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Thread: nth derivative of a function

  1. #1
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    nth derivative of a function

    I need to find that $\displaystyle n^{th}$ derivative of the function $\displaystyle f(x)=\frac{1}{(2-x)^2}$. I'm not really sure how to proceed. I wrote down a few derivatives, but I can't find a pattern. Is their a general method or formula for $\displaystyle n^{th}$ derivatives? Thanks.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by adkinsjr View Post
    I need to find that $\displaystyle n^{th}$ derivative of the function $\displaystyle f(x)=\frac{1}{(2-x)^2}$. I'm not really sure how to proceed. I wrote down a few derivatives, but I can't find a pattern. Is their a general method or formula for $\displaystyle n^{th}$ derivatives? Thanks.
    $\displaystyle f^{\prime}(x)=\frac{2}{(2-x)^3}$

    $\displaystyle f^{\prime\prime}(x)=\frac{6}{(2-x)^4}$

    $\displaystyle f^{\prime\prime\prime}(x)=\frac{24}{(2-x)^5}$

    ...

    So it follows that $\displaystyle f^{(n)}(x)=\frac{(n+1)!}{(2-x)^{n+2}}$.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by skeeter View Post
    Chris forgot a (-1) factor ... easy fix.

    $\displaystyle f^{\prime}(x)=\frac{-2}{(2-x)^3}$

    $\displaystyle f^{\prime\prime}(x)=\frac{6}{(2-x)^4}$

    $\displaystyle f^{\prime\prime\prime}(x)=\frac{-24}{(2-x)^5}$

    ...

    $\displaystyle f^{(n)}(x)=\frac{(-1)^n(n+1)!}{(2-x)^{n+2}}$.
    Don't forget to chain the denominator! $\displaystyle \frac{\,d}{\,dx}(2-x)=-1$! Hence, all the negatives are cancelled!

    So

    $\displaystyle f^{\prime}(x)=\frac{-2}{(2-x)^3}(-1)=\frac{2}{(2-x)^3}$

    $\displaystyle f^{\prime}(x)=\frac{-6}{(2-x)^{4}}(-1)=\frac{6}{(2-x)^4}$

    etc...
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  4. #4
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    my error ... sorry.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by skeeter View Post
    my error ... sorry.
    However, I see your point since $\displaystyle (2-x)^2=(x-2)^2$...

    If we rewrote the function as $\displaystyle f(x)=\frac{1}{(x-2)^2}$, then you would be correct....
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    However, I see your point since $\displaystyle (2-x)^2=(x-2)^2$...

    If we rewrote the function as $\displaystyle f(x)=\frac{1}{(x-2)^2}$, then you would be correct....
    Ok, that's why I was having difficulty find a pattern. I didn't notice $\displaystyle (2-x)^2=(x-2)^2$.
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  7. #7
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    Quote Originally Posted by adkinsjr View Post
    Ok, that's why I was having difficulty find a pattern. I didn't notice $\displaystyle (2-x)^2=(x-2)^2$.
    It would also make this easier to recognize that $\displaystyle f(x)= \frac{1}{(2-x)^2}= \frac{1}{(x-2)^2}= (x- 2)^{-2}$.

    $\displaystyle f'(x)= -2(x-2)^{-3}$
    $\displaystyle f"(x)= 6(x-2)^{-4}= (-1)^2(2+1)!)(x-2)^{-(2+2)}$
    $\displaystyle f"'(x)= -12(x-2)^{-5}= (-1)^3(3+1)!)(x-2)^{-(3+2)}$
    $\displaystyle f""(x)= 60(x-2)^{-6}= (-1)^4(4+1)!(x-2)^{-(4+2)}$
    Last edited by Chris L T521; Aug 15th 2009 at 02:36 PM. Reason: fixed LaTeX error.
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