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.

First we create a difference quotient for this function: $$\lim\limits_{h \rightarrow 0} \frac{(x+h)(1-(x+h))^{-1} - x(1-x)^{-1}}{h}$$
Now we try to simplify the numerator a bit by making it one fraction as opposed to the difference between two fractions:
$$
\begin{align}
\lim\limits_{h \rightarrow 0} \frac{(x+h)(1-(x+h))^{-1} - x(1-x)^{-1}}{h}&= \lim\limits_{h \rightarrow 0} \frac{1}{h} \frac{(x+h)(1-x)-x(1-x-h)}{(1-x-h)(1-x)}\\
&= \lim\limits_{h \rightarrow 0}\frac{1}{h}\frac{-x^2 +x - hx +h -x +x^2+hx}{1 - x -x + x^2 - h  + hx}\\
&= \lim\limits_{h \rightarrow 0}\frac{1}{h}\frac{h}{x^2 + (h-2)x + 1-h }
\end{align}
$$
Now canceling the $\frac{1}{h}$ we get our answer :
$$\begin{align}
\lim\limits_{h \rightarrow 0}\frac{1}{h}\frac{h}{x^2 + (h-2)x + 1-h }
&= \lim\limits_{h \rightarrow 0}\frac{1}{x^2 + (h-2)x + 1-h }\\
\end{align}
$$
And now evaluating our limit we finally get
$$\begin{align}
\lim\limits_{h \rightarrow 0}\frac{1}{x^2 + (h-2)x + 1-h } &= \frac{1}{x^2 -2x + 1}\\
&= \frac{1}{(x-1)^2} \blacksquare 
\end{align}
$$
