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Math Help - chain rule problems...

  1. #1
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    chain rule problems...

    1. (sin x/ (1+cos x))^2

    My work:

    2(sin x/ (1+cos x)) * ((1+cos x)(cos x) - sin x (-sin x)) / (1+ cos x)^2

    ((cos x +(cos x)^2+(sin x) ^2)/ (1+ cos x)^2) * (2sin x/(1+cos x))

    Long story short the answer was suppose to be (2 sin x/(1+ cosx)^2).

    2. I am suppose to find the 2nd derivative of (1+(1/x))^3

    My work:

    3(1+(1/x))^2 * -x^(-2)

    -(6(-x^(-2))) * (-2x^(-3)... this was the farthest I went since I knew this was not right.
    the answer was suppose to be:
    (6/x^3) * (1+(1/x)) * (1+(2/x))

    Thank you in advance.
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  2. #2
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    Quote Originally Posted by driver327 View Post
    1. (sin x/ (1+cos x))^2

    My work:

    2(sin x/ (1+cos x)) * ((1+cos x)(cos x) - sin x (-sin x)) / (1+ cos x)^2

    ((cos x +(cos x)^2+(sin x) ^2)/ (1+ cos x)^2) * (2sin x/(1+cos x))

    Long story short the answer was suppose to be (2 sin x/(1+ cosx)^2).
    All your calculations are OK. There is only one step missing:

    \dfrac{\cos(x)+\cos^2(x) + \sin^2(x)}{(1+\cos(x))^2} \cdot \dfrac{2\sin(x)}{1+\cos(x)} = \dfrac{\cos(x)+1}{(1+\cos(x))^2} \cdot \dfrac{2\sin(x)}{1+\cos(x)} Now cancel and you'll get the given answer.
    2. I am suppose to find the 2nd derivative of (1+(1/x))^3

    My work:

    3(1+(1/x))^2 * -x^(-2)

    -(6(-x^(-2))) * (-2x^(-3)... this was the farthest I went since I knew this was not right.
    the answer was suppose to be:
    (6/x^3) * (1+(1/x)) * (1+(2/x))
    The first step is OK. Before you are going to determine the 2nd derivative simplify the term:

    f(x)=\left(1+\dfrac1x\right)^3~\implies~f'(x)=3\le  ft(1+\dfrac1x\right)^2 \cdot \left(-\dfrac1x\right)^2 = -3 \cdot \dfrac{(x+1)^2}{x^4}

    Can you take it from here?
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