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I am stuck at what is otherwise a trivial question that could be done with the quotient rule, but the manner in which it is asked is confusing.

Find the derivative of $f(y)$ with respect to $y$ if $f(y) = \displaystyle\frac{(y-1)}{(y+1)}$.

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  • $\begingroup$ Do they want to you do find the derivative without using the quotient rule? Write the function as $f(y)=(y-1)(y+1)^{-1}$ and use the product and chain rules. $\endgroup$ – David Mitra Jun 6 '12 at 18:34
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    $\begingroup$ It's just asking you to find $f'(y)$. There are no tricks. Just use the quotient rule and you're done. $\endgroup$ – Arturo Magidin Jun 6 '12 at 18:35
  • $\begingroup$ but the derivative is $\mathrm d/\mathrm dx$, not y!! (just joking; I do think that teachers of mathematics ought to avoid always using the same conventions, though, or you can get thrown by irrelevant differences) $\endgroup$ – Ben Millwood Jun 6 '12 at 18:48
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That's just $y$ instead of $x$. Recall the quotiente rule:

If $\displaystyle f(y) = \frac{g(y)}{h(y)}$ then $\displaystyle f'(y) = \frac{g'(y)h(y) - g(y) h'(y)}{[h(y)]^2}$.

Now apply it and you're done:
$\displaystyle g'(y) = \frac{d}{dy}[y - 1] = 1$
$\displaystyle h'(y) = \frac{d}{dy}[y + 1] = 1$

So $\displaystyle f'(y) = \frac{1 \cdot (y + 1) - (y - 1) \cdot 1}{(y + 1) ^ 2} = \frac{y + 1 - y + 1}{y ^ 2 + 2y + 1} = \frac{2}{y ^ 2 + 2y + 1}$.

If you cannot use or don't want to use the quotient rule, rewrite it as follows and apply the product rule instead:

If $\displaystyle f(y) = g(y)h(y)$ then $\displaystyle f'(y) = g'(y)h(y) + g(y)h'(y)$

$\displaystyle f(y) = \frac{(y - 1)}{y + 1} = (y - 1) \cdot (y + 1) ^ {-1}$.
So you now have (applying the chain rule in the second step): $\displaystyle f'(y) = \frac{d}{dy}[y - 1] \cdot (y + 1) ^ {-1} + (y - 1) \cdot \frac{d}{dy}[(y + 1) ^ {-1}] = (y + 1) ^{-1} + (y - 1) \cdot 1 \cdot [- (y + 1) ^{-2}] = (y + 1) ^ {-1} - (y - 1) \cdot (y + 1) ^{-2} = \frac{1}{y + 1} - \frac{y - 1}{(y + 1) ^ 2} = \frac{y + 1 - (y - 1)}{(y + 1) ^ 2} = \frac{2}{(y + 1) ^ 2} = \frac{2}{y ^ 2 + 2y + 1}$.

As you can see the two results are equal.

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If you don't want to use any rules at all (except the chain rule maybe), just log it: $$\log f(y)= \log(y-1) - \log(y+1)$$ $$\frac{f'(y)}{f(y)}=\frac{1}{y-1}-\frac{1}{y+1}=\frac{2}{(y-1)(y+1)}$$ $$f'(y)=f(y)\frac{2}{(y-1)(y+1)}=\frac{2}{(y+1)^2}$$

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