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Ok so good so far...you have the 1st derivative right...
Okay...checking your second

Great!

Ok that one is fine...I'm going to assume your others one are correct then lol

Haha, yeah I think they are, a lot of work you're doing ;)

Ok..I see part of it...one sec.

\[f^{(2)}=\frac{2 \cdot 3}{(x+1)^3} , f^{(3)}= - \frac{2 \cdot 3 \cdot 4}{(x+1)^4}\]

You notice factorial is involved?

I just notice the exponents and how the top was changing.

Oh, I didn't actually

So you almost had it :)

But the (-1)^n+1 should be there to control the signs right?

Right! Totally.

Your only expression that doesn't work is n=1
So I would say for your expression n>=2

That is when the pattern starts to occur

And you want for n=2 for the expression to be negative so yeah n+1 works
(-1)^{n+1} is -1 for n=2 :)

So the it's
\[\frac{ (-1)^{n+1} n! }{ (x+1)^{n+1} }\]
\[n\]

n>=2

Yep that is exactly right :)

thing*

Find by induction it says, so isn't this the wrong way?

I think it is weird it says find by induction.

Usually I see show or prove by induction.

Ok, maybe they just mean proove the n:th derivitive by induction

If that is the case, I can help you with that

I assume that's it. We've never talked about finding the derivitive by induction anyway

Recall,
\[f^{(k+1)}(x)=(f^{(k)}(x))'\]

Oh, so proving k+1 is just the derivitive of n=k ?

That makes perfect sense

yep just like
\[f^{(3)}(x)=(f^{(2)}(x))'\]

Or for any known value k

Exactly, I get it!

A million thank you myininaya!

Really great tutor!

too*

One last question, the derivitive of the factorial part, how does that work?

Ohh I see it know, it nothing but a constant to :)

now*

You know what I don't think you actually need to do an even k and odd k

It doesn't matter if k is even or odd, you can still do it as one case :)

Yeah, it works out nicely :)

Awesome! I tag you later on if i get any trouble but i doubt it :) Once again, thank you so much!:D

You did, really valueble!