Question: 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.

Answer: 
$f'(x)\equiv \displaystyle\lim\_{h\to 0} \frac{f(x+h)-f(x)}{h}$.

For your function, the numerator in the limit is $\dfrac{x+h}{1-x-h}-\dfrac{x}{1-x}$ which is equal to $\dfrac{(1-x)(x+h)-(1-x-h)x}{(1-x)(1-x-h)}$ which becomes $\dfrac{x-xh-x^2+h-x+x^2+hx}{(1-x)(1-x-h)}$ or $\dfrac{h}{(1-x)(1-x-h)}$. 

Now, you can find the limit as $h\to0$.
Using the definition of derivatives, we have

$f(x)'=\lim\_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$

Thus, the derivative of $\frac{x}{1-x}$ is

$\large f(x)'=\lim\_{h\rightarrow0}\frac{1}{h}(\frac{x+h}{1-x-h}-\frac{x}{1-x})=\lim\_{h\rightarrow 0}\frac{1}{h}(\frac{(x+h)(1-x)-x(1-x-h)}{(1-x-h)(1-x)})=\lim\_{h\rightarrow0}\frac{1}{x^2-2x+1+h(1-x)}=\frac{1}{(x-1)^2}$