Answer to Question #339808 in Differential Equations for Niu

Question #339808

ACTIVITY IN BASIC CALCULUS

QUOTIENT RULE


  I. Find the derivative of the following functions below using the quotient rule. Show your complete solution.


  1. "\\frac{x^2-7x}{2x^3+5x^2}"
  2. "\\frac{3x^2+7x-14}{x^3-4x^2}"
  3. "\\frac{2x^3+8x^2}{x^2+6x^2-10}"




II. Create your own given problem involving quotient rule and solve. Show your complete solution. Do not copy the given example below.


1. Example must have two different terms in numerator, and three different terms in denominator

eg. (do not copy)


y= "\\frac{8x^2-3x}{x^2+6x^2-10}"


2. Example must have three different terms in numerator, and three different terms in denominator

eg. (do not copy)


y= "\\frac{x^2+8x^2-3x}{2x^3+6x^2-10}"


1
Expert's answer
2022-05-12T03:54:51-0400

"(\\frac{x^2-7x}{2x^3+5x^2})'=\\frac{(x^2-7x)'(2x^3+5x^2)-(2x^3+5x^2)'(x^2-7x)}{(2x^3+5x^2)^2}=\\frac{(2x-7)(2x^3+5x^2)-(6x^2+10x)(x^2-7x)}{(2x^3+5x^2)^2} \\\\\n\n(\\frac{3x^2+7x-14}{x^3-4x^2})'=\\frac{(3x^2+7x-14)'(x^3-4x^2)-(x^3-4x^2)'(3x^2+7x-14)}{(x^3-4x^2)^2}=\\frac{(6x+7)(x^3-4x^2)-(3x^2-8x)(3x^2+7x-14)}{(x^3-4x^2)^2} \\\\\n\n(\\frac{2x^3+8x^2}{x^2+6x^2-10})'=\\frac{(2x^3+8x^2)'(x^2+6x^2-10)-(x^2+6x^2-10)'(2x^3+8x^2)}{(x^2+6x^2-10)^2}=\\frac{(6x^2+16x)(7x^2-10)-14x(2x^3+8x^2)}{(7x^2-10)^2}"

Part 2

1. "(\\frac{x+1}{x^2+x+1})'=\\frac{(x+1)'(x^2+x+1)-(x^2+x+1)'(x+1)}{(x^2+x+1)^2}=\\frac{(x^2+x+1)-(2x+1)(x+1)}{(x^2+x+1)^2}"

2. "(\\frac{x+1-x^2}{x^2+x+1})'=\\frac{(x+1-x^2)'(x^2+x+1)-(x^2+x+1)'(x+1-x^2)}{(x^2+x+1)^2}=\\frac{(1-2x)(x^2+x+1)-(2x+1)(x+1-x^2)}{(x^2+x+1)^2}"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS