1. ## Simple derivative question.

F(x) = (x+1)/(x-1) a=3

Compute derivative at x = a using the limit definition and find an equation of the tangent line.

I get up to:

[(3+h)+1/(3+h)-1] - 2
____________________
h

but I don't know where to go from there, help!

2. Originally Posted by coach2uf
F(x) = (x+1)/(x-1) a=3

Compute derivative at x = a using the limit definition and find an equation of the tangent line.

I get up to:

[(3+h)+1/(3+h)-1] - 2
____________________
h

but I don't know where to go from there, help!
I'll start at the start. First of all, you should make life as easy for yourself as possible.

So write $\frac{x + 1}{x - 1}$ as $\frac{(x - 1) + 2}{x - 1} = 1 + \frac{2}{x - 1}$.

I'll do it for general a. If you prefer, you can replace all the a's with 3 .......

Note that if f(x) = 1, then $\frac{f(a + h) - f(a)}{h} = 0$ since f(x) = 1 for all values of x including x = a + h and x = a.

So you only have to worry about getting $\lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h}$ for $f(x) = \frac{2}{x - 1}$:

$f(a + h) - f(a) = \frac{2}{a + h - 1} - \frac{2}{a - 1}$

Put over a common denominator:

$= \frac{2(a - 1)}{(a + h - 1)(a - 1)} - \frac{2(a + h - 1)}{(a - 1)(a + h - 1)} = \frac{2(a - 1) - 2 (a + h - 1)}{(a + h - 1)(a - 1)}$

Simplify:

$= \frac{-2h}{(a + h - 1)(a - 1)}$.

So $\frac{f(a + h) - f(a)}{h} = \frac{-2}{(a + h - 1)(a - 1)}$.

Now take the limit

$\lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h} = \lim_{h \rightarrow} \frac{-2}{(a + h - 1)(a - 1)} = \frac{-2}{(a - 1)^2}$.

When a = 3 the value this limit is equal to $-\frac{1}{2}$.

This means that the gradient of the tangent line is $-\frac{1}{2}$.