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Math Help - Calculus: Differentiation

  1. #1
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    Angry Calculus: Differentiation

    I just started Calculus and here is a problem I just cannot seem to get the hold of:
    "Use the relation 1/(a2 - x2) = 1/2a [1/(a+x) + 1/(a-x)] to find the nth derivative of 1/(a2-x2​)
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Calculus: Differentiation

    This forum is for introductions, the calculus forum would be the place for this topic. I know you are new here, so I am just offering friendly advice.

    I would look at the first few derivatives to see if a pattern emerges. We need only work on the factor involving x.

    \frac{d}{dx}\left(\frac{1}{a+x}+\frac{1}{a-x} \right)=-\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}

    \frac{d}{dx}\left(-\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2} \right)=2\left(\frac{1}{(a+x)^3}+\frac{1}{(a-x)^3} \right)

    \frac{d}{dx}\left(2\left(\frac{1}{(a+x)^3}+\frac{1  }{(a-x)^3} \right) \right)=6\left(-\frac{1}{(a+x)^4}+\frac{1}{(a-x)^4} \right)

    We may now state the induction hypothesis P_n:

    \frac{d^n}{dx^n}\left(\frac{1}{a^2-x^2} \right)=\frac{n!}{2a}\left(\frac{(-1)^n}{(a+x)^{n+1}}+\frac{1}{(a-x)^{n+1}} \right)

    Now, differentiate both sides to try to obtain P_{n+1}.
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    Re: Calculus: Differentiation

    Thanks
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