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Math Help - The derivative of the arcsec function

  1. #1
    Member dokrbb's Avatar
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    The derivative of the arcsec function

    I have got a funny result for a problem:

    given the function f(x) = 5arcsec (6x)

    by differentiating I got f'(x) = \frac{5}{6x*sqrt((6x)^2 - 1)} *{6} = \frac{30}{6x*sqrt(36x^2 - 1)} and the answer is considered a wrong one...

    , but by calculating f(5) for what I obtained gave me a correct answer, is that a coincidence or I miss something?

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    Re: The derivative of the arcsec function

    That "6x" is wrong. You are using the chain rule but you need to multiply by the derivative of 6x, 6, not the 6x itself.
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    Re: The derivative of the arcsec function

    do you mean under the square root too, or not...

    the right answer would be like this f'(x) = \frac{5}{x^2*sqrt(36 - (1/x^2))}

    is it a kind of factorization, and why do I have to take the x^2 from the square root?

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    Re: The derivative of the arcsec function

    Quote Originally Posted by dokrbb View Post
    I have got a funny result for a problem:

    given the function f(x) = 5arcsec (6x)

    by differentiating I got f'(x) = \frac{5}{6x*sqrt((6x)^2 - 1)} *{6} = \frac{30}{6x*sqrt(36x^2 - 1)} and the answer is considered a wrong one...

    , but by calculating f(5) for what I obtained gave me a correct answer, is that a coincidence or I miss something?

    Thanks
    \displaystyle \begin{align*} y &= 5\,\textrm{arcsec}\,{(6x)} \\ \frac{y}{5} &= \textrm{arcsec}\,{(6x)} \\ \sec{\left( \frac{y}{5} \right) } &= 6x \\ \left[ \cos{\left( \frac{y}{5} \right) } \right] ^{-1}  &= 6x \\ \frac{d}{dx} \left\{ \left[ \cos{ \left( \frac{y}{5} \right) }  \right] ^{-1} \right\} &= \frac{d}{dx} \left( 6x \right) \\ \frac{1}{5} \left[ -\sin{ \left( \frac{y}{5} \right) } \right] \left\{ -\left[ \cos{ \left( \frac{y}{5} \right) } \right] ^{-2} \right\} \frac{dy}{dx} &= 6 \\ \frac{1}{5} \tan{ \left( \frac{y}{5} \right) } \sec{ \left( \frac{y}{5} \right) } \, \frac{dy}{dx} &= 6 \\ \frac{1}{5} \, \sqrt{ \sec ^2{ \left( \frac{y}{5} \right) }  -1 } \, \sec{ \left( \frac{y}{5} \right) } \,\frac{dy}{dx} &= 6 \\ \frac{1}{5} \, \sqrt{ \left( 6x \right) ^2 - 1 }  \left( 6x \right) \frac{dy}{dx} &= 6 \\ \frac{dy}{dx} &= \frac{30}{6x\,\sqrt{ 36x^2 - 1 }} \\ \frac{dy}{dx} &= \frac{5}{x\,\sqrt{ 36x^2 - 1 }} \end{align*}

    I completely agree with the answer that you gave.
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    Re: The derivative of the arcsec function

    Quote Originally Posted by HallsofIvy View Post
    That "6x" is wrong. You are using the chain rule but you need to multiply by the derivative of 6x, 6, not the 6x itself.
    The OP DID multiply by the 6 you are referring to, it's after the first fraction...
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    Re: The derivative of the arcsec function

    Quote Originally Posted by Prove It View Post
    \frac{dy}{dx} &= \frac{30}{6x\,\sqrt{ 36x^2 - 1 }} \\ \frac{dy}{dx} &= \frac{5}{x\,\sqrt{ 36x^2 - 1 }}

    I completely agree with the answer that you gave.
    thanks a lot for this detailed solution,

    although, the program we are entering our answers "asked" me imperatively to do a step further and factor x^2 under the radical; it considered this answer as not correct until I did the following:

    \frac{dy}{dx} &= \frac{5}{x^2\,\sqrt{ 36 - (1/x^2) }}

    IMO it's kind of useless... maybe there is a reason(I'm serious, actually - it's not sarcasm), from a deep mathematical point of view and me, as a novice, I am not aware of... what do you think?
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    Re: The derivative of the arcsec function

    Quote Originally Posted by dokrbb View Post
    thanks a lot for this detailed solution,

    although, the program we are entering our answers "asked" me imperatively to do a step further and factor x^2 under the radical; it considered this answer as not correct until I did the following:

    \frac{dy}{dx} &= \frac{5}{x^2\,\sqrt{ 36 - (1/x^2) }}

    IMO it's kind of useless... maybe there is a reason(I'm serious, actually - it's not sarcasm), from a deep mathematical point of view and me, as a novice, I am not aware of... what do you think?
    I think the version your program wants is uglier, bringing in more fractions for no apparent reason... It's also wrong considering that \displaystyle \sqrt{x^2} = |x|, NOT x.
    Thanks from dokrbb
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