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I have a problem on my homework.
What is the derivative of $5/x$ ?
And can you please explain.

Thank you.

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closed as off-topic by TravisJ, Rory Daulton, Eric Wofsey, Leucippus, mixedmath Oct 19 at 0:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TravisJ, Rory Daulton, Eric Wofsey, Leucippus, mixedmath
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nothing , I don't know where to start from. – southpaw93 Sep 27 '12 at 15:48
We don't know where to start either. Do you know any derivative rules? Do you have to find the derivative using only the definition of derivative? – David Mitra Sep 27 '12 at 15:51

3 Answers 3

up vote 1 down vote accepted

Use the power rule as normal and use the following result:

$\frac{d}{dx}(\frac{a}{x}) =a\times\frac{d}{dx}(\frac{1}{x}) $

This yields $ \frac{d}{dx}(\frac{5}{x})=5\times\frac{d}{dx}(\frac{1}{x}) =5\times \frac{d}{dx}(x^{-1}) $ $5\times(-1)x^{-2}=-\frac{5}{x^2}$.

Hope that helps.

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First, it helps to rewrite fractions using negative exponents, so $\frac{5}{x}=5x^{-1}$. Now you can use the power rule to calculate the derivative with $n=-1$.

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thanks for the hint , I realized it now . – southpaw93 Sep 27 '12 at 15:50

You should evaluate $\lim_{h\to 0} \dfrac{ f(x+h)-f(x)}{h}$.

If you run into trouble with calculation, show us your work and we can proceed from there

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This is a comment, not an answer... – Patrick Da Silva Sep 27 '12 at 15:54
I would say that this is an answer that has as much effort behind it as the OP. – mixedmath Sep 27 '12 at 16:47
Can't you just give the example to other users and not post things wrong? You have 22.4k rep, no one expects you to do "as much effort behind it as the OP" when OP is clearly a new user. I mean, I'm actually doing a comment so you decide but that's my opinion. – Patrick Da Silva Sep 28 '12 at 1:04
@Patrick: I read the OP's comment to his question where he said that he didn't know where to begin. I think my answer gives that start. And I would be happy if others were willing to offer additional assistance if the OP asked, as the latter half of my answer does. In short, I defend this answer as an answer, not as a comment. – mixedmath Sep 28 '12 at 1:08

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