What am I doing wrong in this derivative?

Sep 2015
5
0
United States
I have been working on derivatives, but now the homework problem's "show example" skips to immediately finding the derivative and every time I am getting a different answer. The original equation is y=((x-3)/(x+1))^2. The derivative is (8x-24)/(1+x)^3. My answer was (2x-6)/(x+1)^3. I am not understanding where the multiplication of 4 comes into play, this section is on the chain rule and that's what I assumed I had applied but clearly something is wrong. I can no longer use this problem, the website creates a similar problem when I've used the "solve this".
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$y= \left(\dfrac{x-3}{x+1}\right)^2$

$y' = 2\left(\dfrac{x-3}{x+1}\right) \cdot \dfrac{(x+1)-(x-3)}{(x+1)^2}$

$y' = 2\left(\dfrac{x-3}{x+1}\right) \cdot \dfrac{4}{(x+1)^2}$

$y' = \dfrac{8(x-3)}{(x+1)^3}$
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
skeeter is using the "chain rule" and the "quotient rule"

$y= \left(\dfrac{x-3}{x+1}\right)^2$
So y= f(g(x)) where $f(u)= u^2$ and $u= g(x)= \frac{x-2}{x+ 1}$.

By the "chain rule", $df/dx= (df/du)(du/dx)$. It should be immediate that $df/du= 2u= 2\frac{x- 2}{x+ 1}$.

$y' = 2\left(\dfrac{x-3}{x+1}\right) \cdot \dfrac{(x+1)-(x-3)}{(x+1)^2}$
and by the "quotient rule" $du/dx= \frac{ (x+ 1) d(x- 3)/dx- (x- 3)d(x+ 1)/dx}{(x+ 1)^2}$.

Of course, both d(x- 3)/dx and d(x+ 1)/dx are 1 so $du/dx= \frac{(x+1)- (x- 3)}{(x+ 1)^2}= \frac{4}{(x+1)^2}$

$y' = 2\left(\dfrac{x-3}{x+1}\right) \cdot \dfrac{4}{(x+1)^2}$

$y' = \dfrac{8(x-3)}{(x+1)^3}$