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Math Help - fine the derivative

  1. #1
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    fine the derivative

    d/dx [(tanx)^{secx}]

    do i use substition or the chain rule?
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  2. #2
    zeg
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    use chain rule

    and the fact that \frac{d}{dx} a^x = ln(a)a^x
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  3. #3
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    Quote Originally Posted by zeg View Post
    and the fact that \frac{d}{dx} a^x = ln(a)a^x
    so that is equal to ln(tanx)tanx^{secx}, what am i suppose to do now
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  4. #4
    zeg
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    my post was incorrect, sorry

    that is for when a is a constant in this case, it is a function of x, so here's my stab at it:
    y = tan(x)^{sec(x)}
    ln(y) = sec(x)\cdot ln(tan(x))
    now take your derivative, d/dx
    (ln(y))^{\prime} = (sec(x))^{\prime}\cdot ln(tan(x)) + sec(x)\cdot (ln(tan(x)))^{\prime}
    \frac{y^{\prime}}{y} = sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)

    so

    y^{\prime} = [sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)]\cdot tan(x)^{sec(x)}

    that seems messy though. maybe there's some nice simplifications in there somewhere.
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  5. #5
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    Quote Originally Posted by zeg View Post
    that is for when a is a constant in this case, it is a function of x, so here's my stab at it:
    y = tan(x)^{sec(x)}
    ln(y) = sec(x)\cdot ln(tan(x))
    now take your derivative, d/dx
    (ln(y))^{\prime} = (sec(x))^{\prime}\cdot ln(tan(x)) + sec(x)\cdot (ln(tan(x)))^{\prime}
    \frac{y^{\prime}}{y} = sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)

    so

    y^{\prime} = [sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)]\cdot tan(x)^{sec(x)}

    that seems messy though. maybe there's some nice simplifications in there somewhere.

    will this work, since d/dx(p^x) = p^xlnp, then we have tanx^{secx}lntanx = tanx^{secx}/tanx but i dont know what tanx^{secx} is
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  6. #6
    zeg
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    i'm not sure how to simplify that any further

    i'm not sure how to simplify that any further. Maybe use the Euler relations?

    sin(x) = \frac{e^{ix} - e^{-ix}}{2i}
    cos(x) = \frac{e^{ix} + e^{-ix}}{2}

    if you can simplify tan(x)^{sec(x)} with these relations, then I would do that first, then take the derivative. sorry I can't be more helpful!
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  7. #7
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    Quote Originally Posted by zeg View Post
    i'm not sure how to simplify that any further. Maybe use the Euler relations?

    sin(x) = \frac{e^{ix} - e^{-ix}}{2i}
    cos(x) = \frac{e^{ix} + e^{-ix}}{2}

    if you can simplify tan(x)^{sec(x)} with these relations, then I would do that first, then take the derivative. sorry I can't be more helpful!

    thx, you have gave a great help already, don't worry about it
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  8. #8
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    Quote Originally Posted by zeg View Post
    that is for when a is a constant in this case, it is a function of x, so here's my stab at it:
    y = tan(x)^{sec(x)}
    ln(y) = sec(x)\cdot ln(tan(x))
    now take your derivative, d/dx
    (ln(y))^{\prime} = (sec(x))^{\prime}\cdot ln(tan(x)) + sec(x)\cdot (ln(tan(x)))^{\prime}
    \frac{y^{\prime}}{y} = sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)

    so

    y^{\prime} = [sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)]\cdot tan(x)^{sec(x)}

    that seems messy though. maybe there's some nice simplifications in there somewhere.

    I think you made a mistake here, (lntan(x))' = csc x sec x, so that is equal to cscx sec^2x
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