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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.

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$f'(x)\equiv \displaystyle\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$.

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 $h\to0$.

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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 $$\begin{align} \lim\limits_{h \rightarrow 0}\frac{1}{x^2 + (h-2)x + 1-h } &= \frac{1}{x^2 -2x + 1}\\ &= \frac{1}{(x-1)^2} \blacksquare \end{align} $$

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

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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

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

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$$f(x) = \frac{x}{1-x} $$

$$f(x+h) = \frac{x+h}{1-(x +h)}$$

$$f(x+h) - f(x) = \frac{(x+h)(1 -x) - x(1 -x -h)}{(1-x-h)(1-x)}$$

$$f(x+h) - f(x) = \frac{h}{(1-x-h)( 1-x)}$$

$$ f'(x) = \lim_{h \to 0} \frac{f(x+ h) - f(x) }{h}$$

$$ \lim_{h \to 0} \ \frac{h }{h(1-x-h)( 1-x)} $$

Now , divide out h then let the other h (inside parenthesis) go to zero.

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