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I'm trying to find the derivative of $$f(x)=\frac{4+(1/x)}{(x+4)}$$
I applied the quotient rule and I got as far as $(-4-(2/x)-(4/x^2))/(x+4)^2$.

The final answer is $(-4x^2+2x+4)/(x^2(x+4)^2))$ But I do not know how to get there. If someone could show me step by step how to get there, It would be greatly appreciated, thanks!

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    $\begingroup$ multiply out the $x^2$ in the numerator. $\endgroup$ – Prahlad Vaidyanathan Jun 23 '14 at 4:21
  • $\begingroup$ Or, take what you have so far and multiply the top and the bottom by $x^{2}$. Notice that that's a logical thing to try since it will cancel out your denominators in the numerator of the big fraction. $\endgroup$ – coolpapa Jun 23 '14 at 4:29
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To simplify fractions within fractions, just multiply by the least common multiple of the mini-fractions: \begin{align*} f'(x) &= \frac{(x + 4)(\frac{-1}{x^2}) - (4 + \frac{1}{x})(1)}{(x + 4)^2} \\ &= \frac{(\frac{-1}{x} - \frac{4}{x^2}) - (4 + \frac{1}{x})}{(x + 4)^2} \\ &= \frac{-4 - \frac{2}{x} - \frac{4}{x^2}}{(x + 4)^2} \qquad(\leftarrow \text{you got up to here})\\ &= \frac{-4 - \frac{2}{x} - \frac{4}{x^2}}{(x + 4)^2} \cdot \frac{x^2}{x^2} \\ &= \frac{-4x^2 - 2x - 4}{x^2(x + 4)^2} \end{align*} The answer's signs seem to be incorrect.

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  • $\begingroup$ I'm guessing the answer was meant to be typed $-(4x^2+2x+4)$. $\endgroup$ – mathematics2x2life Jun 23 '14 at 4:33
  • $\begingroup$ It's a bit weird that the one who answered first got nothing, and I don't think a straight forward question like this needs three answers that none of them add anything new. $\endgroup$ – Gigili Jun 23 '14 at 5:15
  • $\begingroup$ I'm sorry I could've sworn I clicked the "check". $\endgroup$ – Kenshin Jun 23 '14 at 5:37
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You can make life a bit easier on yourself by simplifying the expression first, then using the quotient rule. $$ \frac{4+\frac{1}{x}}{x+4}=\frac{\frac{1}{x}(4x+1)}{x+4}=\frac{4x+1}{x(x+4)} $$ Now we apply the quotient rule: $$ \frac{d}{dx}\left(\frac{4+\frac{1}{x}}{x+4}\right)=\frac{d}{dx}\left(\frac{4x+1}{x(x+4)}\right)=\frac{4(x^2+4x)-(2x+4)(4x+1)}{(x(x+4))^2}=\frac{-4x^2-2x-4}{(x(x+4))^2} $$ But of course, $$ \frac{-4x^2-2x-4}{(x(x+4))^2}=-\frac{4x^2+2x+4}{(x(x+4))^2} $$ which is what I believed you meant to type.

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You have the following

$$\frac{-4 - \frac{2}{x} - \frac{4}{x^2}}{(x+4)^4}$$

The aim here is to get rid of the nasty fractions from the first two terms and you can do this by multiplying the fractions by the highest power of $x$ which is $x^2$. So you will have

\begin{align} \require{cancel} & \frac{-4x^2-\frac{2}{x^\cancel{1}}\cdot x^\cancel{2} - \frac{4}{\cancel{x^2}}\cancel{x^2}}{x^2(x+4)^2}\\ &=\frac{-4x^2-2x-4}{(x+4)^2}\\ &= \frac{-(4x^2+2x+4)}{(x+4)^2} \end{align}

In general with problems such as these, as I said earlier, you want to get rid of the highest power in the denominator of the fraction and you can do this by multiplying by the highest power to the top and bottom.

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