) is the numerator and \( v(x) \) is the denominator.
  • The Quotient Rule states that the derivative, \( f'(x) \), is given by:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]
  • It's important to differentiate both \( u(x) \) and \( v(x) \), and then correctly apply the formula.
  • Make sure to square \( v(x) \) at the bottom, which is key to the rule.
This method applies directly to our initial function \( f(w) = \frac{w^3-w}{w} \), where the rule simplifies the derivative calculation.
Derivative
The derivative is a fundamental concept in calculus, and it measures how a function changes as its input changes. When we compute a derivative, we are essentially finding the "rate of change" of a function's output with respect to its input.
  • For a simple power function, like \( w^2 \), the derivative is found by bringing down the exponent and multiplying it by the base raised to a power reduced by one.
  • For instance, the derivative of \( w^3 \) is \( 3w^2 \), and the derivative of a constant or a linear term like \( -w \) is \(-1\).
In our example function \( f(w) = \frac{w^3 - w}{w} \), after simplification, we get \( w^2 - 1 \) as the simplified function, whose derivative is straightforward: \( 2w \), since it's just a polynomial.
Simplifying Algebraic Expressions
Simplifying algebraic expressions involves reducing the expression to its simplest form by combining like terms and eliminating any unnecessary complexity. This step can often make calculus operations like differentiation much easier.
  • In the given exercise, before applying calculus operations to the fraction \( \frac{w^3 - w}{w} \), we can simplify it.
  • This is done by factoring out common terms, resulting in \( w \cdot (w^2 - 1)/w \).
  • By cancelling the common \( w \), we are left with a simpler expression, \( w^2 - 1 \).
Once simplified, differentiating \( w^2 - 1 \) becomes a straightforward process, yielding the derivative \( 2w \). Simplification can thus serve as an effective strategy, especially for more complex calculus problems.

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Most popular questions from this chapter

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