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Math Help - A partial differentiation

  1. #1
    Lex
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    A partial differentiation

    Hi,

    I am trying to differentiate
    (N*k/3)*[T-(Q/6.7)*arctan(6.7T/Q)]
    with respect to T.

    Apparently the answer is
    (N*k/3)*[(6.7T/Q)^2 / 1 + (6.7T/Q)^2]

    I am getting
    (N*k/3)*[1/(1+(6.7T/Q)^2)]

    Could someone give me breakdown of how this is done and perhaps point out where I have gone wrong?

    Thanks

    (P.S. sorry for not being able to use MATH)
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  2. #2
    o_O
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    Is this it: F = \frac{Nk}{3}\left[ {\color{red}T} - \frac{Q}{6.7}\arctan \left(\frac{6.7{\color{red}T}}{Q}\right)\right] ?

    It's kind of hard to tell from what you've written. Anyway, assuming that it is, to differentiate with respect to T, treat every other variable as constants.

    So:  \frac{\partial F}{\partial T} = \frac{Nk}{3} \left[ {\color{red}1} - \frac{Q}{6.7}\underbrace{\left( \frac{\frac{6.7}{Q}}{1+ \left(\frac{6.7T}{Q}\right)^2}\right) }_{\displaystyle \text{Since: } \frac{d}{dx} \arctan ({\color{red}u}) = \frac{{\color{red}u}'}{1+{\color{red}u}^2}}\right]
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Lex View Post
    Hi,

    I am trying to differentiate
    (N*k/3)*[T-(Q/6.7)*arctan(6.7T/Q)]
    with respect to T.

    Apparently the answer is
    (N*k/3)*[(6.7T/Q)^2 / 1 + (6.7T/Q)^2]

    I am getting
    (N*k/3)*[1/(1+(6.7T/Q)^2)]

    Could someone give me breakdown of how this is done and perhaps point out where I have gone wrong?

    Thanks

    (P.S. sorry for not being able to use MATH)
    Hi Lex,

    \left(N\frac{k}{3}\right)\frac{d}{dT}\left[T-\left(\frac{Q}{6.7}\right)Tan^{-1}\left(\frac{6.7T}{Q}\right)\right]

    \frac{d}{dx}Tan^{-1}\frac{x}{a}=\frac{a}{a^2+x^2}

    a=\frac{Q}{6.7},\ x=T

    The derivative is

    \left(N\frac{k}{3}\right)\left[1-a\frac{a}{x^2+a^2}\right] =\left(N\frac{k}{3}\right)\left[\frac{x^2+a^2-a^2}{x^2+a^2}\right]

    =\left(N\frac{k}{3}\right)\left[\frac{T^2}{T^2+\left(\frac{Q}{6.7}\right)^2}\right]=\left(N\frac{k}{3}\right)\left[\frac{\left(\frac{6.7T}{Q}\right)^2}{1+\left(\frac  {6.7T}{Q}\right)^2}\right]
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  4. #4
    Lex
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    Thanks to all.

    It appears that my calculus was correct, but it was the rearrangement that eluded me.
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