IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

yogeshmangaj

If f(y)=(1+x)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)........(1+x²ⁿ)
then f'(0)=?

Also, what would be the answer if 2n would be 2ⁿ ?

I intentionally haven't given the choices.

Ans: [ 1         ]
[click and move ur mouse thru the brackets]
================
Well guys, let me tell I ain't no expert in this stuff !!
I found tht question in some crap video,
and posted it here coz i cud'nt understand it myself.

I think the question's got to be 2ⁿ
So the soln is:

f(y)=(1+x)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)........(1+x^{₂^n})

Multiply n divide by (1-x) v get

(1-x)(1+x)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)........(1+x^{₂^n})
-----------------------------------------------------------
                             (1-x)

(1-x)(1+x)=(1-x²)

(1-x²)(1+x²)=(1-x⁴)

And so on....
finally we get

f(x)= (1-x^{₂^n})(1+x^{₂^n})    (1-x^{₂₊₂^n})
         ------------------- = ---------
   (1-x)                   (1-x)
something like tht
n then diff we get 1

[ is that correct ??? ]

---------------------------------
I know i've put lots of colors in there, but i like it tht way!
N thanks for all ur answers !!

sup159

can u post the sol

akhil_o

(1+x)(1+x²)(1+x⁴)...
upto 2ⁿ
so there are n terms
so the coefficient of x=1+1+1+1...1 (n-1 times)=n-1
so expansion is

1+(n-1)x+....
taking derivative
f'(x)=(n-1)+.....
we dont take other terms as they will become 0 at f'(0)
so putting x=0
f'(0)=n-1

sandeepramesh

f(x) = (x-1)f(x)/(x-1) = x^32 - 1/x-1

f'(0) = lim x->0 f(x) = 1

I think you have given 2^n only and not 2n in the question? Huh?

sugeet

I guess ans is 2(1+ 2+3 +4+.................+n)
so n*(n+1) am I right for da first one or do tell me , I am not at all confident , jus a try

hsbhatt

f(x) = (1+x)(1+x²)(1+x⁴)...(1+x²ⁿ)

If you use product rule for differentiation

f'(x) = (1+x²)(1+x⁴)...(1+x²ⁿ)+ x g(x)

as the differentiation of every other factor produces an x term. e.g. d/dx (1+x²) = 2x.

Hence f'(0) = 1.

Ditto for 2ⁿ case

sandeepramesh

but hasnt the person given a 2^n case only as the question? Huh?

hsbhatt

He says 2n and also 2ⁿas far as I can see

sugeet

but how do we get xg(x)??????

sandeepramesh

he' meant the original question to be 2n and the replacement to be 2^n but in the original question, he hasnt given 1+x^6 at all, and he's written for 2^n only

budokai_tenkaichi

its understood

yahan koi math ki dictionary leke nahin baitha hai

hsbhatt

yeah, he has given for 2ⁿ case actually. Anyway procedure is the same for the 2n case (nice flip-flop no?

)

ashish_banga

in the question LHS is a function of y and RHS is a function of x

so answer should be 0

sugeet

i got it thax

Ask & Discuss Questions with Community & Experts