if xⁿ/ (1+x) find f '' (1)?

Question

if f(x) = xⁿ/ (1+x) find f '' (1)

the xⁿ is actually supposed to be x squared .. but i didn't know how to do that on the computer..

the answer in the back of the book turned out to be 1/4.
we are supposed to use (f/g)' = (g)(f')-(f)(g') / g^2

Answer 1

Need a little more info to help you! What is xⁿ/ (1+x) ? Is it f(x)? Is it f ' (x) ?

Answer 2

x^2/(x+1)

I will teach you some little tricks that have been useful to me

First add and sustract 1

[(x^2 -1)+1]/(x+1) =(x^2 -1)/(x+1) + 1/(x+1)

Since x^2 -1 = (x+1)(x-1), the first term is x-1, very easy to take its derivatives. The first one is 1, the second one is 0

So, your functions 2nd derivative is equals to the 2nd derivative of 1/(x+1)=(x+1)^-1

Now consider this:

(u^n)' = n u^(n-1) u'

So, applying this formula, and considering that(x+1)' is 1, you can find in a very easy way its first and its second derivative

[1/(x+1)]' = [(x+1)^-1]' = -(x+1)^-2 = -1/(x+1)^2

And its second derivative is:

[-(x+1)^-2]' = -(-2)(x+1)^-3 = 2/(x+1)^3


So, this is your second derivative. Now just plug 1 in this expression. And yes, the answer is 2/8 = 1/4.

Ana

Answer 3

1^n/1+1, n can be any number and still be 1, so the final answer is 1/2.

f(1), this means you have to put a 1 where the x are, example; f(x)=x+3 f(2), f(x)= 2+3, f(x)=5

Answer 4

if f(x) = 1, then x^2 / (1+x) = 1/2

(1)^2 --> 1
------- ---
1+(1) --> 2

Answer 5

what do u really want?
f '' or just the f ' ?

i found 1/4 for the first diferenciate (f ')
i couldnt resolve the f '', its too hard

look:
(x squared will be called XS ok?)

f '=[(1+x)(1/2*XS) - XS] / (x+1)²

put the number 1 on every X and u will find 1/4

im sorry about my poor english, im not from USA!, if u really want the f'' email me


more: if u want extra explanations,
for example why (XS)'=(1/2XS) email me too!
ill be glad to help, and dont mind these ppl who doesnt even know what f' means!

Answer 6

This is the function of an elipse, so it turns into,
x^(n-1).x/1+x, which is again
x^(n-1)(ax^2+bx+c),
f'x=x^(n-1)(a.2x+b)+
(ax^2+bx+c).(n-1)x^(n-2)
f''x = x^(n-1).(2a)+
(2ax+b).(n-1).x^(n-2)+
(ax^2+bx+c).(n-1).(n-2).
x^(n-3)
+(n-1).x^(n-2).(2ax+b)
here a=sq.rt.(1+0.9^2), b=0.9/a and c= 0.9a
choose an n value here 2 as you mentioned and substitute x= 1 in the f''x equation given above to get the value of f''(1)
that is f''(1)=9.44. may be close to 10.

using your formula
f'x=((1+x).(n.x^n-1)-(x^n))/(1+x)^2
f''x=(1+x)^2.((1+x).
n.n-1.x^n-2)+
n.x^n-1-(n.X^n-1)-
((1+x).(n.x^n-1)-(x^n)).
2.(1+x)//(1+x)^4=1
Please varify by substituting.
Both produce similar answer only thing is if we take 'a' to be close to 1 instead of squrt. 1+0.9^2
we will get similar answer in my version.
**E&OE

Answer 7

Wouldn't you learn more if you found it?

Answer 8

fx=x^n/1+x
fx'=nx^n-1/[1+x]+x^n*-1/[1+x]^2
=nx^n-1/[1+x]-x^n/[1+x]^2
fx"=n*[n-1]x^n-2/[1+x]-2nx^n-1/[1+x]^3
-nx^n-1/[1+x]^2+2x^n/[1+x]^3
f"(1)=n[n-1]/2-2n/8-n/4+2/8
=n[n-1]/2-n/4-n/4+1/4
=n^2/2-n/2-n/2+1/4
=1/2*n^2-n+1/4