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Math Help - finding a derivative

  1. #1
    Member Chokfull's Avatar
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    finding a derivative

    Find f'(a)

     <br />
f(x)=\frac {1} {\sqrt{x+2}}<br />

    So I use the formula f'(a)=\lim_{h\to0}\frac {f(a+h)-f(a)} {h} and get \frac {\frac {1} {\sqrt{a+h+2}}-\frac {1} {\sqrt{a+2}}} {h}. My Maple 13 calculator says this cannot be simplified unless you throw in the limit as h approaches 0, which gets

     <br />
-\frac {1} {2(a+2)^{3/2}}<br />

    How does it reach this answer? Thanks to anyone who helps!
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    Quote Originally Posted by Chokfull View Post
    Find f'(a)

     <br />
f(x)=\frac {1} {\sqrt{x+2}}<br />

    So I use the formula f'(a)=\lim_{h\to0}\frac {f(a+h)-f(a)} {h} and get \frac {\frac {1} {\sqrt{a+h+2}}-\frac {1} {\sqrt{a+2}}} {h}. My Maple 13 calculator says this cannot be simplified unless you throw in the limit as h approaches 0, which gets

     <br />
-\frac {1} {2(a+2)^{3/2}}<br />

    How does it reach this answer? Thanks to anyone who helps!
    Maple 13 calculator (whatever that is) knows squat about good tricks in mathematics: throw it away.
    After you simplify the expression for f'(a) within the limit you get:

    [Sqrt(a+2) - Sqrt(a+h+2)]/[h*sqrt(a+2)sqrt(a+h+2)] .

    Multiply the above by the numerator's conjugate (the same expression as the numerator but with a + sign instead the - one) and simplify. You get an expression perfectly well defined and continuous when h --> 0 and thus you can directly substitute h = 0 and get what you want.

    Tonio
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    hi
    compute {f}' which gives you {f}'(x)= -\frac{1}{2(x+2)^{\frac{3}{2}}} then plug a.
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  4. #4
    Member Chokfull's Avatar
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    Quote Originally Posted by tonio View Post
    Maple 13 calculator (whatever that is) knows squat about good tricks in mathematics: throw it away.
    After you simplify the expression for f'(a) within the limit you get:

    [Sqrt(a+2) - Sqrt(a+h+2)]/[h*sqrt(a+2)sqrt(a+h+2)] .

    Multiply the above by the numerator's conjugate (the same expression as the numerator but with a + sign instead the - one) and simplify. You get an expression perfectly well defined and continuous when h --> 0 and thus you can directly substitute h = 0 and get what you want.

    Tonio
    Yes, then divide by \frac {h} {h} to remove its being inthe denominator. Thanks!
    Quote Originally Posted by Raoh View Post
    hi
    compute {f}' which gives you {f}'(x)= -\frac{1}{2(x+2)^{\frac{3}{2}}} then plug a.
    The original problem was finding the derivative, with \lim_{x\to a}, so that didnt really help, sorry. but thanks for the effort
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