Solving $\frac{x}{1-x}$ using definition of derivative I was trying to find the equation of the tangent line for this function. I solved this using the quotient rule and got $\frac{1}{(x-1)^2}$ but I can't produce the same result using definition of derivatives. Can someone show me how to do it? I tried looking it up on wolfram alpha but I can't get it to produce the result using definition of derivatives.

$$f(x) = \frac{x}{1-x} $$
$$f(x+h) = \frac{x+h}{1-(x +h)}$$
$$f(x+h) - f(x) = \frac{(x+h)(1 -x) - x(1 -x -h)}{(1-x-h)(1-x)}$$
$$f(x+h) - f(x) = \frac{h}{(1-x-h)( 1-x)}$$
$$ f'(x) = \lim_{h \to 0} \frac{f(x+ h) - f(x) }{h}$$
$$ \lim_{h \to 0} \ \frac{h }{h(1-x-h)( 1-x)} $$ 
Now , divide out h then let the other h (inside parenthesis) go to zero.
