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

I can't figure it out, what is the derivative of $$f(x)=\frac{x}{x-1}$$I have tried it many different ways and I just can't seem to get it figured out.

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Does it help to rewrite the numerator as $(x-1)+1$? –  Daniel Fischer Jul 3 '13 at 1:36
Do you know how to use the quotient rule? What are you having trouble with specifically? –  Javier Jul 3 '13 at 1:38
I am given that function, and have to find the equation of the tangent line that passes through the point (-1,5) –  cschurman Jul 3 '13 at 1:53
thanks for all of the help guys!I figured the problem out! I appreciate it! –  cschurman Jul 3 '13 at 2:13
@cschurman: Note that the given point is not on the curve. It will turn out there are two tangent lines. –  André Nicolas Jul 3 '13 at 2:22

Of course, you can use the quotient rule, but with this function, we can express it in a manner which makes computing the derivative fairly straightforward:

You can use polynomial long division, or simply subtract and add $1$ to the numerator to split the function into a sum:

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

Now we can simply compute using the power rule $(\dagger)$ $$f'(x) = \frac{d}{dx}\left(\frac 1{(x-1)}\right) = \frac{d}{dx}(x - 1)^{-1} = -{(x - 1)^{-2}} = \frac{-1}{(x-1)^2}$$

$(\dagger)$ Note that we are technically using the chain rule and power rule to compute $\frac{d}{dx}\left(\frac{1}{x - 1}\right)$, setting $u = x - 1,\; \frac {du}{dx} = 1$ and computing $\frac{d}{dx}\left(u^{-1}\right)= u'\cdot \frac{du}{dx}$. It just so happens that $\frac {du}{dx} = 1$.

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Much easier approach +1 –  Amzoti Jul 4 '13 at 0:16

In addition to amWhy's answer, we have a quotient (fraction) so you should think right away to use the quotient rule (assuming you have learned that). If you know the quotient rule, this problem is as simple as plugging in the relevant parts. Feel free to ask if that is not simple to you, as it probably wasn't simple to me when I first started.

Another way would be to use the product rule by rewriting this as $$f(x) = \frac{x}{x-1} = x(x-1)^{-1}.$$ Then, we have $$f'(x) = (1)(x-1)^{-1} + x \cdot (-1)(x-1)^{-2} = \frac{x-1 - x}{(x-1)^2} = \frac{-1}{(x-1)^2}.$$

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Using the quotient rule.

I like to remember it like: Low D-high minus high D-low over lowlow, the "d" meaning take the derivative.

$$f(x)=\frac{x}{x-1}$$ $$f'(x)=\frac{(x-1)\frac{d}{dx}(x) -[ x\frac{d}{dx}(x-1)]}{(x-1)^2}$$ Since $$\frac{d}{dx}(x) = 1$$ and $$\frac{d}{dx}(x-1) = 1$$ we have: $$f'(x) = \frac{(x-1)(1) - x(1)}{(x-1)^2}$$ simplifying: $$\frac{(x-1) - x}{(x-1)^2} = \frac{x - 1 - x}{(x-1)^2} = -\frac{1}{(x-1)^2}$$

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