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Chapter 3: Problem 84

Special Quotient Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\).

Short Answer

Expert verified
Question: Find two nonconstant functions, \(f(x)\) and \(g(x)\), such that the derivative of their quotient is equal to the quotient of their derivatives. Answer: One example of such functions is \(f(x) = e^{ax} - ce^{ax}\) and \(g(x) = e^{ax}\), where \(a\) and \(c\) are constants.

Step by step solution

01

Quotient Rule

Recall the quotient rule for differentiation: 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: \(h^{\prime}(x) = \frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{g^2(x)}\) So, in our case, we want to find functions \(f(x)\) and \(g(x)\) such that: \(\frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{g^2(x)} = \frac{f^{\prime}(x)}{g^{\prime}(x)}\)
02

Finding Conditions for Special Quotient Rule

Now we want to find the conditions on \(f(x)\) and \(g(x)\) that would make the above equation true. This can be done by cross-multiplying the denominators, which gives us: \(f^{\prime}(x)g(x)g^{\prime}(x) - f(x)g^{\prime}(x)g^{\prime}(x) = f^{\prime}(x)g^2(x)\) We can simplify this by canceling out a common factor of \(f^{\prime}(x)g^{\prime}(x)\): \(g(x) - f(x) = \frac{f^{\prime}(x)}{g^{\prime}(x)}g^2(x)\) We can now separate the functions \(f(x)\) and \(g(x)\) by using the following substitution: \(g(x) - f(x) = k(x) \implies f(x) = g(x) - k(x)\), where \(k(x)\) is some function.
03

Finding Examples of Functions

Now we plug in our substitution \(f(x) = g(x) - k(x)\) into the conditions we found in Step 2: \(\frac{g^{\prime}(x) - k^{\prime}(x)}{g^{\prime}(x)}g^2(x) = k(x)\) This condition tells us that the function \(k(x)\) is such that its derivative is proportional to the function itself. One function that has this property is an exponential function, for example, \(k(x) = c\cdot e^{ax}\), where \(a\) and \(c\) are constants. The derivative of this function is: \(k^{\prime}(x) = cae^{ax}\) Now let's choose a function for \(g(x)\), for example, \(g(x) = e^{ax}\). Then, the derivative of \(g(x)\) is: \(g^{\prime}(x) = ae^{ax}\) Thus, we have found a pair of functions \(f(x) = g(x) - k(x) = e^{ax} - ce^{ax}\) and \(g(x) = e^{ax}\) that satisfy the special quotient rule: \(\frac{\frac{d}{dx}(e^{ax} - ce^{ax})}{\frac{d}{dx}(e^{ax})} = \frac{e^{ax} - ce^{ax}}{e^{ax}}\)

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus and mathematical analysis. It involves finding the rate at which a function changes at any given point. This process is known as finding the derivative of a function. In simple terms, differentiation allows us to understand how a small change in the independent variable (usually denoted as 'x') affects the dependent variable (usually 'y').

The basic idea of differentiation is to compute the slope of the tangent line at any point on a curve. This slope represents the instantaneous rate of change of the function at that point. In practice, differentiation can be applied to various functions using different rules such as the power rule, the product rule, and the quotient rule.

In this specific exercise, the focus is on the quotient rule, a method used to differentiate functions that are ratios of two other functions. This rule helps determine the derivative of a quotient, providing an understanding of how to handle division in differentiation.
Derivative
The derivative of a function is a measure that captures how a function's output value changes in response to changes in its input value. Mathematically, the derivative of a function at a point provides the slope of the tangent to the curve defined by that function.

Let's consider a function expressed as \(f(x)\). Its derivative, represented by \(f'(x)\), tells us the rate of change of \(f(x)\) with respect to \(x\). In calculus, this rate of change is crucial for interpreting physical phenomena like velocity, where the rate of change of distance with respect to time is involved. The sign and magnitude of the derivative provide insights:
  • A positive derivative indicates that the function is increasing at that point.
  • A negative derivative means the function is decreasing.
  • A zero derivative signals potential local maximum or minimum points, implying a flat tangent line.

The process of finding derivatives can leverage several rules and formulas to make calculations straightforward for complex functions, such as employing the product rule, chain rule, and, significantly for this exercise, the quotient rule.
Exponential Function
Exponential functions are mathematical functions of the form \(f(x) = a \cdot e^{bx}\), where \(e\) is a mathematical constant approximately equal to 2.71828. These functions model growth or decay processes, like population growth or radioactive decay.

In calculus and differentiation, exponential functions have a unique property: the derivative of the exponential function \(e^{ax}\) is \(a \cdot e^{ax}\). This means that the rate of change of the function is directly proportional to its current value, a phenomenon tied to exponential growth.

For the given exercise, the use of exponential functions plays a crucial role. By setting \(k(x) = c \cdot e^{ax}\) and \(g(x) = e^{ax}\), a solution is crafted that meets the condition for the special quotient rule. Here, the derivative of an exponential function mirrors the simplicity of using exponential terms to precisely satisfy specific mathematical properties needed in differentiating quotients.

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Most popular questions from this chapter

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this implies that the rule is true for \(k=n+1 .\) (Hint: Write \(e^{(n+1) x}\) as the product of two functions, and use the Product Rule.)

Calculate the following derivatives using the Product Rule. $$\begin{array}{lll} \text { a. } \frac{d}{d x}\left(\sin ^{2} x\right) & \text { b. } \frac{d}{d x}\left(\sin ^{3} x\right) & \text { c. } \frac{d}{d x}\left(\sin ^{4} x\right) \end{array}$$ d. Based upon your answers to parts (a)-(c), make a conjecture about \(\frac{d}{d x}\left(\sin ^{n} x\right),\) where \(n\) is a positive integer. Then prove the result by induction.

A rope passing through a capstan on a dock is attached to a boat offshore. The rope is pulled in at a constant rate of \(3 \mathrm{ft} / \mathrm{s}\) and the capstan is \(5 \mathrm{ft}\) vertically above the water. How fast is the boat traveling when it is \(10 \mathrm{ft}\) from the dock?

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.

The Witch of Agnesi The graph of \(y=\frac{a^{3}}{x^{2}+a^{2}},\) where \(a\) is a constant, is called the witch of Agnesi (named after the 18th-century Italian mathematician Maria Agnesi). a. Let \(a=3\) and find an equation of the line tangent to \(y=\frac{27}{x^{2}+9}\) at \(x=2\) b. Plot the function and the tangent line found in part (a).
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