Question

It was proved that D(xr)=rxr−1 for any positive integer r. Use this result and the Quotient Rule to prove that D(xr)=rxr−1 for any negative integer r.

Answer 1

Quotient rule: \[D \left( \frac{ f }{ g } \right)=\frac{ g*f' - f*g' }{ g² }\]

Answer 2

So... let r be a positive integer... so -r is a negative integer, aye? :)

Answer 3

On second thought, this is confusing... Let's let r be a negative integer instead...

Answer 4

\[D(x^r)=rx^r⁻¹\]

Answer 5

sorry the first post looks confusing

Answer 6

Yeah... so... any idea how to go about this?

Answer 7

I'm thinking how to get two functions g and f out of the original function

Answer 8

Well, you'll find that most work in Maths involves reducing a problem into a mess of previously solved problems...
So now we're to find the derivative of x^r, given that r is a negative integer, aye? So...
\[\huge \frac{d}{dx}x^r \ \ \ \ \ r\in\mathbb{Z}^-\]

Answer 9

so... I can use the power rule, but that doesn't really prove anything new?

Answer 10

x^r= 1/x^-r ........... so now I have quotient?

Answer 11

...when r is a negative integer ofcourse

Answer 12

You seem to have a good idea how to do it :)
So, now you have
\[\huge \frac{d}{dx}x^r=\frac{d}{dx}\frac1{x^{-r}}\]And so you have a quotient... ripe for quotient rule :D
Get to it, champ :)

Answer 13

Not forgetting of course, that since r was negative, -r is positive, so you can use the power rule...

Answer 14

\[D \left( \frac{ 1 }{ x^-r } \right)=\frac{ x⁻^r*0-1*r*x⁽⁻^r⁻¹⁾ }{ x^-²^r }\] ?

Answer 15

whoops missing a minus sign on the numerator

Answer 16

Rectify it :)

Answer 17

\[\frac{ -1*-rx⁻^r⁻¹ }{ x⁻²^r }=\frac{ -1*-r*\left( \frac{ x⁻^r }{ x¹ } \right) }{ x⁻^r*x⁻^r }\] is this right?

Answer 18

Hang on... let me just zoom it in :D
\[\huge \frac{ -1*-rx⁻^r⁻¹ }{ x⁻²^r }=\frac{ -1*-r*\left( \frac{ x⁻^r }{ x¹ } \right) }{ x⁻^r*x⁻^r }\]

Answer 19

Yes, it's right. Now simplify it.

Answer 20

Ok I can basically see it will become rx^(r-1), but I'm not sure what the rules are for dividing quotients, or simplifying them in this situation

Answer 21

Dividing quotients? What do you mean by that?

Answer 22

when there is \[\left( \frac{ x⁻^r }{ x¹ } \right) / (x⁻^r*x⁻^r)\]

Answer 23

what cancels out what

Answer 24

can I multiply by the inverse?

Answer 25

Ahh... Well...
Given this...
\[\huge \frac{\frac{a}{b}}{c}=\frac{a}{bc}\]should clear it up.

Answer 26

\[=\frac{ r*x⁻^r }{ x*x⁻^r*x⁻^r }=\frac{ r }{ x*x⁻^r }=\frac{ r }{ x⁻^r⁺¹ }\] ??

Answer 27

=\[\frac{ r }{ x^r⁻¹ }\]

Answer 28

This is good. You've done it :)

Answer 29

If you doubt yourself, do remember one thing about exponents...
\[\huge a^m=\frac1{a^{-m}}\]

Answer 30

wait, but how do I get to the result rx^r-1 ?

Answer 31

Oh wait...

Answer 32

You stop at...
\[\huge =\frac{ r\times x⁻^r }{ x\times x⁻^r\times x⁻^r }=\frac{ r }{ x\times x⁻^r }=\frac{ r }{ x⁻^r⁺¹ }\]

Answer 33

The last part has...
\[\huge \frac{r}{x^{-r+1}}\]

Answer 34

Right?

Answer 35

aye

Answer 36

Well, it's the same as...
\[\huge \frac{r}{x^{1-r}}\]and you can see it now, right? :)

Answer 37

Actually, much more better would be...
\[\huge \frac{r}{x^{-r+1}}=\frac{r}{x^{-(r-1)}}\]

Answer 38

You know where this is going? :)

Answer 39

y^-z=1/y^z works in reverse too?

Answer 40

Of course :)

Answer 41

wow thanks a lot for helping me solve this one!

Answer 42

No problem :)
You practically solved it yourself, anyway :D