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Math Help - n+1 th derivative

  1. #1
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    Arrow n+1 th derivative

    Hi,

    can someone please help me with the derivative of the following equation.

    f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

    I kind of know the answer for this but just don't know how to get there.(need steps).

    f^(n+1).(x) = −[(2(n+1)!/4(n+1).(n+1)!.(2(n+1) − 1)].{(1 − x)^1/2−(n+1)}

    thanks in advance
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  2. #2
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    Quote Originally Posted by ammayi View Post
    Hi,

    can someone please help me with the derivative of the following equation.

    f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

    I kind of know the answer for this but just don't know how to get there.(need steps).
    Well, there are two ways to do this. Firstly, you can substitute (n+1) for n, which is what you've done here:

    f^(n+1).(x) = −[(2(n+1)!/4(n+1).(n+1)!.(2(n+1) − 1)].{(1 − x)^1/2−(n+1)}
    And which we can simplify:

    f^{(n+1)}(x)=-\frac{(2n+2)!}{(4n+4)(n+1)!(2n+1)}(1-x)^{(\frac{1}{2}-n-1)}

    =-\frac{(2n)!(2n+1)(2n+2)}{2(2n+2)(n+1)!(2n+1)}(1-x)^{(\frac{1}{2}-n-1)}=-\frac{(2n)!}{2(n+1)!}(1-x)^{(\frac{1}{2}-n-1)}

    =-\frac{(2n)!}{n!2(n+1)}(1-x)^{(\frac{1}{2}-n-1)}=-\frac{(2n)!}{n!(2n+2)}(1-x)^{(\frac{1}{2}-n-1)}

    To check our work, we can take the derivative of f^n(x), like so:

    f^{(n+1)}(x)=\frac{d}{dx}[-\frac{(2n)!}{4n!n(2n-1)}(1-x)^{(\frac{1}{2}-n)}]=-\frac{(2n)!}{4n!n(2n-1)}\frac{d}{dx}[(1-x)^{(\frac{1}{2}-n)}]

    =\frac{(2n)!}{4n!n(2n-1)}(\frac{1}{2}-n)(1-x)^{(\frac{1}{2}-n-1)}=[\frac{(2n)!}{4n!n(2n-1)2}-\frac{(2n)!n}{4n!n(2n-1)}](1-x)^{(\frac{1}{2}-n-1)}

    =[\frac{(2n)!}{4n!n(2n-1)2}-\frac{2(2n)!n}{4n!n(2n-1)2}](1-x)^{(\frac{1}{2}-n-1)}=\frac{(2n)!-2(2n)!n}{4n!n(2n-1)2}(1-x)^{(\frac{1}{2}-n-1)}

    =\frac{(2n)!(1-2n)}{4n!n(2n-1)2}(1-x)^{(\frac{1}{2}-n-1)}=-\frac{(2n)!}{n!8n}(1-x)^{(\frac{1}{2}-n-1)}

    And we can see that this does not check out. So, something is either wrong with my work or with the original functions. I believe it is the latter.

    thanks in advance
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  3. #3
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    Quote Originally Posted by ammayi View Post
    Hi,

    can someone please help me with the derivative of the following equation.

    f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

    I kind of know the answer for this but just don't know how to get there.(need steps).

    f^(n+1).(x) = −[(2(n+1)!/4(n+1).(n+1)!.(2(n+1) − 1)].{(1 − x)^1/2−(n+1)}

    thanks in advance
    f^(n).(x) = −[(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2−n}

    To differentiate, multiply by the power, \frac{1}{2} - n, and substract 1 off the power. Don't forget to also multiply by -1:

    f^(n+1).(x) = \left(\frac{1}{2} - n\right) [(2n)!/4n.n!.(2n − 1)].{(1 − x)^1/2− n - 1}
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  4. #4
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    Arrow thanks

    Hi, thanks a lot for your help but i forgot to mention that its an induction problem. so i have to show that value i get by substituting (n+1) in the equ and the value i get by differentiating nth derivative, are the same.

    Now..i am not sure if the way i am approaching this questions is right.
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