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Let \(h(x)=f(g(x))\) and \(k(x)=g(g(x)) .\) Use the table to compute the following derivatives. a. \(h^{\prime}(1)\) b. \(h^{\prime}(2) \quad\) c. \(h^{\prime}(3)\) d. \(k^{\prime}(3)\) e. \(k^{\prime}(1)\) f. \(k^{\prime}(5)\) $$\begin{array}{lrrrrr} x & 1 & 2 & 3 & 4 & 5 \\ \hline f^{\prime}(x) & -6 & -3 & 8 & 7 & 2 \\ g(x) & 4 & 1 & 5 & 2 & 3 \\ g^{\prime}(x) & 9 & 7 & 3 & -1 & -5 \end{array}$$

Short Answer

Expert verified
a. \(h^{\prime}(1) = 63\) b. \(h^{\prime}(2) = -42\) c. \(h^{\prime}(3) = 6\) d. \(k^{\prime}(3) = -15\) e. \(k^{\prime}(1) = -9\) f. \(k^{\prime}(5) = -15\)

Step by step solution

01

a. Find \(h^{\prime}(1)\)

We have, \(h(x)=f(g(x))\). Using the chain rule, we have \(h^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Now we need to find \(f^{\prime}(g(1))\) and \(g^{\prime}(1)\). From the table, we have \(g(1) = 4\) and \(g^{\prime}(1) = 9\). Thus, we have \(h^{\prime}(1) = f^{\prime}(4) \cdot 9\). From the table, we have \(f^{\prime}(4) = 7\). So, \(h^{\prime}(1) = 7 \cdot 9 = 63\).
02

b. Find \(h^{\prime}(2)\)

Using the chain rule, we have \(h^{\prime}(2) = f^{\prime}(g(2)) \cdot g^{\prime}(2)\). From the table, we have \(g(2) = 1\) and \(g^{\prime}(2) = 7\). Thus, we have \(h^{\prime}(2) = f^{\prime}(1) \cdot 7\). From the table, we have \(f^{\prime}(1) = -6\). So, \(h^{\prime}(2) = -6 \cdot 7 = -42\).
03

c. Find \(h^{\prime}(3)\)

Using the chain rule, we have \(h^{\prime}(3) = f^{\prime}(g(3)) \cdot g^{\prime}(3)\). From the table, we have \(g(3) = 5\) and \(g^{\prime}(3) = 3\). Thus, we have \(h^{\prime}(3) = f^{\prime}(5) \cdot 3\). From the table, we have \(f^{\prime}(5) = 2\). So, \(h^{\prime}(3) = 2 \cdot 3 = 6\).
04

d. Find \(k^{\prime}(3)\)

We have, \(k(x)=g(g(x))\). Using the chain rule, we have \(k^{\prime}(x) = g^{\prime}(g(x)) \cdot g^{\prime}(x)\). Now we need to find \(g^{\prime}(g(3))\) and \(g^{\prime}(3)\). From the table, we have \(g(3) = 5\) and \(g^{\prime}(3) = 3\). Thus, we have \(k^{\prime}(3) = g^{\prime}(5) \cdot 3\). From the table, we have \(g^{\prime}(5) = -5\). So, \(k^{\prime}(3) = -5 \cdot 3 = -15\).
05

e. Find \(k^{\prime}(1)\)

Using the chain rule, we have \(k^{\prime}(1) = g^{\prime}(g(1)) \cdot g^{\prime}(1)\). From the table, we have \(g(1) = 4\) and \(g^{\prime}(1) = 9\). Thus, we have \(k^{\prime}(1) = g^{\prime}(4) \cdot 9\). From the table, we have \(g^{\prime}(4) = -1\). So, \(k^{\prime}(1) = -1 \cdot 9 = -9\).
06

f. Find \(k^{\prime}(5)\)

Using the chain rule, we have \(k^{\prime}(5) = g^{\prime}(g(5)) \cdot g^{\prime}(5)\). From the table, we have \(g(5) = 3\) and \(g^{\prime}(5) = -5\). Thus, we have \(k^{\prime}(5) = g^{\prime}(3) \cdot (-5)\). From the table, we have \(g^{\prime}(3) = 3\). So, \(k^{\prime}(5) = 3 \cdot (-5) = -15\).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. A composite function is simply a function made by combining other functions, like nesting one function inside another. For example, if you have a function like \( h(x) = f(g(x)) \), this means you apply \( g(x) \) first, then \( f(x) \). The chain rule helps us find the derivative of such functions.The formula for the chain rule is:
  • If \( h(x) = f(g(x)) \), then the derivative \( h'(x) = f'(g(x)) \cdot g'(x) \).
You start by taking the derivative of the outer function \( f \), evaluated at \( g(x) \), and multiply it by the derivative of the inner function \( g \). This makes it possible to "chain" the differentiation of nested functions. By using the chain rule, you can effectively handle more complex derivative calculations.
Derivative
A derivative in calculus represents the rate at which a function is changing at any given point. It's essentially the "slope" of the function at a particular point on a curve. For functions in calculus, derivatives provide a way to understand how a function behaves, such as increasing or decreasing, at specific points.When you have a function \( y = f(x) \), the derivative is often written as \( f'(x) \) or \( \frac{dy}{dx} \). This notation indicates how \( y \) changes with a small change in \( x \).Derivatives are crucial because they allow us to:
  • Determine the slope of a curve at any point.
  • Find local maxima and minima of a function, which are important in optimization problems.
  • Analyze the concavity, or curvature, of a function.
By understanding derivatives, you gain insights into the behavior and characteristics of functions.
Composite Function
A composite function is formed when one function is applied to the result of another function. It is like nesting functions, applying one inside the other. For example, if \( f(x) \) and \( g(x) \) are two functions, the composite function \( h(x) = f(g(x)) \) is formed.Composite functions are essential in calculus because they allow us to create more complex functions from simpler ones. Understanding how to work with them is critical when using the chain rule for differentiation. You can visualize it like:
  • Imagine \( g(x) \) as being processed first, then \( f \) takes the result of \( g(x) \).
In practice, composite functions often appear in more advanced mathematical scenarios, like physics and engineering problems, making them fundamental in applied mathematics.
Differentiation Table
A differentiation table is a useful tool when calculating derivatives, especially for exercises involving multiple functions evaluated at different points. A table format helps organize various values, such as derivatives of functions and their inner functions, to simplify these computations.In the original exercise, the table provided values for \( f'(x) \), \( g(x) \), and \( g'(x) \) at specific points. This information enabled easy and quick reference to apply the chain rule effectively. Here's why differentiation tables are helpful:
  • They present data neatly, making it easier to apply different rules and operations.
  • They are an effective way to manage values in solving calculus problems accurately.
  • They assist in keeping track of multiple calculations and steps systematically.
Using a differentiation table can significantly enhance your ability to work through complex derivative problems efficiently and accurately.

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Most popular questions from this chapter

Suppose you own a fuel-efficient hybrid automobile with a monitor on the dashboard that displays the mileage and gas consumption. The number of miles you can drive with \(g\) gallons of gas remaining in the tank on a particular stretch of highway is given by \(m(g)=50 g-25.8 g^{2}+12.5 g^{3}-1.6 g^{4},\) for \(0 \leq g \leq 4\). a. Graph and interpret the mileage function. b. Graph and interpret the gas mileage \(m(g) / \mathrm{g}\). c. Graph and interpret \(d m / d g\).

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

An observer stands \(20 \mathrm{m}\) from the bottom of a 10 -m-tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of \(\pi \mathrm{rad} / \mathrm{min},\) and the observer's line of sight with a specific seat on the wheel makes an angle \(\theta\) with the ground (see figure). Forty seconds after that seat leaves the lowest point on the wheel, what is the rate of change of \(\theta ?\) Assume the observer's eyes are level with the bottom of the wheel.

Horizontal tangents The graph of \(y=\cos x \cdot \ln \cos ^{2} x\) has seven horizontal tangent lines on the interval \([0,2 \pi] .\) Find the approximate \(x\) -coordinates of all points at which these tangent lines occur.

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \left(\sec ^{4} x \tan ^{2} x\right)$$

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