I've been stuck with this problem for a long time and can not figure out what I am doing wrong. Here is the question:
Given the function f(x) = (2x)/(x+1)^2 find the tangent line at point (0,0).
My problem lies in taking the derivative of the function. Using the quotient rule combined with the chain rule, I get
f`(x) = (2(x+1)-4x(x+1))/(x+1)^4. This is correct and at x=0 the slope of the tangent line is 2.
My problem is when I tried (several times) to mimic the result by using the derivative formula. No matter how many times I tried, I got the result 1/0 = 0.
Here is what I did.
f`(x) = f(((x+h)-f(x)))/h when h approaches 0.
since I want the derivative where x = 0, I replaced this with:
f`(x) = (f(h) - f(0)) / h
since f(0) = 0
f`(x) = f(h)/h
so we get:
(2h/(h+1)^2 ) / h as h approaches zero.
I know this is wrong but I cannot understand where I erred. I would greatly appreciate an explanation of where I went wrong and how to properly calculate this derivative using the formula.
$$\frac{2h/(h+1)^2}{h} = 2 \frac{1}{(h+1)^2} \to 2 \quad \text{as $h \to 0$}$$