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

Short Answer

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

Step by step solution

01

Apply the quotient rule

The quotient rule for differentiation is: $$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$$ In our case, \(u(x) = f(x)\), \(v(x) = x + 2\), \(u'(x) = f'(x)\), and \(v'(x) = 1\). So, applying the quotient rule, we get: $$\frac{d}{dx}\left(\frac{f(x)}{x+2}\right) = \frac{f'(x)(x+2) - f(x)}{(x+2)^2}$$
02

Evaluate the derivative at x = 4

Now, we will substitute the value \(x=4\) and find the corresponding values of \(f(x)\) and \(f'(x)\) from the table. Looking at the table, when \(x=4\), we have: - \(f(4) = 2\) - \(f'(4) = 1\) Let's put these values into our derivative formula: $$\left.\frac{d}{d x}\left(\frac{f(x)}{x+2}\right)\right|_{x=4} = \frac{1(4+2) - 2}{(4+2)^2}$$
03

Simplify and solve

Now, we will simplify the expression and compute the result: $$\left.\frac{d}{d x}\left(\frac{f(x)}{x+2}\right)\right|_{x=4} = \frac{1(6) - 2}{(6)^2} = \frac{4}{36} = \frac{1}{9}$$ So, the derivative of \(\frac{f(x)}{x+2}\) evaluated at the point \(x=4\) 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
Understanding the quotient rule is essential when dealing with calculus problems involving the division of two functions. It is a procedure used to find the derivative of a quotient of two differentiable functions. When given a function that is formulated as a fraction where both the numerator and denominator are themselves functions, we can't simply differentiate the top and the bottom separately. Instead, we apply the quotient rule.

The quotient rule says that for two functions, let’s say \(u(x)\) as the numerator and \(v(x)\) as the denominator, the derivative of the division \(u(x)/v(x)\) is given by:
\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\]
This formula might look intimidating at first, but once you start applying it step by step, it becomes more manageable. It's about taking the derivative of the top function \(u'(x)\), multiplying it by the bottom function \(v(x)\), then subtracting the product of the top function and the derivative of the bottom function \(v'(x)\), all over the square of the bottom function.
Derivative Evaluation
After finding the general form of the derivative using rules like the quotient rule, the next step is often to evaluate the derivative at a particular point. This involves substituting the value of \(x\) into the derivative you've computed to find the slope of the tangent line at that point on the graph of your function. It's like taking a snapshot of the instantaneous rate of change at a specific moment.

In the provided exercise, we evaluate the derivative \(\frac{d}{dx}(\frac{f(x)}{x+2})\) at the point \(x=4\). After applying the quotient rule to get the derivative, we replace \(x\) with 4 and also look up the values of \(f(4)\) and \(f'(4)\) from the given table. By plugging these numbers into our derivative formula, we obtain the derivative's value at the desired point.

This practice of evaluation is powerful as it gives concrete information about the behavior of functions at specific points, which is crucial in many applied math problems, such as in physics for velocity and acceleration calculations.
Calculus Problems
Calculus problems can often seem daunting due to their abstract nature and reliance on a solid understanding of multiple mathematical concepts. To succeed, it's important to break down the problem into manageable steps and understand the rules and theorems that apply, such as the quotient rule or the product rule for derivatives.

The first step is usually to identify the kind of problem you're faced with: is it a differentiation problem, an integration problem, a limit problem, or something else? Each type of problem has its own set of strategies and applicable rules. For example, in differentiation problems, it's crucial to recognize which rule to apply in finding the derivative, be it the chain rule, product rule, or quotient rule.

Working through problems step by step, as seen in the textbook solution, is a great way to ensure that you understand the underlying concepts and know how to apply them. Practice is also key — the more problems you work through, the more familiar you'll become with the types of questions that can be asked and the methods needed to solve them. Also, remember to check your work for errors by reviewing each step, preferably using a different method to confirm your results if possible.

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Most popular questions from this chapter

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

A study conducted at the University of New Mexico found that the mass \(m(t)\) (in grams) of a juvenile desert tortoise \(t\) days after a switch to a particular diet is described by the function \(m(t)=m_{0} e^{0.004 t},\) where \(m_{0}\) is the mass of the tortoise at the time of the diet switch. If \(m_{0}=64\) evaluate \(m^{\prime}(65)\) and interpret the meaning of this result. (Source: Physiological and Biochemical Zoology, 85,1,2012 )

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u \end{array}\right.$$ a. Show that \(\lim _{y \rightarrow u} H(v)=0\) b. For any value of \(u,\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u)$$ c. Show that $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right)$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\)

\- Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at \(9.00 ?\) (Hint: Use the Law of Cosines.)

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