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Thread: nth derivative

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    Default nth derivative

    can any one check if I am correct?
    Find the 5th derivative of



    I got 1440x+45/8(x^(7/2))
    Last edited by Main; 02-25-2010 at 01:51 AM.
    Oh Hai Dar

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    1440x+((64((96(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x)+((-24(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2-(96(1/(2sqrt(x)))^2*sqrt(x)))/(16(sqrt(x))^4)*4(sqrt(x))^2+48(1/(2sqrt(x)))^3+(-192(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x))*1/(2sqrt(x))*(sqrt(x))^3+((96(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*1/(2sqrt(x))+((96(1/(2sqrt(x))))/(4(sqrt(x))^2)*(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)+(96(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2-768(1/(2sqrt(x))*1/(2sqrt(x))*sqrt(x)))/(16(sqrt(x))^4)*1/(2sqrt(x)))*sqrt(x)+8((-24(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2-(96(1/(2sqrt(x)))^2*sqrt(x)))/(16(sqrt(x))^4)*1/(2sqrt(x))*sqrt(x)+((8(-24(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x)+((-24(-2(1/(2sqrt(x))))/(4(sqrt(x))^2))*4(sqrt(x))^2+192(1/(2sqrt(x))*1/(2sqrt(x))*sqrt(x)))/(16(sqrt(x))^4)*4(sqrt(x))^2-((96(1/(2sqrt(x)))^3+192(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x))))*16(sqrt(x))^4-64((-24(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2-(96(1/(2sqrt(x)))^2*sqrt(x)))*1/(2sqrt(x))*(sqrt(x))^3)/(256(sqrt(x))^8)*4(sqrt(x))^2+144(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)*(1/(2sqrt(x)))^2+(-192(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*1/(2sqrt(x))+((-192(1/(2sqrt(x))))/(4(sqrt(x))^2)*(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)+((-192(-2(1/(2sqrt(x))))/(4(sqrt(x))^2))*4(sqrt(x))^2+1536(1/(2sqrt(x))*1/(2sqrt(x))*sqrt(x)))/(16(sqrt(x))^4)*1/(2sqrt(x)))*sqrt(x))*16(sqrt(x))^4-((192((12(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2+48(1/(2sqrt(x)))^2*sqrt(x))*1/(2sqrt(x))*1/(2sqrt(x))*(sqrt(x))^2+(64((12(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2+48(1/(2sqrt(x)))^2*sqrt(x))*(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)+64(8(12(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x)+(12(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2-96(1/(2sqrt(x))*1/(2sqrt(x))*sqrt(x)))/(16(sqrt(x))^4)*4(sqrt(x))^2+48(1/(2sqrt(x)))^3+96(-2(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x))*1/(2sqrt(x)))*(sqrt(x))^3)))*256(sqrt(x))^8-2048(((96(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x)+((-24(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2-(96(1/(2sqrt(x)))^2*sqrt(x)))/(16(sqrt(x))^4)*4(sqrt(x))^2+48(1/(2sqrt(x)))^3+(-192(1/(2sqrt(x))))/(4(sqrt(x))^2)*1/(2sqrt(x))*sqrt(x))*16(sqrt(x))^4-(64((12(1/(2sqrt(x))))/(4(sqrt(x))^2)*4(sqrt(x))^2+48(1/(2sqrt(x)))^2*sqrt(x))*1/(2sqrt(x))*(sqrt(x))^3))*1/(2sqrt(x))*(sqrt(x))^7)/(65536(sqrt(x))^16)
    This is what my math programme says @_@. Anyone feels like shortening it XD? I'm glad that I'm done with those, i never found any use for them xD....

    ~caused

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    what program was that lol
    Oh Hai Dar

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    I get 315/(32 x^(9/2))+1440x

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    Wolfram Alpha got this: d^6/dx^6(2 x^6+3 sqrt(x)) = 1440-2835/(64 x^(11/2))
    :-)

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    Quote Originally Posted by Method View Post
    Wolfram Alpha got this: d^6/dx^6(2 x^6+3 sqrt(x)) = 1440-2835/(64 x^(11/2))
    That's the 6th derivative.

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    Hmm, guess I read sixth by accident. In that case, senrath's answer is correct.
    :-)

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    Quote Originally Posted by Main_Ftw View Post
    what program was that lol
    It's amazing german software :].

    Called Kurvenprofi. xD

    ~caused

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    god..program now days...

    ty senrath.
    Oh Hai Dar

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    Quote Originally Posted by caused View Post
    I'm glad that I'm done with those, i never found any use for them xD....~caused
    Except that calculus runs the whole of the modern world

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    Quote Originally Posted by silentwolf View Post
    Except that calculus runs the whole of the modern world
    That is true. Though in my projects i rarely found any use for functions higher than x^2 .

    ~caused

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    Quote Originally Posted by Main_Ftw View Post
    can any one check if I am correct?
    Find the 5th derivative of



    I got 1440x+45/8(x^(7/2))
    If you put a little thought into it as to what it should actually look like, you would see that x^(7/2) wouldn't work out correctly. If you have x^(1/2) as your function, to take the derivative you obviously subtract 1 from the exponent each derivative. Considering that 1/2 - 1 = 1/2 - 2/2, you should be able to see that (2/2)*5 (5th derivate) = 10/2, so you could easily know that you have to end with an exponent on the x of 1/2 - 10/2 = -9/2, or 1/(x^(9/2)).

    Just a little thinking.

    "Failure is the opportunity to begin again more intelligently" (Henry Ford)


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    yeah. I there is kind of a pattern now, after doing checking and rechecking for like 3 hours -_- (My TI even graphed it wrong).
    Oh Hai Dar

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    12x^5 + 1.5 x ^-1/2 (1st)
    60 x^4 - .75 x^-1.5 (2nd)
    240 x^3 + 1.125 x^-2.5 (3rd)
    720 x^2 - 2.8125 x^-3.5 (4th)
    1400 x + 9.84375 x ^ 4.5 (5th)

    which is 1400x + 9.84375/SQRT(x^9)

    so senrath is correct

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