trouble with the quotient rule (1 Viewer)

[hunter in use]

(2x-9)^3
_______
5x+1

please anyone.... i have no idea

 

[insert-username]

(2x-9)^3
_______
5x+1

please anyone.... i have no idea

Firstly, notation. u' = derivative of u, and v' = derivative of v. The derivative of the quotient u/v is: (vu' - uv')/v²

So: [(5x+1).3(2x-9)².2 - (2x-9)³.5]/(5x+1)²

= [6(5x+1)(2x-9)² - 5(2x-9)³]/(5x+1)²

= (2x-9)²[6(5x+1) - 5(2x-9)]/(5x+1)² (taking (2x-9)² out as a common factor)

= (2x-9)²[20x + 51]/(5x+1)² (expanding the inside brackets)


I_F

 

[vafa]

An alternative method:
y=(2x-9)^3/(5x+1)

obviously (2x-9)^3=e^(3ln(2x-9))
5x+1=e^(ln(5x+1))
hence
y=e^(3ln(2x-9)-ln(5x+1))

dy/dx=((6/(2x-9))-(5/(5x+1)))e^(3ln(2x-9)-ln(5x+1))
dy/dx=(((30x+6-10x+45)/((2x-9)(5x+1)))((2x-9)^3/(5x+1))

finally you from previous step you get:

dy/dx=(20x+51)(2x-9)^2/(5x+1)^2

note:
this method is not interesting for some people but the you can use this method when you have a messy fraction and you know that you will do some mistakes by doing the normal way. i am not saying that you need to use my way. I am saying that you can count on this method as well and then that will be your choice to select your preferred way to differentiate.

 

[Kezalicious]

I know you already have the answer but I thought you might like an alternative way of remembering and looking at the quotient rule.

bottom x derivative top - top x derivative botom
_______________________________________
bottom^2

i.e top and bottom are your functions

 

[Mc_Meaney]

Quotient rule - 1 big v. vu-uv
v^2

 

[Mountain.Dew]

a more visual way of remembering how to do quotient rule:

using ur example:

u = (2x-9)^3 - outside
u' = 6(2x-9)^2 - inside
v=5x+1 - inside
v' 5 - outside

the method: (INSIDE - OUTSIDE) / (v^2)

so, d((2x-9)^3/(5x+1))/dx = [6(5x+1)(2x-9)<SUP>2</SUP> - 5(2x-9)<SUP>3</SUP>]/(5x+1)<SUP>2</SUP>

= [via simplification] (2x-9)<SUP>2</SUP>[20x + 51]/(5x+1)<SUP>2</SUP>
<SUP></SUP>
hope it helps, M.D.