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 $h\to0$.
$$
f'(x) = \lim\_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
f'(x) = \lim\_{h \rightarrow 0} \frac{\frac{x+h}{1-(x+h)}-\frac{x}{1-x}}{h} \\
f'(x) = \lim\_{h \rightarrow 0} \frac{x-x^2+h-hx-x+x^2+xh}{h(1-x-h)(1-x)} \\
f'(x) = \lim\_{h \rightarrow 0} \frac{h}{h(1-x-h)(1-x)} \\
f'(x) = \lim\_{h \rightarrow 0} \frac{1}{(1-x-h)(1-x)} =\frac{1}{(1-x)^2}\\
$$