nth derivative of cos^2 x = (a) 2^(n-1) ...

nth derivative of `cos^2 x` = (a) `2^(n-1) cos((npi)/2 + 2x)` (b) `2^(n-1) cos((npi)/2 - 2x)` (c) `2^(n+1) cos((npi)/2 + 2x)` (d) `2^(n+2) cos((npi)/2 + 2x)`

Similar Questions

Explore conceptually related problems

nth derivative of cos^(2)x=(a)2^(n-1)cos((n pi)/(2)+2x)(b)2^(n-1)cos((n pi)/(2)-2x)(c)2^(n+1)cos((n pi)/(2)+2x) (d) 2^(n+2)cos((n pi)/(2)+2x)

If n is integer then show that (1 + i)^(2n) + (1 - i)^(2n) = 2 ^(n+1) cos . (npi)/2 .

Prove that cos x+^(n)C_(1)cos2x+^(n)C_(2)cos3x+....+^(n)C_(n)cos(n+1)x=2^(n)*(cos^(n)x)/(2)*cos((n+2)/(2))x

If n is a positive integer, show that (1 + i)^(n) + (1 - i)^(n) = 2 ^((n+2)/2) cos ((npi)/4) .

If cos^2x+cos^2 2x+cos^2 3x=1 then (a) x=(2n+1)pi/4,\ n in I (b) x=(2n+1)pi/2,\ n in I (c) x=npi+-\ pi/6,\ n in I (d) none of these

General solution of the equatin 1+sin^4 2x=cos^2 6x is (A) (npi)/3 (B) (npi)/2 (C) npi (D) none of these

Prove that .^(n)C_(0) + .^(n)C_(5) + .^(n)C_(10) + "….." = (2^(n))/(5) (1+2cos^(n)'(pi)/(5)cos'(npi)/(5)+2cos'(pi)/(5)cos'(2npi)/(5)) .

A general solution of tan^2theta+cos2theta=1 is (n in Z) npi=pi/4 (b) 2npi+pi/4 npi+pi/4 (d) npi