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Chapter 3: Problem 77

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{f(x)}{(x+2)}\right]\right|_{x=4}$$

Short Answer

Expert verified
Answer: \(\frac{1}{9}\)

Step by step solution

01

(Quotient Rule)

The quotient rule states that if we have a function \(h(x) = \frac{f(x)}{g(x)}\), then the derivative of h(x) with respect to x is given by: \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\). In our case, \(g(x) = x+2\), and so we'll apply the quotient rule to calculate the derivative of \(\frac{f(x)}{(x+2)}\). #Step 2: Differentiate the function#
02

(Differentiate \(\frac{f(x)}{x+2}\))

Using the quotient rule, we have: $$\left.\frac{d}{d x}\left[\frac{f(x)}{(x+2)}\right]\right|_{x=4} = \frac{f^{\prime}(x)(x+2) - f(x)(1)}{((x+2))^2}$$ Now we need to evaluate this derivative at x=4. #Step 3: Evaluate the derivative at x=4#
03

(Substitute x = 4)

Plugging x=4 into the expression we found in Step 2, we get: $$\frac{f^{\prime}(4)(4+2) - f(4)(1)}{((4+2))^2}$$ #Step 4: Use the table to find the function values at x=4#
04

(Find \(f(4)\) and \(f^{\prime}(4)\) from the table)

From the given table, we know that \(f(4) = 2\) and \(f^{\prime}(4) = 1\). Substitute these values into the expression: $$\frac{(1)(6) - (2)(1)}{(6)^2}$$ #Step 5: Simplify the expression#
05

(Simplify)

Finally, we can simplify the expression to get the value of the derivative at x=4: $$\frac{6 - 2}{36} = \frac{4}{36} = \frac{1}{9}$$ So the derivative of \(\frac{f(x)}{x+2}\) evaluated at x=4 is \(\frac{1}{9}\).

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Key Concepts

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

Quotient Rule
When dealing with derivatives of functions that are divided by each other, the quotient rule is your go-to tool in calculus. It鈥檚 especially useful when you have a function in the form of a quotient: one function divided by another, such as \( h(x) = \frac{f(x)}{g(x)} \). The quotient rule states that the derivative, \( h'(x) \), is given by:
\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
This formula might look complex, but it's just a systematic approach to differentiate a quotient. Here are the steps simplified:
  • Differentiate the numerator function \( f(x) \) to get \( f'(x) \).
  • Leave the denominator function \( g(x) \) as it is and multiply it by \( f'(x) \).
  • Now, differentiate the denominator \( g(x) \) to get \( g'(x) \).
  • Multiply the numerator \( f(x) \) with \( g'(x) \).
  • Subtract the second product from the first.
  • Finally, divide the result by the square of the denominator \( g(x) \).
Understanding the quotient rule is essential for analyzing functions expressed as ratios, streamlining calculations, and avoiding algebraic missteps.
Function Evaluation
Function evaluation involves substituting specific values into a function to find outputs or results. When you have an expression or function like \( f(x) \), evaluating it at a specific point means to replace \( x \) with the given number. For example, evaluating \( f(x) \) at \( x = 4 \) requires finding \( f(4) \).
Using the exercise, we need to find values from a given table at the specified \( x \) value. Let's do that:
  • Look up or calculate \( f(4) \) and \( f'(4) \) using the information provided.
  • Since the table gives \( f(4) = 2 \) and \( f'(4) = 1 \), these are crucial for further calculations.
Efficient and straightforward function evaluation is critical because it provides the necessary values for simplifying more complex expressions or solving problems.
Simplifying Expressions
In calculus, you're often faced with complex expressions that need simplification to make them manageable or to find a solution. Take the derivative of \( \frac{f(x)}{(x+2)} \) evaluated at \( x=4 \). We end up with an expression like \( \frac{6 - 2}{(6)^2} \) after substitution and application of the quotient rule.
Let's break this down:
  • Combine like terms: \( 6 - 2 = 4 \).
  • Calculate the denominator: square of 6 is 36.
  • Simplify the fraction: \( \frac{4}{36} \).
  • Finally, reduce the fraction to \( \frac{1}{9} \).
This step-by-step simplification ensures accurate calculation. Shrinking complex expressions into simpler forms is essential for clarity and for drawing correct conclusions in mathematical solutions.
Calculus Problem Solving
Calculus problem solving is about combining knowledge, rules, and techniques to tackle calculus problems effectively. Start by determining what's being asked. Identify the relevant rules: here, the quotient rule is key.
Step-by-step approaches are critical:
  • Use the correct differentiation formula for the problem (e.g., quotient rule).
  • Evaluate functions and derivatives at specific values using given data.
  • Substitute these values into the formula.
  • Simplify the resulting expressions carefully, as simplification errors can lead to wrong answers.
Example problems can weave together various calculus concepts and require systematic thinking to solve. By applying the right strategies, calculus problems become more manageable and less intimidating, honing problem-solving skills effectively.

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Most popular questions from this chapter

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right);\) \(\left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli)

Find the derivative of the inverse of the following functions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\) $$f(x)=x^{2}+1, \text { for } x \geq 0 ;(5,2)$$

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}(x f(x))\right|_{x=3}$$

A lighthouse stands 500 m off of a straight shore, the focused beam of its light revolving four times each minute. As shown in the figure, \(P\) is the point on shore closest to the lighthouse and \(Q\) is a point on the shore 200 m from \(P\). What is the speed of the beam along the shore when it strikes the point \(Q ?\) Describe how the speed of the beam along the shore varies with the distance between \(P\) and \(Q\). Neglect the height of the lighthouse.

Proof of \(\frac{d}{d x}(\cos x)=-\sin x\) Use the definition of the derivative and the trigonometric identity $$ \cos (x+h)=\cos x \cos h-\sin x \sin h $$ to prove that \(\frac{d}{d x}(\cos x)=-\sin x\)
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