Question Number 104783 by 175mohamed last updated on 23/Jul/20

find (d^n y/dx^n ) for    f(x)^ =(1/(√(1−x)))

Answered by Dwaipayan Shikari last updated on 23/Jul/20

f(x)=(1/(√(1−x)))  f^′ (x)=(1/2) (1/((1−x)^(3/2) ))  f′′(x)=(3/4) (1/((1−x)^(5/2) ))  f′′′(x)=((15)/8) (1/((1−x)^(7/2) ))  f^( n) (x)=((Π_(n=1) ^n (2n−1))/2^n ) (1/((1−x)^(n+(1/2)) ))

Answered by mathmax by abdo last updated on 24/Jul/20

f(x) =(1−x)^(−(1/2))  ⇒f^((1)) (x) =(1/2)(1−x)^(−(3/2))   f^((2)) (x) =−(3/2^2 )(1−x)^(−(5/2))  ⇒f^((3)) (x) =((3.5)/2^3 )(1−x)^(−(7/2))    ⇒f^((n)) (x) =(((−1)^(n−1) 1.3.5....(2n−1))/2^n ) (1−x)^(−(1/2)−n)  ⇒  f^((n)) (x)  =(((−1)^(n−1) 1.3.5....(2n−1))/(2^n (1−x)^(n+(1/2)) ))

Commented bybubugne last updated on 24/Jul/20

why (−1)^(n−1)  ?

Commented byabdomathmax last updated on 24/Jul/20

the sign of f^((n)) changes....

Commented bybubugne last updated on 25/Jul/20

Does the sign of f^((n))  change?  f(x)=(1−x)^(−(1/2))  ⇒ f^((1)) (x)=(−1)(−(1/2))(1−x)^(−(3/2))   f^((1)) (x)= + (1/2) (1−x)^(−(3/2))  ⇒ f^((2)) (x)= (1/2) (−1)(−(3/2))(1−x)^(−(5/2))   f^((2)) (x)= + (3/4) (1−x)^(−(5/2)) ⇒ f^((3)) (x)= (3/4) (−1)(−(5/2))(1−x)^(−(7/2))   f^((3)) (x)= + ((15)/8) (1−x)^(−(7/2))   f^((n)) (x) = + ((Π_(k=1) ^n (2k−1))/2^n ) (1−x)^(−((2n+1)/2))