1. ## 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

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

It should be easy now
Right?

3. thank you but do you know how do do it with quotient rule? my teacher emphasizes this.

4. Originally Posted by calculus0
thank you but do you know how do do it with quotient rule? my teacher emphasizes this.
$\displaystyle u = -1 \: \rightarrow \: u' = 0$

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

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

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

$\displaystyle y' = -\frac{v'u}{v^2}$. Remember your signs and factors

5. 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

6. 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:

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

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

I have put in that form to make it easier to see what cancels out

7. think i got it...e^(i*pi)

thanks so much