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Chapter 8: Problem 5

i Show that $$ \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{3-4 x}{1+x^{2}}\right)=\frac{4 x^{2}-6 x-4}{\left(1+x^{2}\right)^{2}} $$ ii Determine \(\int \frac{4 x^{2}-6 x-4}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x\)

Short Answer

Expert verified
- The derivative is \ \( \frac{4x^2 - 6x - 4}{(1+x^2)^2} \), and the integral is \ \( \frac{3-4x}{1+x^2} + C \).

Step by step solution

01

- Use the quotient rule

To find the derivative of \ \( f(x) = \frac{3-4x}{1+x^2} \) \ use the quotient rule which states that \ \[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]. Here, let \ \( u = 3 - 4x\) and \ \( v = 1 + x^2\).
02

- Find the derivatives of u and v

Calculate the derivatives of \ \( u \) and \ \( v\) separately: \ \( u' = -4 \) and \ \( v' = 2x\).
03

- Apply the quotient rule

Using the quotient rule, substitute the values: \ \[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{3-4x}{1+x^2} \right) = \frac{-4(1+x^2) - (3-4x)(2x)}{(1+x^2)^2}\].
04

- Simplify the numerator

Expand and simplify the numerator: \ \[-4(1+x^2) - 2x(3-4x) = -4 - 4x^2 - 6x + 8x^2 = 4x^2 - 6x - 4\].
05

- Write the final derivative

Substitute the simplified numerator back: \ \[ \frac{4x^2 - 6x - 4}{(1+x^2)^2} \]. Hence, it is shown that \ \( \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{3-4x}{1+x^2} \right) = \frac{4x^2 - 6x - 4}{(1+x^2)^2} \).
06

- Begin integration of the derived function

To determine \ \( \int \frac{4x^2 - 6x - 4}{(1+x^2)^2} \ \mathrm{d}x \), observe that the integrand is the derivative obtained above.
07

- Integrate and apply the Fundamental Theorem of Calculus

Since \ \( \frac{3-4x}{1+x^2} \) is an antiderivative of \ \( \frac{4x^2 - 6x - 4}{(1+x^2)^2}\), we have \ \[ \int \frac{4x^2 - 6x - 4}{(1+x^2)^2} \ \mathrm{d}x = \frac{3-4x}{1+x^2} + C \], where \ \( C \) is the constant of integration.

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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 technique used for finding the derivative of a fraction of two functions. Specifically, if you have a function in the form of \( f(x) = \frac{u(x)}{v(x)} \), the Quotient Rule states: \[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \].
This rule is essential when dealing with rates of change for rational functions.
In our exercise, it was used to find the derivative of \( f(x) = \frac{3-4x}{1+x^2} \).
  • First, identify \( u \) and \( v \). Here: \( u = 3 - 4x \) and \( v = 1 + x^2 \).
  • Next, find the derivatives of these functions: \( u' = -4 \) and \( v' = 2x \).
  • Finally, apply the quotient rule formula: \[ \frac{-4(1 + x^2) - (3 - 4x)(2x)}{(1 + x^2)^2} = \frac{4x^2 - 6x - 4}{(1 + x^2)^2} \].
Understanding this rule makes calculating derivatives of division problems straightforward.
Derivative Calculation
Calculating derivatives is a fundamental task in calculus, representing how a function changes as its input changes.
In this exercise, we computed the derivative of \( f(x) = \frac{3-4x}{1+x^2} \) using the Quotient Rule.
  • Firstly, substitute the equations of derived functions: \( u = 3-4x \) and \( u' = -4 \), \( v = 1 + x^2 \) and \( v' = 2x \).
  • Then, plug these into the Quotient Rule: \[ \frac{-4(1 + x^2) - (3 - 4x)(2x)}{(1 + x^2)^2} \].
  • Finally, simplify the numerator and denominator: \( -4 - 4x^2 - 6x + 8x^2 = 4x^2 - 6x - 4 \) gives: \[ \frac{4x^2 - 6x - 4}{(1+x^2)^2} \].
This process lets us see the rate of change of the function concerning x, which has applications in various fields such as physics, engineering, and economics.
Integration
Integration, the inverse process of differentiation, helps find the accumulation of quantities, like areas under a curve.
In this context, we were given:
\[ \int \frac{4x^2 - 6x - 4}{(1 + x^2)^2} \mathrm{~d}x \]
  • We observe that this is the same as the derivative we previously computed: \[ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{3-4x}{1+x^2} \right) = \frac{4x^2 - 6x - 4}{(1 + x^2)^2} \].
  • By the Fundamental Theorem of Calculus, if \( a(x) \) is an antiderivative of \( f(x) \), then: \[ \int f(x) \mathrm{~d}x = a(x) + C \].
  • That means \( \int \frac{4x^2 - 6x - 4}{(1 + x^2)^2} \mathrm{~d}x = \frac{3-4x}{1+x^2} + C \), where \( C \) is a constant of integration.
So, the given integral simplifies to \( \frac{3-4x}{1+x^2} + C \), illustrating how closely linked differentiation and integration are.

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Most popular questions from this chapter

Show that $$ \int \frac{\mathrm{d} u}{u \sqrt{u^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1}\left(\frac{u}{a}\right)+C $$

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