simple quotient rule

• Apr 15th 2010, 06:09 AM
calculus0
simple quotient rule
how does the derivative of

-1/ (x - 1)^2

=

2/ (x - 1)^3

seems simple..but i get x in the numerator :(
• Apr 15th 2010, 06:27 AM
TWiX
$\frac{-1}{(x-1)^2}=-(x-1)^{-2}$

It should be easy now
Right? :D
• Apr 15th 2010, 06:32 AM
calculus0
thank you but do you know how do do it with quotient rule? my teacher emphasizes this.
• Apr 15th 2010, 06:41 AM
e^(i*pi)
Quote:

Originally Posted by calculus0
thank you but do you know how do do it with quotient rule? my teacher emphasizes this.

$u = -1 \: \rightarrow \: u' = 0$

$v = (x-1)^2 \: \rightarrow \: v' = 2(x-1)$

Quotient Rule $y' = \frac{u'v - v'u}{v^2}$

In this case we can get rid of $u'v$ because it equals 0.

$y' = -\frac{v'u}{v^2}$. Remember your signs and factors
• Apr 15th 2010, 07:26 AM
calculus0
im so horrible at this..

after 1/2 the equation disappears multiplying by zero i get:

2
-----------
(x - 1)

which is not equal to the solution

2
------------
(x - 1)^3
• Apr 15th 2010, 07:58 AM
e^(i*pi)
Quote:

Originally Posted by calculus0
im so horrible at this..

after 1/2 the equation disappears multiplying by zero i get:

2
-----------
(x - 1)

which is not equal to the solution

2
------------
(x - 1)^3

The problem seems to be that you're putting v on the denominator instead of v^2:

$v^2 = (x-1)^4$

which gives $y' = \frac{2(x-1)}{(x-1)(x-1)^3}$

I have put in that form to make it easier to see what cancels out
• Apr 15th 2010, 08:20 AM
calculus0
think i got it...e^(i*pi)

thanks so much