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Use a computer algebra system to differentiate the function. \(g(x)=\left(\frac{x+1}{x+2}\right)(2 x-5)\)

Short Answer

Expert verified
The derivative of the function \( g(x) = \left(\frac{x+1}{x+2}\right)(2x-5) \) is \( g'(x) = \frac{2}{(x+2)^2} - \frac{5}{(x+2)^2} + \frac{2x+2}{x+2} \).

Step by step solution

01

Recognizing the Function Composition

The given function \( g(x) = \left(\frac{x+1}{x+2}\right)(2x-5) \) can be seen as a product of two smaller functions \( u(x) = \frac{x+1}{x+2} \) and \( v(x) = 2x-5 \).
02

Applying the Product and Quotient Rules

Differentiating this function will apply both the product and quotient rules. The quotient rule states that for any function of the form \( f(x) = \frac{h(x)}{g(x)} \), its derivative is given by \( f'(x) = \frac{h'(x)g(x) - h(x)g'(x)}{[g(x)]^2} \). The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.
03

Differentiating Function 'u'

The function \( u(x) = \frac{x+1}{x+2} \) follows the quotient rule. To find its derivative \( u'(x) \), calculate \( u'(x) = \frac{1(x+2) - (x+1)(1)}{(x+2)^2} = \frac{1}{(x+2)^2} \).
04

Differentiating Function 'v'

The function \( v(x) = 2x-5 \) is a simple linear function. Its derivative \( v'(x) \) is \( v'(x) = 2 \).
05

Applying the Product Rule

Using the product rule on \( g(x) = u(x)v(x) \), the derivative \( g'(x) = u'(x)v(x) + u(x)v'(x) = \frac{1}{(x+2)^2}(2x-5) + \frac{x+1}{x+2}(2) = \frac{2}{(x+2)^2} - \frac{5}{(x+2)^2} + \frac{2x+2}{x+2} \).

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Key Concepts

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

Derivative of a Function
Understanding how to take the derivative of a function is a fundamental skill in calculus. It is the process of finding the rate at which a function is changing at any given point. The derivative of a function at a certain point essentially gives us the slope of the tangent line to the function's graph at that point.

In mathematical terms, if we have a function represented as \( f(x) \), its derivative is often denoted as \( f'(x) \), \( df/dx \), or \( Df \). The process of finding the derivative is called differentiation. There are several rules and methods for calculating derivatives, depending on the complexity of the function, such as the power rule for simple polynomials, the product rule for products of functions, and the quotient rule for ratios of functions.

For example, if we have a simple linear function like \( f(x) = mx + b \), where \( m \) and \( b \) are constants, the derivative is simply \( f'(x) = m \), because the rate of change is constant. However, for more complicated functions, we would use specific differentiation rules and techniques to take the derivative. The key to finding derivatives is to understand and apply these rules correctly.
Product Rule
The product rule is a vital tool for differentiating functions that are the product of two or more functions. It comes into play when you have a function, say \( h(x) = f(x)g(x) \), which is the multiplication of two sub-functions \( f(x) \) and \( g(x) \).

According to the product rule, the derivative of \( h(x) \) is found by multiplying the derivative of the first function by the second function and adding it to the product of the first function and the derivative of the second function. Mathematically, this is expressed as:
\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]

Let's say we have \( f(x) = x^2 \) and \( g(x) = \text{sin}(x) \), to find the derivative of \( h(x) = x^2\text{sin}(x) \), according to the product rule, you would compute:
\[ h'(x) = (2x)\text{sin}(x) + (x^2)\text{cos}(x) \]

The product rule is essential for finding derivatives of complex functions and is one of the first tools introduced in a calculus course after the basic derivatives are covered.
Quotient Rule
The quotient rule is employed when differentiating a function that is the division of two other functions, written as \( k(x) = \frac{f(x)}{g(x)} \), where \( f(x) \) is the numerator and \( g(x) \) is the denominator. The quotient rule allows us to find the derivative of the ratio of these two functions.

The rule states that the derivative \( k'(x) \) is the denominator \( g(x) \) times the derivative of the numerator \( f'(x) \), minus the numerator \( f(x) \) times the derivative of the denominator \( g'(x) \), all divided by the square of the denominator. In formula form, it is expressed as:
\[ k'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

For instance, if we were to differentiate \( k(x) = \frac{x^3}{x+1} \), using the quotient rule we would get:
\[ k'(x) = \frac{3x^2(x+1) - x^3(1)}{(x+1)^2} \]

The quotient rule is a critical concept for calculus students, helping to differentiate functions that might seem complex at first glance. Understanding both the product rule and quotient rule opens the door for solving a wide array of calculus problems involving derivatives.

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Most popular questions from this chapter

A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

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