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.
 A: $f'(x)\equiv \displaystyle\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$.
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 $h\to0$.
A: 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
$$\begin{align}
\lim\limits_{h \rightarrow 0}\frac{1}{x^2 + (h-2)x + 1-h } &= \frac{1}{x^2 -2x + 1}\\
&= \frac{1}{(x-1)^2} \blacksquare 
\end{align}
$$
A: $$
f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
f'(x) = \lim_{h \rightarrow 0} \frac{\frac{x+h}{1-(x+h)}-\frac{x}{1-x}}{h} \\
f'(x) = \lim_{h \rightarrow 0} \frac{x-x^2+h-hx-x+x^2+xh}{h(1-x-h)(1-x)} \\
f'(x) = \lim_{h \rightarrow 0} \frac{h}{h(1-x-h)(1-x)} \\
f'(x) = \lim_{h \rightarrow 0} \frac{1}{(1-x-h)(1-x)} =\frac{1}{(1-x)^2}\\
$$
A: 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
$\large f(x)'=\lim_{h\rightarrow0}\frac{1}{h}(\frac{x+h}{1-x-h}-\frac{x}{1-x})=\lim_{h\rightarrow 0}\frac{1}{h}(\frac{(x+h)(1-x)-x(1-x-h)}{(1-x-h)(1-x)})=\lim_{h\rightarrow0}\frac{1}{x^2-2x+1+h(1-x)}=\frac{1}{(x-1)^2}$
A: $$f(x) = \frac{x}{1-x} $$
$$f(x+h) = \frac{x+h}{1-(x +h)}$$
$$f(x+h) - f(x) = \frac{(x+h)(1 -x) - x(1 -x -h)}{(1-x-h)(1-x)}$$
$$f(x+h) - f(x) = \frac{h}{(1-x-h)( 1-x)}$$
$$ f'(x) = \lim_{h \to 0} \frac{f(x+ h) - f(x) }{h}$$
$$ \lim_{h \to 0} \ \frac{h }{h(1-x-h)( 1-x)} $$ 
Now , divide out h then let the other h (inside parenthesis) go to zero.
