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The Quotient Rule

Abigail Young

Abigail Young

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Study Guide Overview

This study guide covers the Quotient Rule for finding derivatives of rational functions. It provides the definition and a memory aid (Low D-High, High D-Low...). Several examples demonstrate the application and simplification of the Quotient Rule, including trigonometric and exponential functions. Practice problems and solutions reinforce understanding. Key takeaways and a formula image are included.

#The Quotient Rule 🚀

Hey there, future calculus master! Ready to tackle the Quotient Rule? This is a key tool for differentiating those tricky rational functions, and you'll see it pop up all over the AP exam. Let's make sure you're totally comfortable with it!


# ⛳ Quotient Rule Definition

The Quotient Rule helps us find the derivative of a function that's a ratio of two other functions. Here's the formal definition:

ddx[f(x)g(x)]=g(x)ddxf(x)−f(x)ddxg(x)(g(x))2\frac{d}{dx} \Bigg[\frac{f(x)}{g(x)}\Bigg] = \frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{(g(x))^2}dxd​[g(x)f(x)​]=(g(x))2g(x)dxd​f(x)−f(x)dxd​g(x)​


Memory Aid

Low D-High, High D-Low, Draw the Line and Square Below! 🎶 Think of it like a little song.

  • Low: The denominator function, g(x)
  • D-High: Derivative of the numerator function, f'(x)
  • High: The numerator function, f(x)
  • D-Low: Derivative of the denominator function, g'(x)
  • Square Below: The denominator squared, (g(x))^2

Or, if you prefer a more compact version using u and v:

ddx[f(x)g(x)]=ddx(uv)=vu′−uv′v2\frac{d}{dx} \Bigg[\frac{\textcolor{red}{f(x)}}{\textcolor{green}{g(x)}}\Bigg] = \frac{d}{dx} (\frac{\textcolor{red}{u}}{\textcolor{green}{v}}) = \frac{\textcolor{green}{v}\textcolor{purple}{u'}-\textcolor{red}{u}\textcolor{blue}{v'}}{\textcolor{green}{v^2}}dxd​[g(x)f(x)​]=dxd​(vu​)=v2vu′−uv′​


Key Point: The Quotient Rule only applies when you have a function in the form of a fraction, f(x)g(x)\frac{f(x)}{g(x)}g(x)f(x)​, and both f(x) and g(x) are differentiable. ✅


#✏️ Quotient Rule Walkthrough

Let's see it in action!

Example: Find the derivative of y=x2+x−2x3+6y=\frac{x^2+x-2}{x^3+6}y=x3+6x2+x−2​.

  1. Identify f(x) and g(x):

    • f(x)=x2+x−2f(x) = x^2 + x - 2f(x)=x2+x−2
    • g(x)=x3+6g(x) = x^3 + 6g(x)=x3+6
  2. Find the derivatives:

    • f′(x)=2x+1f'(x) = 2x + 1f′(x)=2x+1
    • g′(x)=3x2g'(x) = 3x^2g′(x)=3x2
  3. Apply the Quotient Rule:

    dydx=(x3+6)(2x+1)−(x2+x−2)(3x2)(x3+6)2\frac{dy}{dx} = \frac{(x^3+6)(2x+1)-(x^2+x-2)(3x^2)}{(x^3+6)^2}dxdy​=(x3+6)2(x3+6)(2x+1)−(x2+x−2)(3x2)​

  4. Simplify (this is where your algebra skills come in!):

    dydx=(2x4+x3+12x+6)−(3x4+3x3−6x2)(x3+6)2\frac{dy}{dx} = \frac{(2x^4+x^3+12x+6)-(3x^4+3x^3-6x^2)}{(x^3+6)^2}dxdy​=(x3+6)2(2x4+x3+12x+6)−(3x4+3x3−6x2)​

    dydx=−x4−2x3+6x2+12x+6(x3+6)2\frac{dy}{dx} = \frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}dxdy​=(x3+6)2−x4−2x3+6x2+12x+6​


Exam Tip

Always double-check your algebra when simplifying. A small mistake can cost you points! Also, remember that you don't always have to fully simplify the derivative, but it's good practice.


#🧮 Quotient Rule: Practice Problems

Time to test your skills! Try these examples before looking at the solutions.

#Quotient Rule: Example 1

Find the derivative of y=ex1+x2y = \frac{e^x}{1+x^2}y=1+x2ex​.


Solution:

  1. f(x)=exf(x) = e^xf(x)=ex, so f′(x)=exf'(x) = e^xf′(x)=ex

  2. g(x)=1+x2g(x) = 1 + x^2g(x)=1+x2, so g′(x)=2xg'(x) = 2xg′(x)=2x

  3. Apply the Quotient Rule:

    dydx=(1+x2)(ex)−(ex)(2x)(1+x2)2\frac{dy}{dx} = \frac{(1+x^2)(e^x) - (e^x)(2x)}{(1+x^2)^2}dxdy​=(1+x2)2(1+x2)(ex)−(ex)(2x)​

  4. Simplify:

    dydx=ex+x2ex−2xex(1+x2)2=ex(1−2x+x2)(1+x2)2=ex(1−x)2(1+x2)2\frac{dy}{dx} = \frac{e^x + x^2e^x - 2xe^x}{(1+x^2)^2} = \frac{e^x(1 - 2x + x^2)}{(1+x^2)^2} = \frac{e^x(1-x)^2}{(1+x^2)^2}dxdy​=(1+x2)2ex+x2ex−2xex​=(1+x2)2ex(1−2x+x2)​=(1+x2)2ex(1−x)2​


#Quotient Rule: Example 2

Find the derivative of y=sin⁡(x)1+cos⁡(x)y = \frac{\sin(x)}{1+\cos(x)}y=1+cos(x)sin(x)​.


Solution:

  1. f(x)=sin⁡(x)f(x) = \sin(x)f(x)=sin(x), so f′(x)=cos⁡(x)f'(x) = \cos(x)f′(x)=cos(x)

  2. g(x)=1+cos⁡(x)g(x) = 1 + \cos(x)g(x)=1+cos(x), so g′(x)=−sin⁡(x)g'(x) = -\sin(x)g′(x)=−sin(x)

  3. Apply the Quotient Rule:

    dydx=(1+cos⁡(x))(cos⁡(x))−(sin⁡(x))(−sin⁡(x))(1+cos⁡(x))2\frac{dy}{dx} = \frac{(1+\cos(x))(\cos(x)) - (\sin(x))(-\sin(x))}{(1+\cos(x))^2}dxdy​=(1+cos(x))2(1+cos(x))(cos(x))−(sin(x))(−sin(x))​

  4. Simplify:

    dydx=cos⁡(x)+cos⁡2(x)+sin⁡2(x)(1+cos⁡(x))2\frac{dy}{dx} = \frac{\cos(x) + \cos^2(x) + \sin^2(x)}{(1+\cos(x))^2}dxdy​=(1+cos(x))2cos(x)+cos2(x)+sin2(x)​

Quick Fact

Remember the identity: sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1sin2(x)+cos2(x)=1. This helps simplify trig derivatives!

<math-block>\frac{dy}{dx} = \frac{\cos(x) + 1}{(1+\cos(x))^2} = \frac{1}{1+\cos(x)}</math-block>

#💫 Closing

You've got this! The Quotient Rule might seem tricky at first, but with practice, it'll become second nature. Remember to take your time, double-check your work, and don't be afraid to break down complex problems into smaller steps. 🎉


Key Concept

The Quotient Rule is essential for finding derivatives of rational functions. Make sure you know the formula and can apply it correctly.


Here's a quick recap:


Quotient Rule


Image Courtesy of OnLine Math Learning


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Question 1 of 9

What is the correct formula for the Quotient Rule when finding the derivative of f(x)g(x)\frac{f(x)}{g(x)}g(x)f(x)​? 🤔

f′(x)g(x)−f(x)g′(x)(g(x))2\frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}(g(x))2f′(x)g(x)−f(x)g′(x)​

f(x)g′(x)−g(x)f′(x)(g(x))2\frac{f(x)g'(x) - g(x)f'(x)}{(g(x))^2}(g(x))2f(x)g′(x)−g(x)f′(x)​

g(x)f′(x)+f(x)g′(x)(g(x))2\frac{g(x)f'(x) + f(x)g'(x)}{(g(x))^2}(g(x))2g(x)f′(x)+f(x)g′(x)​

g(x)f′(x)−f(x)g′(x)(g(x))2\frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}(g(x))2g(x)f′(x)−f(x)g′(x)​