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Chapter 3: Problem 62

Consider the following functions (on the given internal, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=\frac{x}{x+5}$$

Short Answer

Expert verified
#Answer#: The derivative of the inverse function is: $$\frac{d}{dx}f^{-1}(x) = \frac{5x}{(1-x)^2}$$

Step by step solution

01

Express \(x\) in terms of \(y\)

Let \(y=f(x)=\frac{x}{x+5}\). To find the inverse function, we will solve this equation for \(x\). We will start by cross-multiplying: $$x+5y = xy$$ Now, let's move the term containing \(x\) to one side and the term containing \(y\) to the other side: $$x-xy= -5y$$ Factor out \(x\) from the left side of the equation: $$x(1-y) = -5y$$ Now, solve for x: $$x = \frac{-5y}{1-y}$$
02

Write the inverse function

We've expressed \(x\) in terms of \(y\), which gives us the inverse function. Swapping \(x\) and \(y\), we have: $$f^{-1}(x) = \frac{-5x}{1-x}$$
03

Differentiate the inverse function

Now we'll find the derivative of the inverse function with respect to \(x\). We'll apply the quotient rule, which states that: $$\frac{d}{dx}\left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$$ Let \(u(x) = -5x\) and \(v(x) = 1-x\). Then, \(u'(x) = -5\) and \(v'(x) = -1\). Applying the quotient rule, we get: $$\frac{d}{dx}f^{-1}(x) = \frac{-5(1-x) - (-5x)(-1)}{(1-x)^2}$$ Simplifying the expression, we have: $$\frac{d}{dx}f^{-1}(x) = \frac{5x}{(1-x)^2}$$ So, the derivative of the inverse function is: $$\frac{d}{dx}f^{-1}(x) = \frac{5x}{(1-x)^2}$$

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Key Concepts

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

Finding Inverse Functions
Understanding how to find inverse functions is a fundamental skill in calculus and algebra. An inverse function essentially reverses the action of the original function. To find the inverse of a function, you swap the roles of the independent variable (usually denoted as x) and the dependent variable (usually y), and then solve for the new dependent variable in terms of the independent variable.

To approach this, we consider the function y = f(x) and then interchange x and y to obtain x = f-1(y). The next step is to solve the equation for y, giving us the inverse function, f-1(x). It's essential that the function is one-to-one, meaning that for every y, there is exactly one x, to ensure the inverse is also a function.

For the given exercise, the process involved cross-multiplication and factoring to isolate x and express it in terms of y. After manipulation, the inverse function f-1(x) was determined. The ability to accurately find an inverse is an invaluable asset when working with functions and their interactions.
Quotient Rule
The Quotient Rule is a key concept in differential calculus and it applies when you're differentiating a function that is the quotient of two other functions. In simplified terms, if you have a function h(x) = u(x)/v(x), where both u and v are differentiable functions of x, the quotient rule can be employed to find h'(x), the derivative of h with respect to x.

The formula for the quotient rule is:
\(h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\)
When applying this formula, you need to calculate the derivatives of the numerator u(x) and the denominator v(x) separately before plugging them into the formula. This step is crucial for finding the derivative of the quotient of functions. In the exercise, the quotient rule was applied to find the derivative of the inverse function f-1(x), requiring careful application and simplification to ensure accuracy.
Differentiation of Inverse Functions
Differentiation of inverse functions involves finding the derivative of a function that has been obtained as an inverse of another given function. This process often requires utilizing other differentiation rules, such as the quotient rule, as seen in the exercise.

It's essential to understand that the derivative of an inverse function at a point is the reciprocal of the derivative of the original function at the corresponding point. This relationship only holds if the original function is differentiable at that point and its derivative is not zero. The derivative of the inverse function tells us how rapidly the inverse function is changing with respect to x.

In the example provided, after finding the inverse function, the quotient rule was applied to differentiate it. The result is a new function that gives the slope of the tangent line to the inverse function at any point x. This slope function is vital when analyzing the behavior of the inverse function and its rate of change.

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Most popular questions from this chapter

Find \(\frac{d y}{d x},\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\)

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12},$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times at which energy use is a minimum or maximum.

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

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

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 they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$
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