• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

how do i differentiate the following, simple! (1 Viewer)

Joined
Oct 29, 2011
Messages
872
Location
Narnia
Gender
Female
HSC
2013
y = 5 / (x-3)

I was thinking of moving the (x-3) to the top so it become (x-3)^-1

and then use y = u~v + v~u

can someplace please help me.. thanks
please show full working.
 

anwar1506

New Member
Joined
Aug 18, 2012
Messages
16
Gender
Male
HSC
2013
You can use the quotient rule. y'= (u'v - uv')/ v^2

Therefore, y'= 0(x-3) - 1(5) / (x-3)^2
∆ y'= -5/(x-3)^2
 

soloooooo

Well-Known Member
Joined
Feb 13, 2012
Messages
3,311
Gender
Female
HSC
N/A
You are correct in moving it to the numerator like that; then just carry out differentiation normally.
 

Demise

Member
Joined
Apr 26, 2011
Messages
636
Gender
Male
HSC
2012
y= 5 / (x-3) -> 5 . (x-3)^-1
y' = -5 . (x-3)^-2
y' = -5 / (x-3)^2

Guy above already beat me.
 

Peeik

Member
Joined
Mar 12, 2009
Messages
274
Location
Sydney
Gender
Male
HSC
2009
The quickest way is using the chain rule. See below.

 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top