Answer to Question #340336 in Calculus for Niu

Question

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= $\frac{x^2+8x^2-3x}{2x^3+6x^2-10}$

Answer 1

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

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