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.

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