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Find the derivatives of the functions in Exercises 17-28. \(y=\frac{(x+1)(x+2)}{(x-1)(x-2)}\)

Short Answer

Expert verified
The derivative is \( y' = \frac{3x^2 + 12}{(x-1)^2(x-2)^2} \).

Step by step solution

01

Identify the Formula for Derivative of a Quotient

To find the derivative of a quotient, we use the quotient rule. If you have a function of the form \( y = \frac{u(x)}{v(x)} \), its derivative \( y' \) is given by the formula:\[ y' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}.\] In this exercise, identify \( u(x) = (x+1)(x+2) \) and \( v(x) = (x-1)(x-2) \).
02

Differentiate the Numerator

Differentiate \( u(x) = (x+1)(x+2) \). Use the product rule of derivatives, which is:\[ u' = f(x)g'(x) + g(x)f'(x),\] where \( f(x) = x+1 \) and \( g(x) = x+2 \).\[ \begin{align*} f'(x) &= 1, \g'(x) &= 1, \u'(x) &= (x+1) \cdot 1 + (x+2) \cdot 1 = x+1 + x+2 = 2x + 3.\end{align*} \]
03

Differentiate the Denominator

Similarly, differentiate \( v(x) = (x-1)(x-2) \) using the product rule. Here, \( f(x) = x-1 \) and \( g(x) = x-2 \). Thus:\[ \begin{align*} f'(x) &= 1, \g'(x) &= 1, \v'(x) &= (x-1) \cdot 1 + (x-2) \cdot 1 = x-1 + x-2 = 2x - 3.\end{align*} \]
04

Apply the Quotient Rule

Plug \( u(x), u'(x), v(x), v'(x) \) into the quotient rule formula:\[ y' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}.\]Substitute the differentiated results:\[ \begin{align*} y' &= \frac{[(x-1)(x-2)](2x + 3) - [(x+1)(x+2)](2x - 3)}{((x-1)(x-2))^2}.\end{align*} \]
05

Simplify the Expression

Now, simplify the expression step by step. First, expand the terms in the numerator:\[ y' = \frac{[(x^2 - 3x + 2)(2x + 3) - (x^2 + 3x + 2)(2x - 3)]}{((x^2 - 3x + 2)^2)}.\]Expand each part:\[ \begin{align*} (x^2 - 3x + 2)(2x + 3) &= 2x^3 + 3x^2 - 6x^2 - 9x + 4x + 6,\ &= 2x^3 - 3x^2 - 5x + 6,\end{align*}\]\[ \begin{align*} (x^2 + 3x + 2)(2x - 3) &= 2x^3 + 3x^2 - 9x^2 - 9x + 4x - 6, \ &= 2x^3 - 6x^2 - 5x - 6.\end{align*}\]Subtract these results:\[ y' = \frac{(2x^3 - 3x^2 - 5x + 6) - (2x^3 - 6x^2 - 5x - 6)}{(x^2 - 3x + 2)^2}.\]\[ y' = \frac{0x^3 + 3x^2 + 12}{(x-1)^2(x-2)^2}.\]
06

Write the Final Derivative

After simplifying, the derivative of the original function is:\[ y' = \frac{3x^2 + 12}{(x-1)^2(x-2)^2}.\]

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Key Concepts

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

Quotient Rule
In calculus, the Quotient Rule is a fundamental tool for finding the derivative of a function that is the ratio of two other functions. Knowing how to handle these kinds of functions is important because they appear frequently in calculus problems. The formula for the quotient rule is:
  • Let \( y = \frac{u(x)}{v(x)} \)
  • The derivative is \( y' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)
The quotient rule helps break down complex rational functions into simpler parts. In the original exercise, \( u(x) = (x+1)(x+2) \) and \( v(x) = (x-1)(x-2) \). By using the quotient rule, the exercise demonstrates how to apply derivatives to both the numerator and the denominator separately before combining them. This approach allows you to systematically compute derivatives even in more complex fractions.
Product Rule
The Product Rule is used in calculus to find the derivative of a product of two functions. This rule is vital because not all functions are combined through addition or subtraction; some are multiplied together. The formula for the product rule is:
  • If \( u(x) = f(x)g(x) \), then the derivative \( u'(x) = f'(x)g(x) + f(x)g'(x) \).
In applying the product rule, you differentiate each function separately and then combine them according to the rule. This method helps unravel products of functions, making them easier to manage. In the step-by-step solution, the product rule was applied to both the numerator \( (x+1)(x+2) \) and the denominator \( (x-1)(x-2) \). By doing this, we identify the individual components that contribute to the derivative, allowing for precise calculation of derivatives involving multiplication.
Rational Functions
Rational functions are another essential concept often encountered in calculus. These are functions represented by the ratio of two polynomial expressions:
  • A rational function can be expressed as \( y = \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials.
Understanding rational functions is crucial because they encompass a wide range of possibilities where different techniques like the quotient rule and product rule become applicable. Rational functions require special attention to both the numerator and the denominator due to their polynomial nature. In solving for derivatives, as shown in the original exercise, breaking the functions into simpler, manageable parts allows for the computation of the derivative using rules specific to their structure. Notice how the function in the exercise is a rational function with both parts represented as products of terms, highlighting the intertwining of various calculus principles.

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Most popular questions from this chapter

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