# Derivative and Tangent Line Help

• Oct 3rd 2007, 07:19 PM
ebonyscythe
[Solved]Derivative Help
I need help with these two problems:

1. Find the Derivative of f(x)=((2*x)/(x=1))^2

2. Find the points on the graph of the function f(x) = (x+1)/(x-1) where the tangent line(s) to the graph are parallel to the line x+2y = 6.

Can someone walk me through the second one more carefully? I can't find an example of something like it in my notebook. Thanks for any help in advance.
• Oct 3rd 2007, 07:53 PM
Soroban
Hello, ebonyscythe!

Quote:

2. Find the points on the graph of the function $f(x) \:= \:\frac{x+1}{x-1}$
where the tangent line(s) to the graph are parallel to the line $x+2y \:=\: 6$

The line is: . $y \:=\:-\frac{1}{2}x+3.$ . It has slope: $m = -\frac{1}{2}$

The slope of the tangent to $f(x)$ is:
. $f'(x) \;=\;\frac{(x-1)\!\cdot\!1 - (x+1)\!\cdot\!1}{(x-1)^2} \;=\;\frac{-2}{(x-1)^2}$

Since they are parallel, their slopes are equal: . $-\frac{1}{2}\;=\;\frac{-2}{(x-1)^2}$

Multiply by $-2(x-1)^2\!:\;\;(x-1)^2 \:=\:4\quad\Rightarrow\quad x-1\:=\:\pm2\quad\Rightarrow\quad x \:=\:1 \pm 2$

Hence:. $x \:=\:3,\,-1$

The corresponding y-values are: . $y \:=\:2,\,0$

Therefore, the points are: . $(3,\,2),\;(-1,\,0)$

• Oct 4th 2007, 04:45 AM
ebonyscythe
Perfect! Thank you... I was thinking that the concept was harder then that, but you made it crystal clear. Thanks again!

When I worked on question 1., I keep getting a different answer then what my Ti-89 gives me. I seem to have misplaced a negative or something. Here's my work:

f(x) = (2x/(x+1))^2
f '(x) = 2(2x/(x+1))((2x*1 - 2(x+1))/(x+1)^2) -->(Chain rule and quotient rule)
f '(x) = 2((2x/(x+1))*(2x-2(x+1)/(x+1)^2))
f '(x) = 2((4x^2-4x(x+1))/(x+1)^3)
f '(x) = 2((4x^2-4x^2-4x)/(x+1)^3)
f '(x) = -8x/(x+1)^3

And my calculator gives me 8x/(x=1)^3. What did I do wrong?
• Oct 4th 2007, 05:16 AM
ebonyscythe
Never mind! I realized I was using the quotient rule wrong...