IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sboosy

f(x) = x ^(n-1)n

1st derivative is

n(n-1)x ^{[(n-1)n -1]}

this way we get n the derivative as

(n(n-1)-0) (n(n-1) -1) (.....) ( n(n-1) - (n-1) * x ^{(n(n-1) - n)}

x = 1

so we get

(n(n-1)-0) (n(n-1) -1) (.....) ( n(n-1) - (n-1))

of the form

t(t-1)(t-2)(t-3) ..(t-(n-1))

[n(n-1)] ! / [n(n-2)] !

hsbhatt

hey, you've noted down the function wrongly i think.

rockey

is it n!(0) =0

nudge me if im wrong

elastiboysai

edited

akhil_o

I think it is

ⁿC₀(2n)!-ⁿC₁(2n-1)!+ⁿC₂(2n-2)!-....+ⁿC_n(n)!

Binomially expanding the function,

nC₀x²ⁿ-ⁿC₁x^2n-1+ⁿC₂x^2n-2-.....ⁿC_nxⁿ

now after taking n derivatives power of derivative keeps reducing and so we have

2n*(2n-1)...1=2n!, (2n-1)! and so on

hsbhatt

elasti: just verify your answers substituting n=1, n=2 etc. the expression does not become zero at the nth derivative.

akhil also needs to check his working.

elastiboysai

P=(2n!)

ci =nCr

c0P-c1P/2n+c2P/(2n)(2n-1)-c3P/(2n)(2n-1)(2n-2)+...

computer001

is it (n^n)*(n!)

hsbhatt

That's fantastic computer001, just go ahead and post ur solution

computer001

f(x)=(x^n-1)^n
1st derivative=(n^2)((x^n-1)^n-1)*x^n-1
now basically we need to bother only abt the terms in which differentiation of (x-1)^n goes on as in the other terms when u later substitue x=1 they become 0 anyway..in the nth derivative the (x-1)^n term which was there initially becomes(x-1)^0..so this term alond with its coeff and x power sometihng is the only 1 which does not become 0 for x=1..the coeff of this will be n^n*n! as on each diff 'n' is contributed due to the x^n inside the bracket and overall n! is produced due to continuous diff of the initial (...)^n..

computer001

i din do ne calculations so i cant exactly post a 'solution'..this is the way i reasoned it out...hope its clear..

hsbhatt

the answer is right. great job computer!

But, any volunteers for a working that can be easily understood.

Ask & Discuss Questions with Community & Experts