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Math Help - Differentiating using log question

  1. #1
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    Differentiating using log question

    find ds/dt for s = cos^4t (1+t^2)^0.5

    (^4 means to the power 4, so cos to power 4, and there's also root (1 plus t squared)

    Problem is that I take logs of both sides then try to differentiate using the 1/s ds/dt = rule but get the wrong answer, so a worked answer would be much appreciated. Thanks for any help.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by wardj View Post
    find ds/dt for s = cos^4t (1+t^2)^0.5

    (^4 means to the power 4, so cos to power 4, and there's also root (1 plus t squared)

    Problem is that I take logs of both sides then try to differentiate using the 1/s ds/dt = rule but get the wrong answer, so a worked answer would be much appreciated. Thanks for any help.
    s=\cos^4(t)\sqrt{1+t^2}

    \ln(s)=\ln\left(\cos^4(t)\sqrt{1+t^2}\right)

    \ln(s)=4\ln\left(\cos(t)\right)+\frac{1}{2}\ln\lef  t(1+t^2\right)

    Differentiating, we get:

    \frac{1}{s}\cdot\frac{ds}{dt}=\frac{4}{\cos(t)}\cd  ot(-\sin(t))+\frac{1}{2(1+t^2)}\cdot(2t)

    \frac{1}{s}\frac{ds}{dt}=-4\tan(t)+\frac{t}{1+t^2}

    \frac{ds}{dt}=s\cdot\left[-4\tan(t)+\frac{t}{1+t^2}\right]

    Since s=\cos^4(t)\sqrt{1+t^2},

    \color{red}\boxed{\frac{ds}{dt}=\cos^4(t)\sqrt{1+t  ^2}\left[-4\tan(t)+\frac{t}{1+t^2}\right]}

    Hope that this made sense!
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  3. #3
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    Quote Originally Posted by wardj View Post
    find ds/dt for s = cos^4t (1+t^2)^0.5

    (^4 means to the power 4, so cos to power 4, and there's also root (1 plus t squared)

    Problem is that I take logs of both sides then try to differentiate using the 1/s ds/dt = rule but get the wrong answer, so a worked answer would be much appreciated. Thanks for any help.
    \log s = 4\log \cos t + \frac12 \log (1+t^2)

    Differentiate:
    \frac{s'}{s} = -4\frac{\sin t}{\cos t} + \frac12 \frac{2t}{1+t^2}

    \frac{s'}{s} = -4\tan t + \frac{t}{1+t^2}

    \frac{ds}{dt} = -4\sin t \cos^3 t \sqrt{1+t^2} + \frac{t\cos^3 t \sqrt{1+t^2}}{1+t^2}

    \frac{ds}{dt} = -4 \cos^3 t \sin t \sqrt{1+t^2} + \frac{t\cos^3 t }{\sqrt{1+t^2}}
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