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Chapter 5: Problem 6

Find the derivative and state a corresponding integration formula. $$ \frac{d}{d x}\left[\frac{x}{x^{2}+3}\right] $$

Short Answer

Expert verified
The derivative is \( \frac{-x^2 + 3}{(x^2 + 3)^2} \); integrate using: \( \int \frac{-x^2 + 3}{(x^2 + 3)^2} \, dx \).

Step by step solution

01

Identify the Function

We have the function \( f(x) = \frac{x}{x^2 + 3} \). Our task is to find its derivative with respect to \( x \).
02

Apply the Quotient Rule

The derivative of a quotient \( \frac{u}{v} \) is given by the quotient rule: \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \). Here, \( u = x \) and \( v = x^2 + 3 \).
03

Differentiate the Numerator and Denominator

Differentiate the numerator \( u = x \) to get \( \frac{du}{dx} = 1 \). Differentiate the denominator \( v = x^2 + 3 \) to get \( \frac{dv}{dx} = 2x \).
04

Substitute into the Quotient Rule

Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule: \[\frac{d}{dx}\left( \frac{x}{x^2 + 3} \right) = \frac{(x^2 + 3)(1) - x(2x)}{(x^2 + 3)^2}\]
05

Simplify the Expression

Simplify the expression to get: \[= \frac{x^2 + 3 - 2x^2}{(x^2 + 3)^2}\]\[= \frac{-x^2 + 3}{(x^2 + 3)^2}\]
06

State the Integration Formula

The corresponding integration formula is: \[\int \frac{-x^2 + 3}{(x^2 + 3)^2} \, dx = -\int \frac{x^2}{(x^2 + 3)^2} \, dx + \int \frac{3}{(x^2 + 3)^2} \, dx\]This represents the reverse operation of differentiation for this function.

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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 used to find the derivative of a function that is the quotient of two differentiable functions. Whenever you see a fraction where one function is divided by another, you can apply the quotient rule. The formula is:
  • Given two functions, say \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u}{v} \) is:
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\]Here is how it works step-by-step:
  • Differentiate the numerator, \( u(x) \).
  • Differentiate the denominator, \( v(x) \).
  • Substitute these derivatives into the quotient rule formula.
  • Simplify to find the final derivative.
Applying the quotient rule allows us to calculate derivatives for more complex functions. **Note:** This rule is incredibly handy, but ensure both the numerator and denominator functions are differentiable.
Differentiation
Differentiation is the process of finding the derivative, which measures how a function's output changes as input changes. It is essentially the slope of the function at any point along its curve. To differentiate a function means to find its derivative. Here's what happens during differentiation:
  • Identify the function to differentiate.
  • Apply relevant rules like the product rule, quotient rule, or chain rule depending on the function's form.
  • Simplify the derivative expression.
The derivative provides crucial information, such as:
  • Rates of change for real-world processes.
  • Slopes of curves.
  • Behavior of the function, like finding maxima or minima.
Differentiation is a foundational concept in calculus, with numerous applications in science, engineering, and economics, especially in analyzing dynamic systems.
Integration Formula
Integration is the reverse process of differentiation. When you integrate a function, you're essentially finding its antiderivative. For the function we just differentiated, the integration formula looks like this:\[\int \frac{-x^2 + 3}{(x^2 + 3)^2} \, dx = -\int \frac{x^2}{(x^2 + 3)^2} \, dx + \int \frac{3}{(x^2 + 3)^2} \, dx\]Breaking down the integration process:
  • Integration combines areas and finds functions whose derivatives result in the original function.
  • Different terms in a derivative often lead to separate integral expressions.
  • Apply the power rule for integration, or special rules for more complex terms.
Integration has wide-ranging applications, such as calculating areas under curves, solving differential equations, and determining accumulated quantities over time.

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Most popular questions from this chapter

(a) Prove that if \(f\) is an odd function, then $$\int_{-a}^{a} f(x) d x=0$$ and give a geometric explanation of this result. [Hint: One way to prove that a quantity \(q\) is zero is to show that \(q=-q .\) ] (b) Prove that if \(f\) is an even function, then $$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$$ and give a geometric explanation of this result. [Hint: Split the interval of integration from \(-a\) to \(a \text { into two parts at } 0 .]\)

A function \(f(x)\) is defined piecewise on an interval. In these exercises: (a) Use Theorem 5.5 .5 to find the integral of \(f(x)\) over the interval. (b) Find an antiderivative of \(f(x)\) on the interval. (c) Use parts (a) and (b) to verify Part 1 of the Fundamental Theorem of Calculus. $$ f(x)=\left\\{\begin{array}{ll}{\sqrt{x},} & {0 \leq x<1} \\ {1 / x^{2},} & {1 \leq x \leq 4}\end{array}\right. $$

Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{0}^{\pi / 3}(2 x-\sec x \tan x) d x $$

Evaluate the definite integral by expressing it in terms of \(u\) and evaluating the resulting integral using a formula from geometry. $$ \int_{\pi / 3}^{\pi / 2} \sin \theta \sqrt{1-4 \cos ^{2} \theta} d \theta ; u=2 \cos \theta $$

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$V=V_{p} \sin (2 \pi f t)$$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts \((\mathrm{V}),\) the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). \((1 \mathrm{Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the \(r m s\) or root-mean-square value of \(V .\) By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}}$$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f,\) and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral. \(]\) (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{V}\) at a frequency of \(60 \mathrm{Hz}\). What is the peak voltage at such an outlet?
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