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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

Reply 1

ghostwalker

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

(edited 10 years ago)

Reply 2

tazmaniac97

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 = nkx^n-1

is there any way we could use this rule?

Reply 3

Mr M

Study Forum Helper

Original post by tazmaniac97

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.

Reply 4

tazmaniac97

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 :smile:

I forgot to use negative indices and attempted to differentiate like this: (1/r)¹ :facepalm:

this is what tiredness does to you :tongue:

Thanks

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