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tazmaniac97
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I didn't do a level maths and this came up in one of my lecture slides, how is this true?

the derivative of 1/r is -1/r^2
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ghostwalker
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(Original post by tazmaniac97)
I didn't do a level maths and this came up in one of my lecture slides, how is this true?

the derivative of 1/r is -1/r^2
Assuming you don't want to derive it from scratch, use the quotient rule.

But if you're doing a maths degree, try it from scratch \displaystyle\lim_{h\to 0}\frac{f(r+h)-f(r)}{h} where f(r) = 1/r
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tazmaniac97
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(Original post by ghostwalker)
Assuming you don't want to derive it from scratch, use the quotient rule.

But if you're doing a maths degree, try it from scratch \displaystyle\lim_{h\to 0}\frac{f(r+h)-f(r)}{h} where f(r) = 1/r
what's the quotient rule? And to be honest which ever way is easier, I've been teaching myself and this is the rule I'm familiar with:

y = kxn, then dy/dx = nkxn-1

is there any way we could use this rule?
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Mr M
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(Original post by tazmaniac97)
what's the quotient rule? And to be honest which ever way is easier, I've been teaching myself and this is the rule I'm familiar with:

y = kxn, then dy/dx = nkxn-1

is there any way we could use this rule?
Yes.

\displaystyle y=\frac{1}{r}=r^{-1}

Now use your rule.

If you don't know what is going on then you need to study negative indices.
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tazmaniac97
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(Original post by Mr M)
Yes.

\displaystyle y=\frac{1}{r}=r^{-1}

Now use your rule.

If you don't know what is going on then you need to study negative indices.
Oh I got the answer

I forgot to use negative indices and attempted to differentiate like this: (1/r)1 :facepalm: this is what tiredness does to you

Thanks
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