Thread: Derivative and Tangent Line 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.

Hello, ebonyscythe!

2. Find the points on the graph of the function
where the tangent line(s) to the graph are parallel to the line

The line is: . . It has slope:


The slope of the tangent to is:
.

Since they are parallel, their slopes are equal: .

Multiply by

Hence:.

The corresponding y-values are: .


Therefore, the points are: .

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?

Never mind! I realized I was using the quotient rule wrong...